TITLE: How to find two orthogonal vectors $a$ and $b$ such that $a$ is the projection of vector U on V and $a+b=U$
QUESTION [0 upvotes]: Given that U=<2,6> and V=<9,2> find two orthogonal vectors  $a$ and $b$ such that $a$ is the projection of vector U on V and $a+b=U$

I first wanted to know how to even do this problem and understand it.

I don't understand the "orthogonal vectors" part, I was wondering if this was referring to the legs the are formed when you project two vectors?
If not how does one get two orthogonal vectors out of this problem?
And also assuming that it is the legs, then I must do $Proj_{v}{u}$ right? And if I do that then I only get one vector, then what do I do? And also if this is the case, then is the fact that $a+b=U $ now useless?

REPLY [0 votes]: $a$ and $b$ orthogonal means that $\langle a, b \rangle =0$
$a$ is the projection of $U$ on $V$ gives that $a=\lambda V$

Then $\langle b, V \rangle=0$ and $\langle U, V \rangle=\langle a, V \rangle$. Then $\lambda \Vert V \Vert ^2 = \langle U, V \rangle = 30$. This gives $\lambda = \frac{6}{17}$.
Then $a=(\frac{54}{17}, \frac{12}{17})$ and $b=U-a=(\frac{-20}{17},\frac{90}{17})$


TITLE: Describe the rings: a) $\mathbb{Z}[x] / (x^2 - 3, 2x + 4)$, b) $\mathbb{Z}[i]/ (2 + i)$
QUESTION [2 upvotes]: Describe each of following the rings:
a) $\mathbb{Z}[x] / (x^2 - 3, 2x + 4)$
b) $\mathbb{Z}[i]/ (2 + i)$
a) Well, $\mathbb{Z}[x]$ is the set of all polynomials with integer coefficients and $(x^2 - 3, 2x + 4)$  is the ideal generated by two polynomials. I know that
$\mathbb Z[x] / (x^2 - 3, 2x + 4) = \{ r + (x^2 - 3, 2x + 4) : r \in \mathbb{Z}[x] \}$
b) I know that $\mathbb{Z}[i]$ is the set of all polynomials with the complex number $i$ as its variable. $\mathbb{Z}[i]/ (2 + i) = \{ r + (2 + i) : r \in \mathbb{Z}[i] \}$.
I already know that b) is $\mathbb Z$ modulo 5. What would a) be?

REPLY [5 votes]: b) The link didn't contain what I consider the cleanest approach to this problem, and this is a good warm-up for a), so I thought I'd answer it again.  The trick is to think of $\mathbb{Z}[i]$ as $\mathbb{Z}[x]/(x^2+1)$.  Then we can compute the quotient in two steps: $\mathbb{Z}[x]/(x^2+1,x+2) = \mathbb{Z}/((-2)^2+1) = \mathbb{Z}/(5)$.
a) We can write $(x^2-3,2x+4) = (x^2-3,2x+4,2x^2-6)=(x^2-3,2x+4,(-4)x-6)$ $=(x^2-3,2x+4,4x+6-2(2x+4))$ $=(x^2-3,2x+4,-2) = (x^2+1,2)$.  You can rewrite this in terms of $\mathbb{Z}[i]$ if you like, but I think it's better to make the substitution $y=x+1$ to get the lovely expression $\mathbb{F}_2[y]/(y^2)$.
Hopefully this was useful, and hopefully you know that your grader probably also uses StackExchange and cares whether you properly cite your sources.


TITLE: Show that a functional is a distribution
QUESTION [4 upvotes]: Consider the following functional
$$
  \langle u , \phi \rangle = \int_0^{\infty} \phi(t) \frac{\mathrm{d}t}{t^{\alpha}}
$$
I want to show that it is a functional in $\mathcal{D'}{(\mathbb{R})}$. Because of the compact support of $\phi$, the indefinite integral could be made definite by an upper bound $C$.
$$
  \langle u , \phi \rangle = \int_0^{C} \phi(t) \frac{\mathrm{d}t}{t^{\alpha}}
$$
So, first linearity is clear because of this property from the integral.
$$
 \langle u , a \phi + b \psi \rangle
  =  \int_0^{C} (a \phi(t) + b \psi) \frac{\mathrm{d}t}{t^{\alpha}}
  =  a \int_0^{C} \phi(t)\frac{\mathrm{d}t}{t^{\alpha}} + b \int_0^{C} \psi(t)\frac{\mathrm{d}t}{t^{\alpha}}
  =  a\langle u ,  \phi \rangle + b\langle u, b\psi \rangle
$$
For continuity i have to consider a sequence $\phi_k \to \phi$, but thats were i stuck, i have no idea how to show that $\langle u, \phi_k \rangle \to \langle u, \phi \rangle$? Do you have any hints for me ?

REPLY [4 votes]: Any locally integrable function $f\colon\mathbb{R}\to\mathbb{R}$ (that is, $\int_K|f(x)|\,dx<\infty$ for all compact $K\subset\mathbb{R}$) defines a distribution on $\mathbb{R}$ through
$$
\langle u_f,\phi\rangle=\int_{-\infty}^{\infty}f(x)\,\phi(x)\,dx\quad\forall \phi\in\mathcal{D}(\mathbb{R}).
$$
Since $\phi$ has compact support and $f$ is locally integrable, $u_f$ is well defined. Linearity is obvious. As for continuity, suppose $\phi_n\to\phi$ in $\mathcal{D}$. This means in particular that there exists a compact $K\subset\mathbb{R}$ such that the support of $\phi_n$ is contained in $K$ for all $n$ and that $\phi_n$ converges uniformly to $\phi$ on $K$. Since $f$ is integrable on $K$, it follows that
$$
\lim_{n\to\infty}\langle u_f,\phi_n\rangle=\lim_{n\to\infty}\int_Kf(x)\,\phi_n(x)\,dx=\int_Kf(x)\,\phi(x)\,dx=\langle u_f,\phi\rangle.
$$
In your question $f(x)=\chi_{(0,\infty)}(x)x^{-\alpha}$, where $\chi_A$ is the characteristic function of a set $A$. This function is locally integrable if an only if $\alpha<1$. If $\alpha\ge1$, as Davide's comment shows, is not the functional is not defined.
On the other hand, your functional defines a distribution on $(0,\infty)$, since $x^{-\alpha}$ is localy integrable on $(0,\infty)$.