TITLE: Does a complete set of commuting observables always exist?
QUESTION [18 upvotes]: The Wikipedia article for complete set of commuting observables (CSCO) outlines a method for finding a CSCO. However, it does not suggest why such a set should exist. On one hand, I think it is very possible that it might not in general, as most systems are not integrable. But on the other hand, we normally talk of symmetries giving rise to degeneracies, so if there is some degeneracy, we should be able to find a commuting observable that distinguishes between the degenerate states.
Can anyone provide a proof that CSCOs exist in general/a counterexample of some system that does not permit a CSCO?

REPLY [15 votes]: Sometimes textbooks on introductory quantum theory give the impression that every self-adjoint operator represents an observable, and many simple models do have that property. Such simple models always have a CSCO. However, the general principles of quantum theory don't require that every self-adjoint operator be an observable. In fact, some models don't even have any CSCOs.
So the short answer to the question is: counterexamples do exist. I'll start by reviewing some natural counterexamples that actually arise in physics, but the math is deep. To compensate for that, I'll finish the answer with a contrived counterexample in which the math is elementary.
 Natural (but difficult) counterexamples
Models exist in which the algebra of observables is a type III factor, which cannot have a CSCO in any representation on a separable Hilbert space. I'll explain what this jargon means in the next section, but first I'll give the examples.
The counterexamples come from relativistic quantum field theory (RQFT). At least in one axiomatic system for RQFT, typical free-field RQFTs (in continuous spacetime) can be shown to have the property that the algebra of observables associated with a bounded region of spacetime is a type-III von Neumann algebra. Below, I'll prove that such an algebra cannot have a CSCO. By choosing an arbitrarily-large but bounded region of spacetime and calling that the whole model, we obtain a physically-natural example of a model with no CSCOs. More information about this is given in ref 1. Here's an excerpt from the abstract:

In relativistic quantum field theory (RQFT), ... factors of type III occur naturally. The same holds true in quantum statistical mechanics of infinite systems.

And an excerpt from pages 3-4:

Investigations on the foundations of relativistic quantum field theory (RQFT) revealed already more than 40 years ago that the algebras generated by observables that are localized in bounded regions of space-time cannot be type I and that they are, in fact, generically of type III. This is far from being obvious but turns out to be a consequence of general physical requirements of RQFT. Another case where type I is excluded is the statistical physics of infinite systems with nonzero density [11].

Free-field RQFTs are not the only counterexamples. Asymptotically-free models like quantum chromodynamics (QCD) presumably also have this property, but that has not yet been proven: we do not yet have a proof that QCD satisfies the axioms of RQFT in continuous spacetime from which the type-III property is deduced. Counterexamples that are physically natural and mathematically tractable (but still not easy) can be found in non-relativistic quantum physics. A few are listed on page 273 in ref 2.
 Technical background
Here's some math background to explain what all that jargon means:

A von Neumann algebra can be defined as a set $M$ of operators on a Hilbert space that is self-contained with respect to adjoints and that has the property $(M')'=M$, where $M'$ denotes the set of all operators that commute with $M$. This isn't the usual definition of a von Neumann algebra, but it's equivalent to the usual definition, thanks to the double commutant theorem.

A factor is a von Neumann algebra $M$ in which operators proportional to the identity operator are the only ones that commute with everything else in $M$.

A factor is called type I if it is equivalent to the algebra of all operators on some Hilbert space. By the way, throughout this answer, "operator" means "linear operator that is defined on the whole Hilbert space". Although it is often convenient to represent observables by operators that are only defined on part of the Hilbert space, this is never necessary.

A factor $M$ is called type III if it has this property: for every non-zero projection operator $P\in M$, there is another operator $W\in M$ such that $W^*W = 1$ and $WW^* = P$. The asterisk here denotes the operator adjoint, which is more commonly denoted $W^\dagger$ in the physics literature.

 Proof that a type III algebra cannot have a CSCO
The counterexamples mentioned above are counterexamples because, in quantum physics, a type III factor cannot have a CSCO. To prove this, remember that quantum physics requires that the Hilbert space be separable — it must have a countable orthonormal basis. On such a Hilbert space, the algebra generated by a CSCO (via the double commutant, so that it is topologically complete) must include projection operators that project onto one-dimensional subspaces. Let $P$ be such a projection operator. The claim that the algebra is type III means that it also includes an operator $W$ for which $W^*W = 1$ and $WW^* = P$. Multiplying the first equation on the left by $W$ gives $PW=W$, and multiplying this on the left by $W^*$ gives $W^*PW=1$, which is impossible if $P$ projects onto a one-dimensional subspace. Therefore, a CSCO cannot exist in a (faithful representation of a) type III factor on a separable Hilbert space.
 An easier (but contrived) counterexample
The examples mentioned above all involve an infinite-dimensional Hilbert space, and they are mathematically difficult. Can we contrive easier examples that use a finite-dimensional Hilbert space? The answer is yes.
I'll use the Heisenberg picture, where all time-dependence is carried by the observables. (This was also implicit in the natural counterexamples mentioned above.) At any given time $t$, the model has only five observables: $A_1(t),A_2(t),A_3(t),A_4(t)$ and the time-indepenent Hamiltonian $H$. Choose the four self-adjoint operators $A_k(0)$ to have these properties:
$$
 A_j(0)A_k(0)=-A_k(0)A_j(0)\neq 0
\tag{1}
$$
and
$$
 \big(A_k(0)\big)^2 = 1.
\tag{2}
$$
Below, I'll show that such operators do exist. The time-dependence of the observables is defined by
$$
 A_k(t) \equiv e^{iHt}A_k(0) e^{-iHt}.
\tag{3}
$$
If we take the Hamiltonian to be $H=0$, then $A_k(t)=A_k(0)$, so the model has only five observables altogether, if we count the (trivial) Hamiltonian.
To prove that such a model cannot have a CSCO, first note that the smallest Hilbert space on which such a system of operators can be faithfully represented is a four-dimensional Hilbert space. Equation (2) says that each observable has no more than two eigenvalues, namely $+1$ and $-1$, and equation (1) says that none of the observables commute with each other (except for the trivial observable $H$), so every set of commuting observables has the form $(A_k(0),H)$ for some $k$. Such a pair cannot be a CSCO, because it has only two eigenvalue-pairs, namely $(1,0)$ and $(-1,0)$. This is not enough to discern all four of the state-vectors that any complete basis in this Hilbert space must have, so a CSCO cannot exist.
That completes the proof, except that we still need to show that operators with the properties (1)-(2) actually exist. To show this, first define the $2\times 2$ self-adjoint matrices
\begin{gather}
X = \left(\begin{matrix}
 0&1\\ 1&0
 \end{matrix}\right)
\hspace{1cm}
Y = \left(\begin{matrix}
 0&i\\-i&0
 \end{matrix}\right)
\\
Z = \left(\begin{matrix}
 1&0\\ 0&-1
 \end{matrix}\right)
\hspace{1cm}
I = \left(\begin{matrix}
 1&0\\ 0&1
 \end{matrix}\right),
\tag{4}
\end{gather}
and notice that $X^2=Y^2=Z^2=I$ and that $X,Y,Z$ all anticommute with each other. Then define these self-adoint $4\times 4$ matrices:
\begin{gather}
A_1(0) = \left(\begin{matrix}
 0&X\\ X&0
 \end{matrix}\right)
\hspace{1cm}
A_2(0) = \left(\begin{matrix}
 0&Y\\ Y&0
 \end{matrix}\right)
\\
A_3(0) = \left(\begin{matrix}
 0&Z\\ Z&0
 \end{matrix}\right)
\hspace{1cm}
A_4(0) = \left(\begin{matrix}
 0&iI\\ -iI&0
 \end{matrix}\right).
\tag{5}
\end{gather}
These satisfy the conditions (1) and (2), which shows that operators having those properties do exist.
 A challenge
In the contrived example shown above, the observables don't all commute with each other. This is typically what people mean when they say that a model is "quantum." However, we could also adopt a more narrow definition of "quantum," such as requiring incompatibility with any non-contextual hidden-variables model. I haven't tried to determine whether or not the contrived model above has that property. If it doesn't, then finding a no-CSCO model on a finite-dimensional Hilbert space that also has this extra property might be an interesting challenge.

References:

Yngvason (2004), "The Role of Type III Factors in Quantum Field Theory" (http://arxiv.org/abs/math-ph/0411058)

Ruetsche and Earman, "Interpreting probabilities in quantum field theory and quantum statistical mechanics" (http://pitt.edu/~jearman/EarmanRuetsche2011a.pdf)


TITLE: Integrating by parts - question on a limit
QUESTION [4 upvotes]: I want to integrate this function by parts:
$$\int_0^\infty   x\lambda e^{-\lambda x} \, \mathrm{d}x$$
And I arrive to the following expression:
$$\int_0^\infty   x\lambda e^{-\lambda x} \, \mathrm{d}x = \left. -xe^{-\lambda x}\right|^\infty_0 + \int_0^\infty   e^{-\lambda x} \, \mathrm{d}x$$
How do I evaluate this:
$ \left.-xe^{-\lambda x}\right|^\infty_0$ as it seems to yield $\frac{-\infty}{0}$
Thanks

REPLY [1 votes]: $\lim_{x \to \infty}-xe^{-\lambda x}=\lim_{x \to \infty}\frac{-x}{e^{\lambda x}}$ has the indeterminate form $\frac{-\infty}{\infty}$ which sets up Hopital's Rule.


TITLE: why here $x^2$ is used ? why not $x ?$
QUESTION [2 upvotes]: Does  there exists a function $f : \mathbb{R }\rightarrow \mathbb{R}$ which is differentiable only at the point $0.$?

My attempt :  I found the  answer  here Is there a function $f: \mathbb R \to \mathbb R$ that has only one point differentiable?
But  i didn't understands the  answer , my doubts given below

REPLY [1 votes]: With $f(x)=xp(x),$ the derivative at $0$ would be $\lim\limits_{x\to0}\dfrac{f(x)-f(0)}{x-0}=\lim\limits_{x\to0}\dfrac{xp(x)}x=\lim\limits_{x\to0}p(x),$
which does not exist, since $p(x)$ takes values of $0$ and $1$ for arbitrarily small $x$.


TITLE: How we can find $n$ positive real numbers, $x_1,\ldots,x_n$ such that $x_i>\sum\limits_{j\neq i}x_j$?
QUESTION [0 upvotes]: We let $n\geq 3$.
How we can find $n$ positive real numbers $x_1, x_2, \ldots x_n$, such that
\begin{align}x_1&>\displaystyle\sum_{i\neq 1} x_i, \\ x_2&>\displaystyle\sum_{i\neq 2} x_i,\\ &\vdots \\ x_n&>\displaystyle\sum_{i\neq n} x_i. \end{align}

REPLY [0 votes]: Let note $\displaystyle \bar x=\frac 1n\sum\limits_{i=1}^n x_i\quad$ the average of all values.
Your condition is $x_i > n\bar x -x_i\iff 2x_i > n\bar x\iff x_i>\frac n2 \bar x$
Since $n\ge 3$ then $\frac n2> 1$ and your condition becomes $$\forall i=1..n:\ x_i>\bar x$$
You are requesting that all elements should be strictly above the average, this is impossible !
By definition of the average there are elements below it and elements above it, or at the worst all equal, but not all can be strictly greater that the average.
Just sum up the inequality over all $x_i$ and this gives $n\bar x>n\bar x$ which is an obvious contradiction.


