text:
\section{Proof of Theorem \ref{circ thm}}\label{sec circ thm}
This proof is a slight modification of the proof of Theorem $4.1$ of Krahmer et al. \cite{krahmer2014suprema} and we will directly borrow their notation and estimates. First, we restate their Theorem $3.1$.
\begin{theorem} \label{thm 31}Let $\Ac$ be a set of matrices and let $\xi$ be a random vector whose entries $\xi_j$ are standard normal. Let $d_F,d_{2\rightarrow 2}$ be the Frobenius and spectral norm distance metrics respectively. Set
\begin{align}
&E=\gamma_2(\Ac,\|\cdot\|)(\gamma_2(\Ac,\|\cdot\|)+d_F(\Ac))+d_F(\Ac)d_{2\rightarrow 2}(\Ac),\\
&V=d_{2\rightarrow 2}(\Ac)(\gamma_2(\Ac,\|\cdot\|)+d_F(\Ac)),~\text{and}~U=d^2_{2\rightarrow 2}(\Ac).
\end{align}
Then, for some absolute constants $c_1,c_2>0$ and for all $t>0$,
\[
\Pro(\sup_{\A\in\Ac}|\tn{\A\xi}^2-\E\tn{\A\xi}^2|\geq c_1 E+t)\leq\exp(-c_2\min\{\frac{t^2}{V^2},\frac{t}{U}\}).
\]
\end{theorem}
Theorem \ref{sec circ thm} is a variation of Theorem $4.1$ of \cite{krahmer2014suprema}. In light of Theorem \ref{thm 31}, we simply need to adapt the estimates developed during the proof of Theorem $4.1$ of \cite{krahmer2014suprema} for our purposes. We are interested in a fixed submatrix of size $m\times s$ compared to all $s$-column submatrices for fixed $m$-rows. This makes our set $\Ac$ a subset of their set and also makes their estimates an upper bound on our estimates. Following arguments of \cite{krahmer2014suprema}, for some constant $c_3>0$, we have
\[
d_F(\Ac)=1,~d_{2\rightarrow 2}(\Ac)\leq \sqrt{s/m},~\gamma_2(\Ac,\|\cdot\|)\leq c_3{\sqrt{s/m}\log(2s)\log(2d)}.
\]
To proceed, we will apply Theorem \ref{thm 31}. This will be done in two scenarios depending on whether isometry constant obeys $\delta\leq 1$ or not. Recall that $m_0=c_0s\log^2(2s)\log^2(2d)$. Below, we pick $c_0$ sufficiently large to compensate for $c_1,c_2,c_3$.

\noindent {\bf{$m\geq m_0$ case:}} We have that $\gamma_2(\Ac,\|\cdot\|)\leq c_3{\sqrt{s/m}\log (2s)\log (2d)}\leq 1$ so that $E\leq \sqrt{s/m}+2\gamma_2(\Ac,\|\cdot\|)\leq 3c_3\sqrt{s/m}\log(2s)\log(2d)$. Similarly, $V\leq 2\sqrt{s/m}$ and $U\leq s/m$. In this case, picking large $c_0$, observe that $c_1 E\leq \sqrt{m_0/(4m)}$. With this, we can pick $t=\sqrt{m_0/(4m)}$ to guarantee $c_1E+t\leq \sqrt{m_0/m}$. We have that $t^2/V^2\geq m_0/(16s),~t/U\geq t^2/U\geq m_0/(4s)$. Picking $c_0\geq 16/c_2$, we conclude with the desired probability $\exp(-\log^2(2d)\log^2(2s))$.

\noindent {\bf{$m< m_0$ case:}} In this case, we have
\[
\gamma_2(\Ac,\|\cdot\|)+d_F(\Ac)\leq c'{\sqrt{s/m}\log(2s)\log(2d)},
\]
where $c'=c_3+\sqrt{c_0}$. Hence, we find
\[
E\leq c'c_3(s/m)\log(2s)^2\log(2d)^2+\sqrt{s/m}.
\]
Observe that, we can ensure i) $c_1\sqrt{s/m}\leq \sqrt{m_0/m}/4\leq m_0/(4m)$ and ii) $c_1c'c_3(s/m)\log(2s)^2\log(2d)^2\leq m_0/(4m)$ for sufficiently large constant $c_0$. The latter one follows from the fact that $c'$ grows proportional to $\sqrt{c_0}$ whereas $m_0$ grows proportional to $c_0$. With this, we can pick $t=m_0/(2m)$ which guarantees $c_1E+t\leq \frac{m_0}{m}$.

To find the probability, we again pick $c_0$ to be sufficiently large to guarantee that i) $c_2t/U\geq c_2(m_0/(2m))/(s/m)\geq \log^2(2s)\log^2(2d)$ and ii)
\begin{align*}
t^2/V^2&\geq \frac{(m_0/m)^2}{4(s/m) (c'{\sqrt{s/m}\log(2s)\log(2d)})^2}\\
&= \frac{c_0^2s^2\log^4(2s)\log^4(2d)}{4s^2(c')^2\log^2(2s)\log^2(2d)}\\
&=\frac{c_0^2\log^2(2s)\log^2(2d)}{4(c_3+\sqrt{c_0})^2}\geq \log^2(2s)\log^2(2d)/c_2,
\end{align*}
which concludes the proof by yielding $\exp(-\log^2(2d)\log^2(2s))$ probability of success.


TITLE: How to find the vector that passes through a point and is perpendicular to another vector.
QUESTION [1 upvotes]: Let $ \mathbb{a} = i+4j-3k$ and $b = 7i+20j-12k$ be vectors and $A(2,5,-3)$ be a point. I want find the line $l_ 3$ passing through point $A$ wich is perpendicular to both veotors. How should I do this?

REPLY [0 votes]: The line takes the form $\left\{ \vec{u}+\lambda\vec{v}\mid\lambda\in\mathbb{R}\right\} $.
Here $\vec{u}=\left(2,5,-3\right)$ and $\vec{v}=\left(x,y,z\right)$
is perpendicular to $\left(1,4,-3\right)$ and $\left(7,20,-12\right)$.
That leads to the equations:
$x+4y-3z=0$ and $7x+20y-12z=0$.
They determine the direction of
$\vec{v}$.
When it comes to its length you can make your own choice.