TITLE: How can I show that $a_0x^n+...+a_n$ and $a_nx^n+...a_0$ have the same discriminant?
QUESTION [3 upvotes]: How can I show that $a_0x^n+...+a_n$ and $a_nx^n+...a_0$ have the same discriminant?
You can use two different definition of the discriminant of the polynomial $f(x)=a_nx^n+...a_0$.
The first is $$D(f)=a_n^{2n-2}\prod_{i<k}(\alpha_i-\alpha_j),$$ where $\{\alpha_i\}_{i=1,...n}$ are the roots of the polynomial.
The second is $$D(f)=(-1)^{\frac{n(n-a)}{2}}\frac{1}{a_n}R(f,f')$$ where $R(f,f')$ is the resultant of $f$ and $f'$.

REPLY [3 votes]: The roots of $a_nx^n+\cdots +a_0$ are the reciprocals of the roots of $a_0x^n+\cdots +a_n$.
Therefore $\prod (\alpha_i-\alpha_j)$ is replaced with $\prod (\frac1{\alpha_i}-\frac1{\alpha_j})=\prod \frac{\alpha_j-\alpha_i}{\alpha_i\alpha_j}$. The numerators yield the original discriminant (check that there is no sign change!). The denominators produce $(\prod\alpha_i)^{2(n-1)}=(\frac{a_0}{a_n})^{2n-2}$, so that everything sorts out.


TITLE: On the series $\sum_{k=1}^{+\infty}f(1/k)$
QUESTION [0 upvotes]: Let use assumed that $f$ is a real-valued defined on $[0,1]$ such that $f$, $f'$, and $f''$ are all continuous on $[0,1]$. How can we prove that the series
$$\sum_{k=1}^{+\infty}f(1/k)$$
is convergent, whenever $f(0)=f'(0)=0$. I understand the importance of the assumption that $f$ is continuous and that $f(0)=0$. These ensure that $\lim_{k\to\infty}f(1/k)=0$. I tried using MVT but got confused beacuse $0$ is an endpoint. Where can we apply the continuity of $f'$ and $f''$? Can somebody give a hint on how to start? Thanks.

REPLY [1 votes]: By Mean Value Theorem we have that $$f(1/k)=\frac{1}{k}f^\prime(c_k) $$ for some $c_k\in(0,1/k).$ Similarly, by MVT we also have $$f^\prime(c_k) = c_k f^{\prime \prime}(c^\prime_k) $$ for some $c_k^\prime\in(0,c_k).$  Since $f^{\prime \prime}$ is continuous on $[0,1]$, it is bounded.
Given all this,the result follows.


TITLE: Solving $x + r\ln(x) = c$ using Lambert's W function?
QUESTION [0 upvotes]: A very elementary question, I am sure.
I would like to solve
$x + r \ln(x) = c$
where $c>1$ and $r<1$.
(There is a unique solution for $c>1,$ and the LHS is convex, btw.)
By "solve", I mean not numerically, but expressing the solution in terms of known functions, preferably something built-in in MATLAB.
Now, as far as I see, I cannot use Lambert's functions (either branch) because I get imaginary solutions that way.
Suggestions?  Thanks!

REPLY [1 votes]: Simplify it to $x^r e^x = e^c \Rightarrow \frac{x}{r} e^{\frac{x}{r}} = \frac{e^{\frac{c}{r}}}{r}$. Thus $x = r W(\frac{e^{\frac{c}{r}}}{r})$.


text:
\begin{document}

\address{$^{[1_a]}$ University of Monastir, Faculty of Economic Sciences and Management of Mahdia, Mahdia, Tunisia}
\address{$^{[1_b]}$ Laboratory Physics-Mathematics and Applications (LR/13/ES-22), Faculty of Sciences of Sfax, University of Sfax, Sfax, Tunisia}
\email{\url{kais.feki@hotmail.com}\,;\,\url{kais.feki@fsegma.u-monastir.tn}}

\address{$^{[2]}$ P.G. Department of Mathematics, Utkal University,
Vanivihar, Bhubaneswar 751004, India.}
\email{\url{satyajitsahoo2010@gmail.com}}

\subjclass{Primary 47A12, 46C05; Secondary 47B65, 47A05.}

\keywords{Positive operator, $A$-adjoint operator, numerical radius, operator matrix, inequality.}

\date{\today}

\author[Kais Feki and Satyajit Sahoo] {\Large{Kais Feki}$^{1_{a,b}}$ and \Large{Satyajit Sahoo}$^{2}$ }
\title[Further inequalities for the $\mathbb{A}$-numerical radius of certain $2 \times 2$ operator matrices]{Further inequalities for the $\mathbb{A}$-numerical radius of certain $2 \times 2$ operator matrices}
\maketitle

\begin{abstract}
Let $\mathbb{A}=
\begin{pmatrix}
A & O \\
O & A \\
\end{pmatrix}
$ be a $2\times2$ diagonal operator matrix whose each diagonal entry is a bounded positive (semidefinite) linear operator $A$ acting on a complex Hilbert space $\mathcal{H}$. In this paper, we derive several $\mathbb{A}$-numerical radius inequalities for $2\times 2$ operator matrices whose entries are bounded with respect to the seminorm induced by the positive operator $A$ on $\mathcal{H}$. Some applications of our inequalities are also given.
\end{abstract}

\section{Introduction and Preliminaries}\label{s1}
Throughout this article, $(\mathcal{H},\langle\cdot,\cdot\rangle)$ stands for a complex Hilbert space with associated norm $\|\cdot\|$. If $\mathcal{M}$ is a given linear subspace of $\mathcal{H}$, then $\overline{\mathcal{M}}$ denotes its closure in the norm topology of $\mathcal{H}$. Further, the orthogonal projection onto a closed subspace $\mathcal{S}$ of $\mathcal{H}$ will be denoted by $P_{\mathcal{S}}$. Let $\mathcal{B}(\mathcal{H})$ denote the $C^{\ast}$-algebra of all bounded linear operators acting on $\mathcal{H}$ with the identity operator $I_{\mathcal{H}}$ (or simply $I$ if no confusion arises). If $T\in\mathcal{B}(\mathcal{H})$, then $\mathcal{N}(T), \mathcal{R}(T)$  and $T^*$ are denoted by the kernel, the range and the adjoint of $T$, respectively.  An operator $T\in \mathcal{B}(\mathcal{H})$ is called positive (semi-definite) $\langle Ax, x\rangle\geq 0$, for every $x\in \mathcal{H}$.
For the rest of this paper, by an operator we mean a bounded linear operator. Further, we suppose that $A\in\mathcal{B}(\mathcal{H})$ is a nonzero positive operator which induces the following semi-inner product
$$\langle\cdot,\cdot\rangle_{A}:\mathcal{H}\times \mathcal{H}\longrightarrow\mathbb{C},\;(x,y)\longmapsto \langle x, y\rangle_{A}:=\langle Ax, y\rangle=\langle A^{1/2}x, A^{1/2}y\rangle.$$
Here $A^{1/2}$ denotes the square root of $A$. Notice that the seminorm induced by ${\langle \cdot, \cdot\rangle}_A$ is given by ${\|x\|}_A=\|A^{1/2}x\|$, for all $x\in\mathcal{H}$. It can checked that ${\|\cdot\|}_A$ is a norm on $\mathcal{H}$ if and only if $A$ is injective, and that the seminormed space $(\mathcal{H}, {\|\cdot\|}_A)$ is complete if and only if $\mathcal{R}(A)$ is a closed subspace of $\mathcal{H}$.

Let $T \in \mathcal{B}(\mathcal{H})$. An operator $S\in\mathcal{B}(\mathcal{H})$ is called an $A$-adjoint of $T$ if $\langle Tx, y\rangle_A=\langle x, Sy\rangle_A$ for all $x,y\in \mathcal{H}$ (see \cite{acg1}). Thus, the existence of an $A$-adjoint of $T$ is equivalent to the existence of a solution of the equation $AX = T^*A$. Notice that this kind of equations can be investigated by using a theorem due to Douglas \cite{doug} which establishes the equivalence between the following  statements:
\begin{itemize}
  \item [(i)] The operator equation $TX=S$ has a bounded linear solution $X$.
  \item [(ii)] $\mathcal{R}(S) \subseteq \mathcal{R}(T)$.
  \item [(iii)] There exists a positive number $\lambda$ such that $\|S^*x\|\leq \lambda \|T^*x\|$ for all $x\in \mathcal{H}$.
\end{itemize}
\noindent Moreover, among many solutions of $AX=S$, it has only one, say $Q$, which satisfies $\mathcal{R}(Q) \subseteq \overline{\mathcal{R}(T^{*})}$. Such $Q$ is said the reduced solution of the equation $TX=S$. If we denote by $\mathcal{B}_{A}(\mathcal{H})$, the subspace of all operators admitting $A$-adjoints, then by Douglas theorem, we  have
$$\mathcal{B}_{A}(\mathcal{H})=\left\{T\in \mathcal{B}(\mathcal{H})\,;\;\mathcal{R}(T^{*}A)\subset \mathcal{R}(A)\right\}.$$
If $T\in \mathcal{B}_A(\mathcal{H})$, the reduced solution of the equation
$AX=T^*A$ will be denoted by $T^{\sharp_A}$. We mention here that, $T^{\sharp_A}=A^\dag T^*A$ in which $A^\dag$ is the Moore-Penrose inverse of $A$ (see \cite{acg2}). In addition, if $T \in \mathcal{B}_A({\mathcal{H}})$, then $T^{\sharp_A} \in \mathcal{B}_A({\mathcal{H}})$, $(T^{\sharp_A})^{\sharp_A}=P_{\overline{\mathcal{R}(A)}}TP_{\overline{\mathcal{R}(A)}}$ and $((T^{\sharp_A})^{\sharp_A})^{\sharp_A}=T^{\sharp_A}$. Moreover, If $S\in \mathcal{B}_A(\mathcal{H})$, then $TS \in\mathcal{B}_A({\mathcal{H}})$ and $(TS)^{\sharp_A}=S^{\sharp_A}T^{\sharp_A}.$ Furthermore, for every $T \in \mathcal{B}_A({\mathcal{H}})$, we have
\begin{equation}\label{diez}
\|T^{\sharp_A}T\|_A = \| TT^{\sharp_A}\|_A=\|T\|_A^2 =\|T^{\sharp_A}\|_A^2.
\end{equation}
 An operator $U\in  \mathcal{B}_A(\mathcal{H})$ is called $A$-unitary if $\|Ux\|_A=\|U^{\sharp_A}x\|_A=\|x\|_A$, for all $x\in \mathcal{H}$. It is worth mentioning that, an operator $U\in  \mathcal{B}_A(\mathcal{H})$ is $A$-unitary if and only if $U^{\sharp_A} U=(U^{\sharp_A})^{\sharp_A} U^{\sharp_A}=P_{\overline{\mathcal{R}(A)}}$ (see \cite{acg1}). For an account of the results, we invite the reader to consult \cite{acg1,acg2}.

An operator $T$ is called $A$-bounded if there exists $\lambda>0$ such that $ \|Tx\|_{A} \leq \lambda \|x\|_{A},\;\forall\,x\in \mathcal{H}.$  By applying Douglas theorem, one can easily see that the subspace of all operators admitting $A^{1/2}$-adjoints, denoted by $\mathcal{B}_{A^{1/2}}(\mathcal{H})$, is equal the collection of all $A$-bounded operators, i.e.,
$$\mathcal{B}_{A^{1/2}}(\mathcal{H})=\left\{T \in \mathcal{B}(\mathcal{H})\,;\;\exists \,\lambda > 0\,;\;\|Tx\|_{A} \leq \lambda \|x\|_{A},\;\forall\,x\in \mathcal{H}  \right\}.$$
Notice that $\mathcal{B}_{A}(\mathcal{H})$ and $\mathcal{B}_{A^{1/2}}(\mathcal{H})$ are two subalgebras of $\mathcal{B}(\mathcal{H})$ which are, in general, neither closed nor dense in $\mathcal{B}(\mathcal{H})$. Moreover, we have $\mathcal{B}_{A}(\mathcal{H})\subset \mathcal{B}_{A^{1/2}}(\mathcal{H})$ (see \cite{acg1,acg3}). Clearly, $\langle\cdot,\cdot\rangle_{A}$ induces a seminorm on $\mathcal{B}_{A^{1/2}}(\mathcal{H})$. Indeed, if $T\in\mathcal{B}_{A^{1/2}}(\mathcal{H})$, then it holds that
\begin{equation}\label{semii}
\|T\|_A:=\sup_{\substack{x\in \overline{\mathcal{R}(A)},\\ x\not=0}}\frac{\|Tx\|_A}{\|x\|_A}=\sup\big\{{\|Tx\|}_A\,; \,\,x\in \mathcal{H},\, {\|x\|}_A =1\big\}<\infty.
\end{equation}

Saddi \cite{saddi} in 2012 defined the $A$-numerical radius of an operator $T\in\mathcal{B}(\mathcal{H})$ by
\begin{align*}
\omega_A(T)
&:= \sup \left\{|\langle Tx, x\rangle_A|\,;\;x\in\mathcal{H},\|x\|_A = 1\right\}.
\end{align*}

Faghih-Ahmadi and Gorjizadeh \cite{fg} in 2016 showed that for $T\in\mathcal{B}_{A^{1/2}}(\mathcal{H})$, we have
\begin{equation}\label{newsemi}
\|T\|_A=\sup\left\{|\langle Tx, y\rangle_A|\,;\;x,y\in \mathcal{H},\,\|x\|_{A}=\|y\|_{A}= 1\right\}.
\end{equation}

We notice here that it may happen that ${\|T\|}_A$ and $\omega_A(T)$ are equal to $+ \infty$ for some $T\in\mathcal{B}(\mathcal{H})$ (see \cite{feki01}). However, ${\|\cdot\|}_A$ and $\omega_A(\cdot)$ are equivalent seminorms on $\mathcal{B}_{A^{1/2}}(\mathcal{H})$. More precisely, In 2018, Baklouti et al. \cite{bakfeki01} showed that  for every $T\in \mathcal{B}_{A^{1/2}}(\mathcal{H})$, we have
\begin{equation}\label{refine1}
\tfrac{1}{2} \|T\|_A\leq\omega_A(T) \leq \|T\|_A.
\end{equation}

For the sequel, for any arbitrary operator $T\in {\mathcal B}_A({\mathcal H})$, we write
$$\Re_A(T):=\frac{T+T^{\sharp_A}}{2}\;\;\text{ and }\;\;\Im_A(T):=\frac{T-T^{\sharp_A}}{2i}.$$
Recently, in 2019 Zamani \cite[Theorem 2.5]{zamani1} showed that if $T\in\mathcal{B}_{A}(\mathcal{H})$, then
\begin{align}\label{zm}
\omega_A(T) = \displaystyle{\sup_{\theta \in \mathbb{R}}}{\left\|\Re_A(e^{i\theta}T)\right\|}_A=\displaystyle{\sup_{\theta \in \mathbb{R}}}{\left\|\Im_A(e^{i\theta}T)\right\|}_A.
\end{align}
Notice that \eqref{zm} is also proved in a general context in \cite{bm}. In 2020, the concept of the $A$-spectral radius of $A$-bounded operators was introduced by the first author in \cite{feki01} as follows:
\begin{equation}\label{newrad}
r_A(T):=\displaystyle\inf_{n\geq 1}\|T^n\|_A^{\frac{1}{n}}=\displaystyle\lim_{n\to\infty}\|T^n\|_A^{\frac{1}{n}}.
\end{equation}
Here we want to mention that the proof of the second equality in \eqref{newrad} can also be found in \cite[Theorem 1]{feki01}. Like the classical spectral radius of Hilbert space operators, it was shown in \cite{feki01} that $r_A(\cdot)$ satisfies the commutativity property, i.e.
\begin{equation}\label{commut}
r_A(TS)=r_A(ST),
\end{equation}
for all $T,S\in \mathcal{B}_{A^{1/2}}(\mathcal{H})$. For the sequel, if $A=I$, then $\|T\|$, $r(T)$ and $\omega(T)$ denote respectively the classical operator norm, the spectral radius and the numerical radius of an operator $T$.

An operator $T\in\mathcal{B}(\mathcal{H})$ is called $A$-selfadjoint if $AT$ is selfadjoint, i.e., $AT = T^*A$ and it is called $A$-positive if $AT\geq0$. If $T$ is $A$-positive, we will write $T\geq_{A}0$. In recent years, several results covering some classes of operators on a complex Hilbert space $\big(\mathcal{H}, \langle \cdot, \cdot\rangle\big)$ were extended to $\big(\mathcal{H}, {\langle \cdot, \cdot\rangle}_A\big)$. Of course, the extension is not trivial since many difficulties arise. For instance, as mentioned above, it may happen
that ${\|T\|}_A = \infty$ for some $T\in \mathcal{B}(\mathcal{H})$. Moreover, no operator admits an adjoint operator for the
semi-inner product ${\langle \cdot, \cdot\rangle}_A$. In addition, for $T \in \mathcal{B}_A({\mathcal{H}})$, we have $(T^{\sharp_A})^{\sharp_A}=P_{\overline{\mathcal{R}(A)}}TP_{\overline{\mathcal{R}(A)}}\neq T$. For further details about $A$-numerical radius, interested readers can follow \cite{bakna,bakfeki01,bakfeki04,BPl,ConFe,bfeki,feki03,fekilaa,zamani1,NSD} and the references therein.

In this paper, we consider the ${2\times 2}$ operator diagonal matrix $\mathbb{A}=\begin{pmatrix}
A &O\\
O &A
\end{pmatrix}$. Clearly, $\mathbb{A}\in \mathcal{B}(\mathcal{H}\oplus \mathcal{H})^+$. So, $\mathbb{A}$ induces the following semi-inner product
$$\langle x, y\rangle_{\mathbb{A}}= \langle \mathbb{A}x, y\rangle=\langle x_1, y_1\rangle_A+\langle x_2, y_2\rangle_A,$$
for all $x=(x_1,x_2)\in \mathcal{H}\oplus \mathcal{H}$ and $y=(y_1,y_2)\in \mathcal{H}\oplus \mathcal{H}$.

Recently, several inequalities for the $\mathbb{A}$-numerical radius of $2 \times 2$ operator matrices have been established by Bhunia et al.  when $A$ is a positive injective operators (see \cite{BPN}). Moreover, different upper and lower bounds of $\mathbb{A}$-numerical radius when $A$ is a positive semidefinite operator has been recently investigated by the first author in \cite{feki04}, by Rout et al. in \cite{rout} and by Kittaneh et al. in \cite{KITSAT}. In this article, we will continue working in this direction and we will prove several new $\mathbb{A}$-numerical radius inequalities of certain $2 \times 2$ operator matrices. The inspiration for our investigation comes from \cite{dolat,HirKit,HirKit2}.

 \section{Results}\label{s2}
In this section, we present our results. Throughout this section $\mathbb{A}$ is denoted to be the $2\times 2$ operator diagonal matrix whose each diagonal entry is the positive operator $A$. To prove our first result, the following lemmas are required.

\begin{lemma}(\cite{feki01})\label{ll2020}
Let $T\in\mathcal{B}(\mathcal{H})$ is an $A$-self-adjoint operator. Then,
\begin{equation*}
\|T\|_{A}=\omega_A(T)=r_A(T).
\end{equation*}
\end{lemma}

\begin{lemma}\label{lm5}(\cite{feki02})
Let $\mathbb{T}=\begin{pmatrix}
T_{11}&T_{12} \\
T_{21}&T_{22}
\end{pmatrix}$ be such that $T_{ij}\in \mathcal{B}_{A^{1/2}}(\mathcal{H})$ for all $i,j\in\{1,2\}$. Then, $\mathbb{T}\in \mathcal{B}_{\mathbb{A}^{1/2}}(\mathcal{H}\oplus \mathcal{H})$ and
$$ r_\mathbb{A}\left(\mathbb{T}\right)\leq r\left[\begin{pmatrix}
\|T_{11}\|_A & \|T_{12}\|_A \\
\|T_{21}\|_A & \|T_{22}\|_A
\end{pmatrix}\right].$$
\end{lemma}
\begin{lemma}(\cite{bfeki,feki04})\label{lemma1}
Let $P, Q, R, S\in\mathcal{B}_{A^{1/2}}(\mathcal{H})$. Then, the following assertions hold
\begin{itemize}
\item[(i)] $\omega_{\mathbb{A}}\left[\begin{pmatrix}
P & O \\
O & S
\end{pmatrix}\right] = \max\big\{\omega_{A}(P), \omega_{A}(S)\big\}$.
\item[(ii)] ${\left\|\begin{pmatrix}
P & O \\
O & S
\end{pmatrix}\right\|}_{\mathbb{A}} = {\left\|\begin{pmatrix}
O & P \\
S & O
\end{pmatrix}\right\|}_{\mathbb{A}} = \max\big\{{\|P\|}_{A}, {\|S\|}_{A}\big\}$.
\item[(iii)] If $P, Q, R, S\in\mathcal{B}_{A}(\mathcal{H})$, then ${\begin{pmatrix}
P & Q \\
R & S
\end{pmatrix}}^{\sharp_{\mathbb{A}}} = \begin{pmatrix}
P^{\sharp_A} & R^{\sharp_A} \\
Q^{\sharp_A} & S^{\sharp_A}
\end{pmatrix}$.
\end{itemize}
\end{lemma}
 Now, we are ready to prove our first result which generalizes \cite[Theorem 2.7]{dolat}.
\begin{theorem}\label{thm101}
Let  $P, Q, R, S\in \mathcal{B}_{A}(\mathcal{H})$. Then, for $\lambda\in [0, 1]$, we have
\begin{align*}
&\omega_{\mathbb{A}}\left[\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}\right]\\
&\leq \frac{1}{2}\left(\|P\|_A+2\omega_A(S)+\sqrt{\|\lambda^2 PP^{\sharp_A}+QQ^{\sharp_A}\|_A}+\sqrt{\|(1-\lambda)^2 PP^{\sharp_A}+R^{\sharp_A}R\|_A}\right).
\end{align*}
\end{theorem}
\begin{proof}
Let $\mathbb{T}=\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}$. It is not difficult to see that $\Re_\mathbb{A}(e^{i\theta}\mathbb{T})$ is an $\mathbb{A}$-selfadjoint operator. So, by Lemma \ref{ll2020} we have
$${\left\|\Re_\mathbb{A}(e^{i\theta}\mathbb{T})\right\|}_\mathbb{A}=\omega_\mathbb{A}\Big(\Re_\mathbb{A}(e^{i\theta}\mathbb{T})\Big).$$
So, by applying Lemma \ref{lemma1} (i), we see that
\begin{align*}
&{\left\|\Re_\mathbb{A}(e^{i\theta}\mathbb{T})\right\|}_\mathbb{A}\\
&=\frac{1}{2}\omega_{\mathbb{A}}\left[\begin{pmatrix}
e^{i\theta}P+e^{-i\theta}P^{\sharp_A}&e^{i\theta}Q+e^{-i\theta}R^{\sharp_A} \\
e^{i\theta}R+e^{-i\theta}Q^{\sharp_A}&e^{i\theta}S+e^{-i\theta}S^{\sharp_A}
\end{pmatrix}\right]\\
&\leq\frac{1}{2}\omega_{\mathbb{A}}\left[\begin{pmatrix}
\lambda (e^{i\theta}P+e^{-i\theta}P^{\sharp_A})&e^{i\theta}Q \\
e^{-i\theta}Q^{\sharp_A}&0
\end{pmatrix}\right]\\
&+\frac{1}{2}\omega_{\mathbb{A}}\left[\begin{pmatrix}
(1-\lambda) (e^{i\theta}P+e^{-i\theta}P^{\sharp_A})&e^{-i\theta}R^{\sharp_A} \\
e^{i\theta}R&0
\end{pmatrix}\right]+\frac{1}{2}\omega_{\mathbb{A}}\left[\begin{pmatrix}
0&0 \\
0&e^{i\theta}S+e^{-i\theta}S^{\sharp_A}
\end{pmatrix}\right]\\
&\leq\frac{1}{2}\omega_{\mathbb{A}}\left[\begin{pmatrix}
\lambda (e^{i\theta}P+e^{-i\theta}P^{\sharp_A})&e^{i\theta}Q \\
e^{-i\theta}Q^{\sharp_A}&0
\end{pmatrix}\right]\\
&\;\;+\frac{1}{2}\omega_{\mathbb{A}}\left[\begin{pmatrix}
(1-\lambda) (e^{i\theta}P+e^{-i\theta}P^{\sharp_A})&e^{-i\theta}R^{\sharp_A} \\
e^{i\theta}R&0
\end{pmatrix}\right]+\omega_A(S),
\end{align*}
where the last inequality follows by Lemma \ref{lemma1} (i) together with the triangle inequality. Moreover, it can be observed that
\begin{align*}
\mathbb{A}\begin{pmatrix}
\lambda (e^{i\theta}P+e^{-i\theta}P^{\sharp_A})&e^{i\theta}Q \\
e^{-i\theta}Q^{\sharp_A}&0
\end{pmatrix}&=\begin{pmatrix}
\lambda (e^{i\theta}AP+e^{-i\theta}AP^{\sharp_A})&e^{i\theta}AQ \\
e^{-i\theta}AQ^{\sharp_A}&0
\end{pmatrix}\\
&=\begin{pmatrix}
\lambda \left(e^{i\theta}(P^{\sharp_A})^*A+e^{-i\theta}P^*A\right)&e^{i\theta}(Q^{\sharp_A})^*A \\
e^{-i\theta}Q^*A&0
\end{pmatrix}\\
&=\begin{pmatrix}
\lambda (e^{-i\theta}P^{\sharp_A}+e^{i\theta}P)&e^{i\theta}Q \\
e^{-i\theta}Q^{\sharp_A}&0
\end{pmatrix}^*\mathbb{A}.
\end{align*}
Hence $\begin{pmatrix}
\lambda (e^{i\theta}P+e^{-i\theta}P^{\sharp_A})&e^{i\theta}Q \\
e^{-i\theta}Q^{\sharp_A}&0
\end{pmatrix}$ is $\mathbb{A}$-selfadjoint operator. Similarly one can show that $\begin{pmatrix}
(1-\lambda) (e^{i\theta}P+e^{-i\theta}P^{\sharp_A})&e^{-i\theta}R^{\sharp_A} \\
e^{i\theta}R&0
\end{pmatrix}$ is $\mathbb{A}$-selfadjoint operator. So by applying Lemma \ref{ll2020} we see that
\begin{align*}
{\left\|\Re_\mathbb{A}(e^{i\theta}\mathbb{T})\right\|}_\mathbb{A}
&\leq\frac{1}{2}r_{\mathbb{A}}\left[\begin{pmatrix}
\lambda (e^{i\theta}P+e^{-i\theta}P^{\sharp_A})&e^{i\theta}Q \\
e^{-i\theta}Q^{\sharp_A}&O
\end{pmatrix}\right]\\
&\;\;+\frac{1}{2}r_{\mathbb{A}}\left[\begin{pmatrix}
(1-\lambda) (e^{i\theta}P+e^{-i\theta}P^{\sharp_A})&e^{-i\theta}R^{\sharp_A} \\
e^{i\theta}R&O
\end{pmatrix}\right]+\omega_A(S).
\end{align*}
So, by using \eqref{commut} we infer that
\begin{align*}
{\left\|\Re_\mathbb{A}(e^{i\theta}\mathbb{T})\right\|}_\mathbb{A}
&\leq \frac{1}{2}r_{\mathbb{A}}\left[\begin{pmatrix}
e^{-i\theta}I&O \\
\lambda P&Q
\end{pmatrix}\begin{pmatrix}
\lambda P^{\sharp_A}&e^{i\theta}I \\
Q^{\sharp_A}&O
\end{pmatrix}\right]\\
&\;\;+\frac{1}{2}r_{\mathbb{A}}\left[\begin{pmatrix}
e^{-i\theta}I&O \\
(1-\lambda) P&e^{-2i\theta}R^{\sharp_A}
\end{pmatrix}\begin{pmatrix}
(1-\lambda) P^{\sharp_A}&e^{i\theta}I \\
e^{2i\theta}R&O
\end{pmatrix}\right]+\omega_A(S)
\\
&\leq \frac{1}{2}r_{\mathbb{A}}\left[\begin{pmatrix}
\lambda e^{-i\theta}P^{\sharp_A} & I \\
\lambda^2 PP^{\sharp_A}+QQ^{\sharp_A}& \lambda e^{i\theta}P
\end{pmatrix}\right]\\
&\;\;+\frac{1}{2}r_{\mathbb{A}}\left[\begin{pmatrix}
(1-\lambda) e^{-i\theta}P^{\sharp_A} & I \\
(1-\lambda)^2 PP^{\sharp_A}+R^{\sharp_A}R&(1-\lambda) e^{i\theta}P
\end{pmatrix}\right]+\omega_A(S).
\end{align*}
So, by using Lemma \ref{lm5} we get
\begin{align*}
&{\left\|\Re_\mathbb{A}(e^{i\theta}\mathbb{T})\right\|}_\mathbb{A}\\
&\leq \frac{1}{2}r\left[\begin{pmatrix}
\lambda \|P\|_A & 1 \\
\|\lambda^2 PP^{\sharp_A}+QQ^{\sharp_A}\|_A& \lambda \|P\|_A
\end{pmatrix}\right]\\
&\;\;+\frac{1}{2}r\left[\begin{pmatrix}
(1-\lambda) \|P\|_A & 1 \\
\|(1-\lambda)^2 PP^{\sharp_A}+R^{\sharp_A}R\|_A&(1-\lambda) \|P\|_A
\end{pmatrix}\right]+\omega_A(S)
\\
&=\frac{1}{2}\bigg[\|P\|_A+2\omega_A(S)+\sqrt{\|\lambda^2 PP^{\sharp_A}+QQ^{\sharp_A}\|_A}+\sqrt{\|(1-\lambda)^2 PP^{\sharp_A}+R^{\sharp_A}R\|_A}\bigg].
\end{align*}
So, by taking the supremum over all $\theta\in\mathbb{R}$ in the last inequality and then using \ref{zm} we get desired result.
\end{proof}

The following Lemma is useful in the sequel. Notice that its assertions generalize recent results done by Rout et al. in \cite{rout} for operators in $\mathcal{B}_{A}(\mathcal{H})$.
\begin{lemma}\label{lem100}
	Let $T,S\in \mathcal{B}_{A^{1/2}}(\mathcal{H})$. Then,
\begin{itemize}
  \item [(i)] $ \omega_\mathbb{A}\left[\begin{pmatrix}
	T&S\\
	S&T
	\end{pmatrix}\right]= \max\{\omega_A(T+S),\omega_A(T-S)\}.$
  \item [(ii)] $ \omega_\mathbb{A}\left[\begin{pmatrix}
	T&-S\\
	S&T
	\end{pmatrix}\right]= \max\{\omega_A(T+iS),\omega_A(T-iS)\}.$
\end{itemize}
\end{lemma}
In order to prove Lemma \ref{lem100} we need the following result.
\begin{lemqt}\label{weak}(\cite{bfeki})
Let $T\in \mathcal{B}_{A^{1/2}}(\mathcal{H})$. Then,
\begin{equation*}
\omega_A(U^{\sharp_A}TU)=\omega_A(T),
\end{equation*}
for any $A$-unitary operator $U\in\mathcal{B}_A(\mathcal{H})$.
\end{lemqt}
Now, we state the proof of Lemma \ref{lem100}.
\begin{proof}[Proof of Lemma \ref{lem100}]
\noindent (i)\;Let $\mathbb{U}=\frac{1}{\sqrt{2}}\begin{pmatrix}
	I & I\\
	-I & I
	\end{pmatrix}$. By using Lemma \ref{lemma1} (iii), we see that
$$\mathbb{U}^{\sharp_\mathbb{A}}=\frac{1}{\sqrt{2}}\begin{pmatrix}
	P_{\overline{\mathcal{R}(A)}} &-P_{\overline{\mathcal{R}(A)}}\\
	P_{\overline{\mathcal{R}(A)}} &P_{\overline{\mathcal{R}(A)}}
	\end{pmatrix}.$$
So, by using the fact that $AP_{\overline{\mathcal{R}(A)}}=A$, we can verify that $\|\mathbb{U}x\|_{\mathbb{A}}=\|\mathbb{U}^{\sharp_{\mathbb{A}}}x\|_{\mathbb{A}}=\|x\|_{\mathbb{A}}$ for all $x\in \mathcal{H}\oplus \mathcal{H}$. Hence $\mathbb{U}$ is $\mathbb{A}$-unitary. So, by Lemma \ref{weak} we see that
	\begin{align*}
	\omega_\mathbb{A}\left[\begin{pmatrix}
	T&S\\
	S&T
	\end{pmatrix}\right]
&= \omega_\mathbb{A}\left[\mathbb{U}^{\sharp_\mathbb{A}}\begin{pmatrix}
	T&S\\
	S&T
	\end{pmatrix}\mathbb{U}\right]\\
	&=\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P_{\overline{\mathcal{R}(A)}} &O\\
	O &P_{\overline{\mathcal{R}(A)}}
	\end{pmatrix}
	\begin{pmatrix}
	T-S &O\\
	O&T+S
	\end{pmatrix}\right]\\
	&=\omega_{\mathbb{A}}\left[
	\begin{pmatrix}
	T-S &O\\
	O&T+S
	\end{pmatrix}\right]\\
	&=\max\{\omega_A(T+S),\omega_A(T-S)\},
	\end{align*}
where the last equality follows from Lemma \ref{lemma1} (i).
\par \vskip 0.1 cm \noindent (ii)\;By considering the $\mathbb{A}$-unitary operator $\mathbb{U}=\frac{1}{\sqrt{2}}\begin{pmatrix}
I & iI\\
iI & I
\end{pmatrix}$ and proceeding as above we get the desired result.
\end{proof}

As an application of Theorem \ref{thm101} together with Lemma \ref{lem100}, we state the following corollary.
\begin{corollary}
Let $P,Q\in \mathcal{B}_{A}(\mathcal{H})$. Then for all $\lambda\in [0,1]$, we have
\[\omega_A\left(P\pm Q\right)\leq \min\{\mu,\nu \},\]
where
\begin{align*}
\mu
& =\omega_A(P)+\frac{1}{2}\left(\|P\|_A+\sqrt{\|\lambda^2 PP^{\sharp_A}+QQ^{\sharp_A}\|_A}+\sqrt{\|(1-\lambda)^2 PP^{\sharp_A}+Q^{\sharp_A}Q\|_A}\right),
\end{align*}
and
\begin{align*}
\nu
& =\frac{3}{2}\omega_A(P)+\frac{1}{2}\bigg[\sqrt{\lambda^2 \omega^2_A(P)+\|Q\|_A^2}
+\sqrt{(1-\lambda)^2 \omega^2_A(P)+\|Q\|_A^2}\bigg].
\end{align*}
\end{corollary}
\begin{proof}
Recall first that it has been recently proved in \cite{BPN1} that
\begin{align}\label{ineq106}
&\omega_{\mathbb{A}}\left[\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}\right] \nonumber \\
&\leq \frac{1}{2}\bigg[\omega_A(P)+2\omega_A(S)+\sqrt{\lambda^2 \omega^2_A(P)+\|Q\|_A^2}
+\sqrt{(1-\lambda)^2 \omega^2_A(P)+\|R\|_A^2}\bigg].
\end{align}
So, we get the desired result by applying Theorem \ref{thm101} together with \eqref{ineq106} and Lemma \ref{lem100}
\end{proof}
Now we state the following theorem which generalizes a recent result proved by Rout et al. in \cite{rout} since $\mathcal{B}_{A}(\mathcal{H})$ is in general a proper subspace of $\mathcal{B}_{A^{1/2}}(\mathcal{H})$.
\begin{theorem}\label{them10}
Let $\mathbb{T}=\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}$ be such that $P, Q, R, S\in \mathcal{B}_{A^{1/2}}(\mathcal{H})$. Then,
\begin{align}\label{upper1}
\omega_{\mathbb{A}}\left[\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}\right]
&\leq \max\{\omega_A(P), \omega_A(S)\}+\frac{\omega_A(Q+R)+\omega_A(Q-R)}{2}.
\end{align}
\end{theorem}
\begin{proof}
	Using triangle inequality we have
\begin{align}\label{sho}
\omega_{\mathbb{A}}\left[\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}\right]&\leq \omega_{\mathbb{A}}\left[\begin{pmatrix}
P&O \\
O&S
\end{pmatrix}\right]+\omega_{\mathbb{A}}\left[\begin{pmatrix}
O&Q \\
R&O
\end{pmatrix}\right]\nonumber\\
&=\max\{\omega_A(P), \omega_A(S)\}+\omega_{\mathbb{A}}\left[\begin{pmatrix}
O&Q \\
R&O
\end{pmatrix}\right],
\end{align}
where the last equality follows by Lemma \ref{lemma1} (i). 	Let $\mathbb{U}=\frac{1}{\sqrt{2}}\begin{pmatrix}
	I & -I\\
	I & I
	\end{pmatrix}$. By proceeding as above we prove that $\mathbb{U}$ is $\mathbb{A}$-unitary. So, by Lemma \ref{weak} we see that
	\begin{align*}
	\omega_\mathbb{A}\left[\begin{pmatrix}
	O&Q \\
	R&O
	\end{pmatrix}\right]
&= \omega_\mathbb{A}\left[\mathbb{U}^{\sharp_\mathbb{A}}\begin{pmatrix}
	O&Q \\
	R&O
	\end{pmatrix}\mathbb{U}\right]\nonumber\\
	&=\frac{1}{2}\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P_{\overline{\mathcal{R}(A)}} &O\\
	O &P_{\overline{\mathcal{R}(A)}}
	\end{pmatrix}
	\begin{pmatrix}
	R+Q &-R+Q\\
	R-Q &-R-Q
	\end{pmatrix}\right]\nonumber\\
	&=\frac{1}{2}\omega_{\mathbb{A}}\left[
	\begin{pmatrix}
	R+Q &-R+Q\\
	R-Q &-R-Q
	\end{pmatrix}\right]\nonumber\\
	&\leq\frac{1}{2}\omega_{\mathbb{A}}\left[
	\begin{pmatrix}
	R+Q &O\\
	O &-R-Q
	\end{pmatrix}\right]
+\frac{1}{2}\omega_{\mathbb{A}}\left[
	\begin{pmatrix}
	O &-R+Q\\
	R-Q &O
	\end{pmatrix}\right]
\nonumber\\
	&=\frac{1}{2}\Big(\omega_A(Q+R)+\omega_A(Q-R)\Big),
	\end{align*}
where the last equality follows from Lemma \ref{lem100}. So,
\begin{equation}\label{qq2}
	\omega_\mathbb{A}\left[\begin{pmatrix}
	O&Q \\
	R&O
	\end{pmatrix}\right]\leq \frac{1}{2}\Big(\omega_A(Q+R)+\omega_A(Q-R)\Big).
\end{equation}
Combining \eqref{qq2} together with \eqref{sho} yields to the desired result.
\end{proof}

Our next objective is to present an improvement of the inequality \eqref{upper1}. To do this, we need the following lemma.

\begin{lemma}\label{p4200}(\cite{HORN})	Let $T=[t_{ij}]\in M_n(\mathbb{C})$ be such that $t_{ij}\geq 0$ for all $i, j=1, 2,\ldots, n$. Then
	$$\omega(T)= \frac{1}{2}r([t_{ij}+t_{ji}]).$$
\end{lemma}

\begin{theorem}\label{them100}
Let $\mathbb{T}=\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}$ be such that $P, Q, R, S\in \mathcal{B}_{A}(\mathcal{H})$. Then,
\begin{align*}
&\omega_{\mathbb{A}}\left[\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}\right]\\
&\leq \frac{1}{2}\bigg(\omega_A(P)+\omega_A(S)+\sqrt{(\omega_A(P)-\omega_A(S))^2+(\omega_A(Q+R)+\omega_A(Q-R))^2}\bigg).
\end{align*}
Moreover, the inequality is sharper than the inequality \eqref{upper1}.
\end{theorem}
\begin{proof}
It follows from \cite[Theorem 2.3]{feki04} that
	\begin{align*}
	\omega_\mathbb{A}\left(\mathbb{T}\right)
&\leq\omega\left[\begin{pmatrix}
	\omega_\mathbb{A}\left({P}\right) & \omega_{\mathbb{A}}\begin{pmatrix}
	O&Q \\
	R&O
	\end{pmatrix}\\
	\omega_{\mathbb{A}}\begin{pmatrix}
	O&Q \\
	R&O
	\end{pmatrix} & \omega_\mathbb{A}\left({S}\right)
	\end{pmatrix} \right] \\
	&=r\left[\begin{pmatrix}
	\omega_\mathbb{A}\left({P}\right) & \omega_{\mathbb{A}}\begin{pmatrix}
	O&Q \\
	R&O
	\end{pmatrix}\\
	\omega_{\mathbb{A}}\begin{pmatrix}
	O&Q \\
	R&O
	\end{pmatrix} & \omega_\mathbb{A}\left({S}\right)
	\end{pmatrix}\right]\\
	&=\frac{1}{2}\bigg[\omega_A(P)+\omega_A(S)+\sqrt{(\omega_A(P)-\omega_A(S))^2+4\omega^2_{\mathbb{A}}\begin{pmatrix}
		O&Q \\
		R&O
		\end{pmatrix}}\bigg],
	\end{align*}
where the last equality follows from Lemma \ref{p4200}. So, by applying \eqref{qq2} we get
\begin{align*}
&\omega_{\mathbb{A}}\left[\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}\right]\\
&\leq \frac{1}{2}\bigg[\omega_A(P)+\omega_A(S)+\sqrt{(\omega_A(P)-\omega_A(S))^2+(\omega_A(Q+R)+\omega_A(Q-R))^2}\bigg].
\end{align*}
This proves the desired inequality. Moreover, it can be observed that
\begin{align*}
& \max\{\omega_A(P), \omega_A(S)\}+\frac{\omega_A(Q+R)+\omega_A(Q-R)}{2}\\
 &=\frac{\omega_A(P)+\omega_A(S)+|\omega_A(P)-\omega_A(S)|}{2} + \frac{\omega_A(Q+R)+ \omega_A(Q-R)}{2}\\
&\geq \frac{1}{2}\bigg(\omega_A(P)+\omega_A(S)+\sqrt{(\omega_A(P)-\omega_A(S))^2+(\omega_A(Q+R)+\omega_A(Q-R))^2}\bigg).
\end{align*}
Hence, the proof is complete.
\end{proof}

As a consequence of Theorem \ref{them100} we state the following corollary.
\begin{corollary}\label{cor100}
Let  $P, Q, R, S\in \mathcal{B}_{A}(\mathcal{H})$. Then,
\begin{align*}
r_A(PQ+RS)
&\leq \frac{1}{2}\big[\omega_A(QP)+\omega_A(SR)\big]\\
&+\frac{1}{2}\sqrt{\big[\omega_A(QP)-\omega_A(SR)\big]^2+\big[\omega_A(QR+SP)+\omega_A(QR-SP)\big]^2}.
\end{align*}
\end{corollary}
In order to prove Corollary \ref{cor100}, we need the following lemma.
\begin{lemma}(\cite{feki01})\label{kkkk2020}
If $T\in \mathcal{B}_{A^{1/2}}(\mathcal{H})$, then
\begin{equation*}
r_A(T)\leq \omega_A(T).
\end{equation*}
\end{lemma}
Now we are in a position to establish Corollary \ref{cor100}.
\begin{proof}[Proof of Corollary \ref{cor100}]
It can be observed that
\begin{align*}
    r_A(PQ+RS)
    &=r_{\mathbb{A}}\left[\begin{pmatrix}
PQ+RS&O \\
O&O
\end{pmatrix}\right]\\
    &=r_{\mathbb{A}}\left[\begin{pmatrix}
P&R \\
O&O
\end{pmatrix}\begin{pmatrix}
Q&O \\
S&O
\end{pmatrix}\right]\\
&=r_{\mathbb{A}}\left[\begin{pmatrix}
Q&O \\
S&O
\end{pmatrix}\begin{pmatrix}
P&R \\
O&O
\end{pmatrix}\right]\quad (\text{by }\,\eqref{commut})\\
&=r_{\mathbb{A}}\left[\begin{pmatrix}
QP&QR \\
SP&SR
\end{pmatrix}\right]\\
&\leq \omega_{\mathbb{A}}\left[\begin{pmatrix}
QP&QR \\
SP&SR
\end{pmatrix}\right]\quad(\text{by Lemma } \ref{kkkk2020}).
\end{align*}
So, by applying  Theorem \ref{them100} we reach the required result.
\end{proof}
\begin{remark}
Recently the first author proved in \cite{feki02} that for $P, Q, R, S\in \mathcal{B}_{A}(\mathcal{H})$ it holds
\begin{align}\label{kk2020}
r_A(PQ+RS)
&\leq \frac{1}{2}\big[\omega_A(QP)+\omega_A(SR)\big]\nonumber\\
&+\frac{1}{2}\sqrt{\big[\omega_A(QP)-\omega_A(SR)\big]^2+4\left\Vert QR\right\Vert_A \left\Vert SP\right\Vert_A }.
\end{align}
If $QR=SP$, then it can be seen that the inequality in Corollary \ref{cor100} is sharper than \eqref{kk2020}.
\end{remark}

\begin{remark}
Notice that by letting $Q=S=I$ in Corollary \ref{cor100} we get
\begin{align*}
r_A(P+R)
&\leq \frac{1}{2}\big[\omega_A(P)+\omega_A(R)\big]\\
&+\frac{1}{2}\sqrt{(\omega_A(P)-\omega_A(R))^2+(\omega_A(P+R)+\omega_A(P-R))^2}.
\end{align*}
\end{remark}

To establish further upper bounds for the $\mathbb{A}$-numerical radius of the operator matrix $\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}$, we need the following lemmas.
\begin{lemma}\label{lmm05}(\cite{feki02})
Let $\mathbb{T}=\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}$ be such that $P, Q, R, S\in \mathcal{B}_{A^{1/2}}(\mathcal{H})$. Then, $\mathbb{T}\in \mathcal{B}_{\mathbb{A}^{1/2}}(\mathcal{H}\oplus \mathcal{H})$ and
$$\|\mathbb{T}\|_\mathbb{A}\leq \left\|\begin{pmatrix}
\|P\|_A & \|Q\|_A \\
\|R\|_A & \|S\|_A
\end{pmatrix}\right\|.$$
\end{lemma}
\begin{lemma}\label{a5so}
Let $T,S\in \mathcal{B}_{A}(\mathcal{H})$. Then,
$$\|T^{\sharp_A}S\|_A=\|S^{\sharp_A}T\|_A.$$
\end{lemma}
\begin{proof}
By using the fact that $P_{\overline{\mathcal{R}(A)}}A=AP_{\overline{\mathcal{R}(A)}}=A$ together with \eqref{newsemi}, we see that
\begin{align*}
\|T^{\sharp_A}S\|_A
&=\|S^{\sharp_A}P_{\overline{\mathcal{R}(A)}}TP_{\overline{\mathcal{R}(A)}}\|_A\\
&=\sup\left\{|\langle AP_{\overline{\mathcal{R}(A)}}x, (S^{\sharp_A}P_{\overline{\mathcal{R}(A)}}T)^{\sharp_A}y\rangle|\,;\;x,y\in \mathcal{H},\,\|x\|_{A}=\|y\|_{A}= 1\right\}\\
&=\sup\left\{|\langle S^{\sharp_A}P_{\overline{\mathcal{R}(A)}}Tx, y\rangle_A|\,;\;x,y\in \mathcal{H},\,\|x\|_{A}=\|y\|_{A}= 1\right\}\\
&=\sup\left\{|\langle AP_{\overline{\mathcal{R}(A)}}Tx, Sy\rangle|\,;\;x,y\in \mathcal{H},\,\|x\|_{A}=\|y\|_{A}= 1\right\}\\
&=\sup\left\{|\langle S^{\sharp_A}Tx, y\rangle_A|\,;\;x,y\in \mathcal{H},\,\|x\|_{A}=\|y\|_{A}= 1\right\}\\
&=\|S^{\sharp_A}T\|_A.
\end{align*}
This proves the desired result.
\end{proof}

Now, we are in a position to state the following theorem.
\begin{theorem}\label{theorem:upper 3}
Let $P,Q,R,S\in \mathcal{B}_{A}(\mathcal{H})$. Then,
\[\omega_\mathbb{A}\left[\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}\right]\leq \min\{\mu,\nu \},\]
where
\begin{align*}
\mu
 & =\frac{\sqrt{2}}{2}\sqrt{\|P\|_A^2+\|Q\|_A^2+\sqrt{(\|P\|_A^2-\|Q\|_A^2)^2 +4\|P^{\sharp_A}Q\|_A^2} } \\
 &\quad\quad\quad+\frac{\sqrt{2}}{2}\sqrt{\|R\|_A^2+\|S\|_A^2+\sqrt{(\|R\|_A^2-\|S\|_A^2)^2 +4\|S^{\sharp_A}R\|_A^2} },
\end{align*}
and
\begin{align*}
\nu
& =\frac{\sqrt{2}}{2}\sqrt{\|P\|_A^2+\|R\|_A^2+\sqrt{(\|P\|_A^2-\|R\|_A^2)^2+4\|PR^{\sharp_A}\|_A^2 }} \\
 &\quad\quad\quad+\frac{\sqrt{2}}{2}\sqrt{\|Q\|_A^2+\|S\|_A^2+\sqrt{(\|Q\|_A^2-\|S\|_A^2)^2 +4\|SQ^{\sharp_A}\|_A^2} }.
\end{align*}
\end{theorem}
\begin{proof}
We first prove that
\begin{equation}\label{ffirst}
\omega_\mathbb{A}\left[\begin{pmatrix}
P&Q \\
O&O
\end{pmatrix}\right]\leq \frac{\sqrt{2}}{2}\sqrt{\|P\|_A^2+\|Q\|_A^2+\sqrt{(\|P\|_A^2-\|Q\|_A^2)^2+4\|P^{\sharp_A}Q\|_A^2 }}\,.
\end{equation}
By using \eqref{refine1} together with \eqref{diez} we see that
\begin{align*}
 \omega_\mathbb{A}\left[\begin{pmatrix}
P&Q\\
    O& O
    \end{pmatrix}\right]
    &\leq \left\|\begin{pmatrix}
P&Q\\
    O & O
    \end{pmatrix}\right\|_\mathbb{A}\\
    &=\left\|\begin{pmatrix}
P&Q\\
    O& O
    \end{pmatrix}^{\sharp_\mathbb{A}}
    \begin{pmatrix}
P&Q\\
    O& O
    \end{pmatrix}\right\|_\mathbb{A}^{\frac{1}{2}}\\
    &=\left\|\begin{pmatrix}
    P^{\sharp_A}& O\\
    Q^{\sharp_A}& O
    \end{pmatrix}\begin{pmatrix}
    P&Q\\
    O& O
    \end{pmatrix}\right\|_\mathbb{A}^{\frac{1}{2}}\\
    &=\left\|\begin{pmatrix}
    P^{\sharp_A}P & P^{\sharp_A}Q\\
    Q^{\sharp_A}P& Q^{\sharp_A}Q
    \end{pmatrix}\right\|_\mathbb{A}^{\frac{1}{2}}.
\end{align*}
So, by using Lemma \ref{lmm05} together with Lemma \ref{a5so} we see that
\begin{align*}
 \omega_\mathbb{A}^2\left[\begin{pmatrix}
P&Q\\
    O& O
    \end{pmatrix}\right]
&\leq \left\|\begin{pmatrix}
    \|P^{\sharp_A}P\|_A & \|P^{\sharp_A}Q\|_A\\
    \|Q^{\sharp_A}P\|_A& \|Q^{\sharp_A}Q\|_A
    \end{pmatrix}\right\|\\
 &=\left\|\begin{pmatrix}
    \|P\|_A^2& \|P^{\sharp_A}Q\|_A\\
    \|P^{\sharp_A}Q\|_A& \|Q\|_A^2
    \end{pmatrix}\right\|\\
  &=r\left[\begin{pmatrix}
    \|P\|_A^2& \|P^{\sharp_A}Q\|_A\\
    \|P^{\sharp_A}Q\|_A& \|Q\|_A^2
    \end{pmatrix}\right]\\
  &= \frac{1}{2}\left(\|P\|_A^2+\|Q\|_A^2+\sqrt{(\|P\|_A^2-\|Q\|_A^2)^2+4\|P^{\sharp_A}Q\|_A^2}\right).
\end{align*}
This proves \eqref{ffirst}. Let $\mathbb{U}=\begin{pmatrix}
O&I \\
I&O
\end{pmatrix}.$ In view of Lemma \ref{lemma1} (iii) we have $\mathbb{U}\in \mathcal{B}_{\mathbb{A}}(\mathcal{H}\oplus \mathcal{H})$ and $\mathbb{U}^{\sharp_{\mathbb{A}}}=\begin{pmatrix}
O&P_{\overline{\mathcal{R}(A)}}\\
P_{\overline{\mathcal{R}(A)}}&O
\end{pmatrix}.$ Further, it can be seen that $\mathbb{U}$ is $\mathbb{A}$-unitary operator. So, by using Lemma \ref{weak} together with \eqref{ffirst} we get
\begin{align*}
\omega_{\mathbb{A}}\left[\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}\right]
& \leq\omega_{\mathbb{A}}\left[\begin{pmatrix}
P&Q\\
O &O
\end{pmatrix}\right]+\omega_{\mathbb{A}}\left[\begin{pmatrix}
O &O\\
R&S
\end{pmatrix}\right] \\
 &=\omega_{\mathbb{A}}\left[\begin{pmatrix}
P&Q\\
O &O
\end{pmatrix}\right]+\omega_{\mathbb{A}}\left[\mathbb{U}^{\sharp_{\mathbb{A}}}\begin{pmatrix}
O &O\\
R&S
\end{pmatrix}\mathbb{U}\right] \\
 &=\omega_{\mathbb{A}}\left[\begin{pmatrix}
P&Q\\
O &O
\end{pmatrix}\right]+\omega_{\mathbb{A}}\left[\begin{pmatrix}
P_{\overline{\mathcal{R}(A)}} &O\\
O &P_{\overline{\mathcal{R}(A)}}
\end{pmatrix}
\begin{pmatrix}
S &R\\
O &O
\end{pmatrix}\right]\\
 &=\omega_{\mathbb{A}}\left[\begin{pmatrix}
P&Q\\
O &O
\end{pmatrix}\right]+\omega_{\mathbb{A}}\left[
\begin{pmatrix}
S &R\\
O &O
\end{pmatrix}\right]\\
 & \leq \frac{\sqrt{2}}{2}\sqrt{\|P\|_A^2+\|Q\|_A^2+\sqrt{(\|P\|_A^2-\|Q\|_A^2)^2 +4\|P^{\sharp_A}Q\|_A^2} } \\
 &+\frac{\sqrt{2}}{2}\sqrt{\|R\|_A^2+\|S\|_A^2+\sqrt{(\|R\|_A^2-\|S\|_A^2)^2 +4\|S^{\sharp_A}R\|_A^2} }.
\end{align*}
Now, since $\omega_\mathbb{A}\left[\begin{pmatrix}
P&Q \\
R&S
\end{pmatrix}\right]=\omega_\mathbb{A}\left[\begin{pmatrix}
P^{\sharp_A} & R^{\sharp_A} \\
Q^{\sharp_A} & S^{\sharp_A}
\end{pmatrix}\right]$, then by using similar arguments as above we obtain
\begin{align*}
&\omega_\mathbb{A}\left[\begin{pmatrix}
P&Q \\
R&S
    \end{pmatrix}\right]\\
& \leq\frac{\sqrt{2}}{2}\sqrt{\|P^{\sharp_A}\|_A^2+\|R^{\sharp_A}\|_A^2+\sqrt{(\|P^{\sharp_A}\|_A^2-\|R^{\sharp_A}\|_A^2)^2+4\|(P^{\sharp_A})^{\sharp_A}R^{\sharp_A}\|_A^2 }} \\
 &+\frac{\sqrt{2}}{2}\sqrt{\|Q^{\sharp_A}\|_A^2+\|S^{\sharp_A}\|_A^2+\sqrt{(\|Q^{\sharp_A}\|_A^2-\|S^{\sharp_A}\|_A^2)^2 +4\|(S^{\sharp_A})^{\sharp_A}Q^{\sharp_A}\|_A^2} }\\
 & =\frac{\sqrt{2}}{2}\sqrt{\|P\|_A^2+\|R\|_A^2+\sqrt{(\|P\|_A^2-\|R\|_A^2)^2+4\|PR^{\sharp_A}\|_A^2 }} \\
 &+\frac{\sqrt{2}}{2}\sqrt{\|Q\|_A^2+\|S\|_A^2+\sqrt{(\|Q\|_A^2-\|S\|_A^2)^2 +4\|SQ^{\sharp_A}\|_A^2} }\,.
\end{align*}
Hence, the proof is complete.
\end{proof}

Our next result provides a lower bound for the $\mathbb{A}$-numerical radius of a $2 \times 2$ operator matrix such that its second row consists of zero operators. We mention here that our result improves a recent result proved by Rout et al. in \cite{rout} since $\mathcal{B}_{A}(\mathcal{H})\subseteq\mathcal{B}_{A^{1/2}}(\mathcal{H})$.
\begin{theorem}\label{Theorem 2.6}
	Let $P, Q\in\mathcal{B}_{A^{1/2}}(\mathcal{H})$. Then
\begin{equation}\label{sahoo1}
\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P & Q\\
	O & O
	\end{pmatrix}\right]\geq \frac{1}{2}\max\{\alpha,\beta\},
\end{equation}
where $\alpha=\omega_A(P+ Q)+\omega_A(P- Q)$ and $\beta=\omega_A(P+ iQ)+\omega_A(P- iQ)$.
\end{theorem}
\begin{proof}
	Let $\mathbb{U}=\begin{pmatrix}
	O & I\\I & O
	\end{pmatrix}$.
	By using Lemma \ref{lemma1} (iii), we see that
	$$\mathbb{U}^{\sharp_\mathbb{A}}=\begin{pmatrix}
	P_{\overline{\mathcal{R}(A)}} &O\\
	O &P_{\overline{\mathcal{R}(A)}}
	\end{pmatrix}\mathbb{U}=\begin{pmatrix}
	O & P_{\overline{\mathcal{R}(A)}}\\
	P_{\overline{\mathcal{R}(A)}} & O
	\end{pmatrix}.$$
	So, by using the fact that $AP_{\overline{\mathcal{R}(A)}}=A$, we can verify that $\|\mathbb{U}x\|_{\mathbb{A}}=\|\mathbb{U}^{\sharp_{\mathbb{A}}}x\|_{\mathbb{A}}=\|x\|_{\mathbb{A}}$ for all $x\in \mathcal{H}\oplus \mathcal{H}$. Hence $\mathbb{U}$ is $\mathbb{A}$-unitary. So, using  Lemma  \ref{lem100} (i) we observes that
	\begin{align*}
	\max\{\omega_A(P+Q),\omega_A(P-Q)\}=&\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P&Q\\
	Q &P
	\end{pmatrix}\right]\\
	& \leq\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P&Q\\
	O &O
	\end{pmatrix}\right]+\omega_{\mathbb{A}}\left[\begin{pmatrix}
	O &O\\
	Q&P
	\end{pmatrix}\right] \\
	&=\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P&Q\\
	O &O
	\end{pmatrix}\right]+\omega_{\mathbb{A}}\left[\mathbb{U}^{\sharp_{\mathbb{A}}}\begin{pmatrix}
	O &O\\
	Q&P
	\end{pmatrix}\mathbb{U}\right] \\
	&=\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P&Q\\
	O &O
	\end{pmatrix}\right]+\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P_{\overline{\mathcal{R}(A)}} &O\\
	O &P_{\overline{\mathcal{R}(A)}}
	\end{pmatrix}
	\begin{pmatrix}
	P &Q\\
	O &O
	\end{pmatrix}\right]\\
	&=\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P&Q\\
	O &O
	\end{pmatrix}\right]+\omega_{\mathbb{A}}\left[
	\begin{pmatrix}
	P &Q\\
	O &O
	\end{pmatrix}\right]\\
	=&2\omega_{\mathbb{A}}\left[
	\begin{pmatrix}
	P &Q\\
	O &O
	\end{pmatrix}\right].
	\end{align*}

On the other hand, by considering the $\mathbb{A}$-unitary operator $\mathbb{V}=\begin{pmatrix}
	I & O\\
O & -I
	\end{pmatrix}$ and proceeding as above we see that
	\begin{align*}
	\max\{\omega_A(P+iQ),\omega_A(P-iQ)\}=&\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P&-Q\\
	Q &P
	\end{pmatrix}\right]\quad(\text{by Lemma } \ref{lem100}\;(ii))\\
	&\leq 2\omega_{\mathbb{A}}\left[
	\begin{pmatrix}
	P &Q\\
	O &O
	\end{pmatrix}\right].
	\end{align*}
	Hence we get our desired result.
\end{proof}

Now, in order to prove a lower bound for $\omega_\mathbb{A}\left[\begin{pmatrix}
	P & Q\\
	O & O
	\end{pmatrix}\right]$, we need the following lemma.
\begin{lemma}(\cite{feki004})\label{n1}
Let $T, S\in\mathcal{B}_{A}(\mathcal{H})$. Then,
\begin{align}\label{SK1}
\omega_A(TS\pm ST^{\sharp_A}) \leq2\|T\|_A\,\omega_A(S).
\end{align}
\end{lemma}

In the next result, Lemma \ref{lem100} enables us to present another application of the inequality \eqref{SK1}.
\begin{theorem}\label{app1}
	Let $P, Q\in\mathcal{B}_A(\mathcal{H})$. Then
\begin{equation}\label{sahoo3}
\omega_\mathbb{A}\left[\begin{pmatrix}
	P & Q\\
	O & O
	\end{pmatrix}\right]\geq \frac{1}{2}\max\left\{\omega_A(P+ iQ),\omega_A(P- iQ)\right\}.
\end{equation}
\end{theorem}
\begin{proof}
Let $\mathbb{T}=\begin{pmatrix}
	P & Q\\
	O & O
	\end{pmatrix}$ and $\mathbb{U}=\begin{pmatrix}
	O & -I\\I & O
	\end{pmatrix}$. It is not difficult to verify that $\mathbb{U}$ is $\mathbb{A}$-unitary. Moreover, since $P_{\overline{\mathcal{R}(A)}}X^{\sharp_A}=X^{\sharp_A}P_{\overline{\mathcal{R}(A)}}=X^{\sharp_A}$ for all $X\in\mathcal{B}_A(\mathcal{H})$ (see \cite{faiot}), then it can be verified that
$$\mathbb{U}^{\sharp_{\mathbb{A}}}\mathbb{T}^{\sharp_{\mathbb{A}}}+\mathbb{T}^{\sharp_{\mathbb{A}}}(\mathbb{U}^{\sharp_{\mathbb{A}}})^{\sharp_{\mathbb{A}}}=\begin{pmatrix}
	Q^{\sharp_A} & P^{\sharp_A}\\
	-P^{\sharp_A} & Q^{\sharp_A}
	\end{pmatrix} =\begin{pmatrix}
	Q & -P\\
	P & Q
	\end{pmatrix}^{\sharp_{\mathbb{A}}}.$$
By using Lemma \ref{n1} we see that
$$\omega_{\mathbb{A}}(\mathbb{U}^{\sharp_{\mathbb{A}}}\mathbb{T}^{\sharp_{\mathbb{A}}}+\mathbb{T}^{\sharp_{\mathbb{A}}}(\mathbb{U}^{\sharp_{\mathbb{A}}})^{\sharp_{\mathbb{A}}})
\leq 2\omega_{\mathbb{A}}(\mathbb{T}^{\sharp_{\mathbb{A}}}),$$
which, in turn, implies that
\begin{align*}
\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P & Q\\
	O & O
	\end{pmatrix} \right]
&\geq \frac{1}{2}\omega_{\mathbb{A}}\left[\begin{pmatrix}
	Q & -P\\
	P & Q
	\end{pmatrix}^{\sharp_{\mathbb{A}}} \right] \\
&=\frac{1}{2}\omega_{\mathbb{A}}\left[\begin{pmatrix}
	Q & -P\\
	P & Q
	\end{pmatrix}\right] \\
 &=\frac{1}{2}\max\left\{\omega_A(P+ iQ),\omega_A(P- iQ)\right\},
\end{align*}
where the last equality follows by applying Lemma \ref{lem100}(ii).
\end{proof}

As an application of the above theorem, we can derive the following $A$-numerical radius inequality.
\begin{theorem}\label{f12}
Let $T\in \mathcal{B}_A(\mathcal{H})$. Then
$$\omega_A(T)\leq 2\min\left\{\omega_\mathbb{A}\left[\begin{pmatrix}
	\Re_A(T) & O\\
	\Im_A(T) & O
	\end{pmatrix}\right], \omega_\mathbb{A}\left[\begin{pmatrix}
	O & -i\Im_A(T)\\
	\Re_A(T) & O
	\end{pmatrix}\right] \right\}. $$
\end{theorem}
To prove Theorem \ref{f12}, we need the following lemma.
\begin{lemma}\label{lemf}
Let $T,S\in \mathcal{B}_A(\mathcal{H})$. Then
\begin{align}
 \omega_A(T\pm iS)\leq 2\,\omega_{\mathbb{A}}\left[\begin{pmatrix}
O&T\\
iS &O
\end{pmatrix}\right].
\end{align}
\end{lemma}
\begin{proof}
Let $\mathbb{X}=\begin{pmatrix}
I &I\\
O &O
\end{pmatrix}$ and $\mathbb{Y}=\begin{pmatrix}
O &T\\
S &O
\end{pmatrix}$. It can be observed that
$$\mathbb{X}\mathbb{Y}\mathbb{X}^{\sharp_\mathbb{A}}=\begin{pmatrix}
TP_{\overline{\mathcal{R}(A)}}+ SP_{\overline{\mathcal{R}(A)}} &O\\
O &O
\end{pmatrix},$$
and
\begin{align*}
\|\mathbb{X}\|_{\mathbb{A}}^2=\|\mathbb{X}\mathbb{X}^{\sharp_\mathbb{A}}\|_{\mathbb{A}}
&=\left\|\begin{pmatrix}
2P_{\overline{\mathcal{R}(A)}}&O\\
O &O
\end{pmatrix}\right\|_{\mathbb{A}}\\
 &=2\|P_{\overline{\mathcal{R}(A)}}\|_A=2.
\end{align*}
So, by using the fact that $AP_{\overline{\mathcal{R}(A)}}=A$ we see that
\begin{align*}\label{inff}
\omega_A(T+S)
&=\omega_A(TP_{\overline{\mathcal{R}(A)}}+ SP_{\overline{\mathcal{R}(A)}})\\
&=\omega_\mathbb{A}\left[\begin{pmatrix}
TP_{\overline{\mathcal{R}(A)}}+ SP_{\overline{\mathcal{R}(A)}} &O\\
O &O
\end{pmatrix}\right]\\
& = \omega_\mathbb{A}(\mathbb{X}\mathbb{Y}\mathbb{X}^{\sharp_\mathbb{A}})\quad(\text{by Lemma } \ref{lemma1})\nonumber\\
 &\leq \|\mathbb{X}\|_{\mathbb{A}}^2\omega_\mathbb{A}(\mathbb{Y})\quad(\text{by \cite[Lemma 4.4.]{zamani1}})\nonumber\\
  &= 2\omega_\mathbb{A}(\mathbb{Y}).
\end{align*}
This gives
\begin{equation}\label{c2}
\omega_A(T+S)\leq2\omega_\mathbb{A}\left[\begin{pmatrix}
O &T\\
S &O
\end{pmatrix}\right].
\end{equation}
By replacing $S$ by $-S$ in \eqref{c2} and then using the fact that $\omega_\mathbb{A}\left[\begin{pmatrix}
O &T\\
S &O
\end{pmatrix}\right]=\omega_\mathbb{A}\left[\begin{pmatrix}
O &T\\
-S &O
\end{pmatrix}\right]$ we get
\begin{equation}\label{5arita}
 \omega_A(T\pm S)\leq 2\,\omega_{\mathbb{A}}\left[\begin{pmatrix}
O&T\\
S &O
\end{pmatrix}\right].
\end{equation}
Finally, by replacing $S$ by $iS$ in \eqref{5arita}, we reach the desired results.
\end{proof}
Now we are ready to prove Theorem \ref{f12}.

\begin{proof}[Proof of Theorem \ref{f12}]
Clearly $T$ can written as $T=\Re_A(T)+i\Im_A(T)$ where
$$\Re_A(T):=\frac{T+T^{\sharp_A}}{2}\;\;\text{ and }\;\;\Im_A(T):=\frac{T-T^{\sharp_A}}{2i}.$$
So, $T^{\sharp_A}=[\Re_A(T)]^{\sharp_A}-i[\Im_A(T)]^{\sharp_A}$. Moreover, a short calculation reveals that $(T^{\sharp_A})^{\sharp_A}=[\Re_A(T)]^{\sharp_A}+i[\Im_A(T)]^{\sharp_A}$. So, by applying Theorem \ref{app1} we see that
\begin{align*}
&\omega_\mathbb{A}\left[\begin{pmatrix}
	\Re_A(T) & O\\
	\Im_A(T) & O
	\end{pmatrix}\right]\\
&=\omega_\mathbb{A}\left[\begin{pmatrix}
	[\Re_A(T)]^{\sharp_A} & [\Im_A(T)]^{\sharp_A}\\
	O & O
	\end{pmatrix}\right]\\
&\geq \frac{1}{2}\max\left\{\omega_A\Big([\Re_A(T)]^{\sharp_A}+ i[\Im_A(T)]^{\sharp_A}\Big),\omega_A\Big([\Re_A(T)]^{\sharp_A}- i[\Im_A(T)]^{\sharp_A}\Big)\right\}\\
&=\frac{1}{2}\max\left\{\omega_A(T^{\sharp_A}),\omega_A\left((T^{\sharp_A})^{\sharp_A}\right)\right\}\\
&=\frac{1}{2}\omega_A(T).
\end{align*}
On the other hand, by applying Lemma \ref{lemf} we get
\begin{align*}
&\omega_\mathbb{A}\left[\begin{pmatrix}
	O & -i\Im_A(T)\\
	\Re_A(T) & O
	\end{pmatrix}\right]\\
&=\omega_\mathbb{A}\left[\begin{pmatrix}
	O & [\Re_A(T)]^{\sharp_A}\\
	i[\Im_A(T)]^{\sharp_A} & O
	\end{pmatrix}\right]\\
&\geq \frac{1}{2}\max\left\{\omega_A\Big([\Re_A(T)]^{\sharp_A}+ i[\Im_A(T)]^{\sharp_A}\Big),\omega_A\Big([\Re_A(T)]^{\sharp_A}- i[\Im_A(T)]^{\sharp_A}\Big)\right\}\\
&=\frac{1}{2}\omega_A(T).
\end{align*}
\end{proof}

Based on Lemma \ref{lem100}, the forthcoming theorem which provides a lower bound for $\mathbb{A}$-numerical radius of a $2 \times 2$ operator matrix is also an application of the inequality \eqref{SK1}.
\begin{theorem}
	Let $P, Q,R,S\in\mathcal{B}_A(\mathcal{H})$. Then
\begin{equation}\label{sahoo2}
\omega_\mathbb{A}\left[\begin{pmatrix}
	P & Q\\
	R & S
	\end{pmatrix}\right]\geq \frac{1}{2}\max\{\mu,\nu\},
\end{equation}
where $$\mu=\max\{\omega_A(Q+R+S+P) ,\omega_A(Q+R-S-P) \},$$ and $$\nu=\max\big\{\omega_A\big(Q-R+i(S+P)\big) ,\omega_A\big(Q-R-i(S+P)\big)\big\}.$$
\end{theorem}
\begin{proof}
	Let $\mathbb{U}=\begin{pmatrix}
	O & I\\I & O
	\end{pmatrix}$.
	By using Lemma \ref{lemma1} (iii), we see that
	$$\mathbb{U}^{\sharp_\mathbb{A}}=\begin{pmatrix}
	P_{\overline{\mathcal{R}(A)}} &O\\
	O &P_{\overline{\mathcal{R}(A)}}
	\end{pmatrix}\mathbb{U}=\begin{pmatrix}
	O &P_{\overline{\mathcal{R}(A)}}\\
	P_{\overline{\mathcal{R}(A)}} &O
	\end{pmatrix}.$$
	So, by using the fact that $AP_{\overline{\mathcal{R}(A)}}=A$, we can verify that $\|\mathbb{U}x\|_{\mathbb{A}}=\|\mathbb{U}^{\sharp_{\mathbb{A}}}x\|_{\mathbb{A}}=\|x\|_{\mathbb{A}}$ for all $x\in \mathcal{H}\oplus \mathcal{H}$. Hence $\mathbb{U}$ is $\mathbb{A}$-unitary. Moreover, clearly $(\mathbb{U}^{\sharp_{\mathbb{A}}})^{\sharp_{\mathbb{A}}}=\mathbb{U}^{\sharp_{\mathbb{A}}}$. Now, let $\mathbb{T}=\begin{pmatrix}
	P & Q\\
	R & S
	\end{pmatrix}$. Since, $P_{\overline{\mathcal{R}(A)}}X^{\sharp_A}=X^{\sharp_A}P_{\overline{\mathcal{R}(A)}}=X^{\sharp_A}$ for all $X\in\mathcal{B}_A(\mathcal{H})$ (see \cite{faiot}), then a short calculation shows that
$$\mathbb{U}^{\sharp_{\mathbb{A}}}\mathbb{T}^{\sharp_{\mathbb{A}}}+\mathbb{T}^{\sharp_{\mathbb{A}}}(\mathbb{U}^{\sharp_{\mathbb{A}}})^{\sharp_{\mathbb{A}}}=\begin{pmatrix}
	Q^{\sharp_A}+R^{\sharp_A} & S^{\sharp_A}+P^{\sharp_A}\\
	S^{\sharp_A}+P^{\sharp_A} & Q^{\sharp_A}+R^{\sharp_A}
	\end{pmatrix} =\begin{pmatrix}
	Q+R & S+P\\
	S+P & Q+R
	\end{pmatrix}^{\sharp_{\mathbb{A}}}.$$
By using Lemma \ref{n1}, we see that
$$\omega_{\mathbb{A}}(\mathbb{U}^{\sharp_{\mathbb{A}}}\mathbb{T}^{\sharp_{\mathbb{A}}}+\mathbb{T}^{\sharp_{\mathbb{A}}}(\mathbb{U}^{\sharp_{\mathbb{A}}})^{\sharp_{\mathbb{A}}})
\leq 2\omega_{\mathbb{A}}(\mathbb{T}^{\sharp_{\mathbb{A}}}),$$
which, in turn, implies that
\begin{align*}
\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P & Q\\
	R & S
	\end{pmatrix} \right]
&\geq \frac{1}{2}\omega_{\mathbb{A}}\left[\begin{pmatrix}
	Q+R & S+P\\
	S+P & Q+R
	\end{pmatrix} \right] \\
 &=\frac{1}{2}\max\{\omega_A(Q+R+S+P) ,\omega_A(Q+R-S-P) \}:=\frac{1}{2}\mu,
\end{align*}
where the last equality follows by applying Lemma \ref{lem100} (i). On the other hand, by choosing $\mathbb{U}=\begin{pmatrix}
	O & I\\-I & O
	\end{pmatrix}$ and proceeding as above we obtain
\begin{align*}
\omega_{\mathbb{A}}\left[\begin{pmatrix}
	P & Q\\
	R & S
	\end{pmatrix} \right]
&\geq \frac{1}{2}\omega_{\mathbb{A}}\left[\begin{pmatrix}
	Q-R & -(S+P)\\
	S+P & Q-R
	\end{pmatrix} \right] \\
 &=\frac{1}{2}\max\big\{\omega_A\big(Q-R+i(S+P)\big) ,\omega_A\big(Q-R-i(S+P)\big)\big\}:=\frac{1}{2}\nu,
\end{align*}
where the last equality follows by applying Lemma \ref{lem100} (ii). This completes the proof of the theorem.
\end{proof}

config: arxiv


TITLE: Non-inductive, not combinatorial proof of $\sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$
QUESTION [7 upvotes]: I've seen the identity $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ used here recently.
I checked for proofs here http://www.proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients
I couldn't have figured out the combinatorial proof by myself, and the inductive proof assumes you already know the answer...
So my question is : do you know how to prove directly through computation that $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ ?

REPLY [9 votes]: $\newcommand{\+}{^{\dagger}}
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${\tt\color{#f00}{\mbox{This method is a "direct calculation" which is very}}}$
${\tt\color{#f00}{\mbox{ convenient when we don't know the answer and, in addition,}}}$
${\tt\color{#f00}{\mbox{ we don't have to guess combinations of Newton binomials.}}}$
It's based in the identity:

$$
{m \choose s}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{m} \over z^{s + 1}}\,{\dd z \over 2\pi\ic}
$$

\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 0}^{n}{n \choose k}^{2}} =
\sum_{k = 0}^{n}{n \choose k}
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over
z^{k + 1}}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z}
\sum_{k = 0}^{n}{n \choose k}\pars{1 \over z}^{k}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z}
\pars{1 + {1 \over z}}^{n}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\bbox[10px,border:1px groove navy]{2n \choose n}
\end{align}