Large deviation in relation with Wishard matrix I try to prove the following fact.

Let $A$ be an $m\times n$ matrix with iid standard normal random variable. 
Then $B:=A^{t}A$ is a $Wishart$ matrix with $m$ degrees of freedom and covariane $Id_{m}$.

Then $B_{11}$ is distributed as an inverse chi-squared random variable with $m-n+1 $ degrees of freedom.

The author of a paper I read claims that one can verify easily by linear algebra the $1/B_{11}$ is equal to the squared distance between the first column of $A$ and the linear space spanned by the others. If this is true then it is clear that $B_{11}$ has the given distribution. However how to prove his claim?

I played with the matrix expression for orthogonal projections, but formulas get so complicated, there must e some easier way.
We first show that the squared distance of a vector $v\in\mathbb{R}^{m}$ to a subspace $L\subseteq \mathbb{R}^{m}$ spanned by the linearly independent vectors $v_{1},...,v_{n}$ is given by
$$d^2(v,L)=\frac{G(v_1,\ldots,v_n,v)}{G(v_1,\ldots,v_n),}$$
where 
$$G(x_1,\ldots,x_k)=\det
\begin{pmatrix}
\langle x_1,x_1 \rangle &amp; \ldots &amp; \langle x_1,x_k \rangle\\
\ldots &amp; \ldots &amp; \ldots\\
\langle x_k,x_1 \rangle &amp; \ldots &amp; \langle x_k,x_k \rangle
\end{pmatrix}$$

is a Gram determinant. Note that $(x_{1},...,x_{k})$ is a linear dependent family if and only if $G(x_{1},...,x_{k})=0$. Indeed, if $(x_{1},...,x_{k})$ is a linear dependent family, then there is $(\lambda_{1},...,\lambda_{k})\neq 0$ such that $\sum_{i=1}^{k}\lambda_{i}x_{i}=0$. Then $\sum_{i=1}^{k}\lambda_{i}L_{i}=0,$ where $L_{1},...,L_{k}$ are the lines of Gram matrix of $x_{1},...,x_{k}$. Conversely if $G(x_{1},...,x_{k})=0$, then there is $(\lambda_{1},...,\lambda_{k})\neq 0$ such that $\sum_{i=1}^{k}\lambda_{i}L_{i}=0,$ where $L_{1},...,L_{k}$ are the lines of the Gram matrix of $x_{1},...,x_{k}$ as previously. Then we can see that $\sum_{i=1}^{k}\lambda_{i}x_{i}=0$ is orthogonal to all $x_{i}$, and is thus 0. Linear dependence follows. \

Now let $v=u+k$, where $u\in L$ and $k\in L^{\perp}$. By using the multi-linearity of the determinant, we find that

\begin{align*}
G(v_1,\ldots,v_n,v)&amp;= G(v_1,\ldots,v_n,u)+G(v_1,\ldots,v_n,k)\\
                   &amp;  =G(v_1,\ldots,v_n,u)+        \det \begin{pmatrix}
\langle v_1,v_1 \rangle  &amp; \ldots &amp; \langle v_1,v_n \rangle &amp; 0\\
\ldots                   &amp; \ldots &amp; \ldots                  &amp; \ldots\\
\langle v_n,v_1 \rangle  &amp; \ldots &amp; \langle v_n,v_n \rangle &amp;0\\
*                        &amp; *      &amp; *                       &amp; ||k||^{2} 
\end{pmatrix} \\
                  &amp; =G(v_1,\ldots,v_n,u)+ G(v_1,\ldots,v_n)||k||^{2} 
\end{align*}
Now $G(v_1,\ldots,v_n,u)=0$, because $(v_1,\ldots,v_n,u)$ is a linear dependent family. So we get that 
$$||u||^{2}=d^2(v,L)=\frac{G(v_1,\ldots,v_n,v)}{G(v_1,\ldots,v_n)}.$$\

Let $\Phi$ be an $m\times n$ matrix with independent and identically distributed standard normal random variables. Denote the columns of $\Phi$ by $C_{1}$ to $C_{n}$. Since $m\geq n$ it is clear that the family $(C_{1},...,C_{n})$ is linearly independent.  Then it is easy to see that $(\Phi^{*}\Phi)_{ij}=\left&lt;C_{i},C_{j}\right&gt;$. Let us denote the elements of the inverse Wishart matrix $(\Phi^{*}\Phi)^{-1}=:B$ by $B_{ij} \,(i,j=1,...,n)$. Using the basic formulas for computing the inverse of matrices we find that
\begin{align*}
B_{11}&amp;=\frac{\det
\begin{pmatrix}
\langle C_2,C_2 \rangle &amp; \ldots &amp; \langle C_2,C_n \rangle\\
\ldots &amp; \ldots &amp; \ldots\\
\langle C_p,C_2 \rangle &amp; \ldots &amp; \langle C_p,C_n \rangle
\end{pmatrix}}{\det
\begin{pmatrix}
\langle C_1,C_1 \rangle &amp; \ldots &amp; \langle C_1,C_n \rangle\\
\ldots &amp; \ldots &amp; \ldots\\
\langle C_n,C_1 \rangle &amp; \ldots &amp; \langle C_n,C_n \rangle
\end{pmatrix}}\\
 &amp;= \frac{G(C_{2},...,C_{n})}{G(C_{1},...,C_{n})}
\end{align*}
By what we have seen previously we see that
$$B_{11}=\frac{1}{d(C_{1},L)^2},$$
where $L$ is the vector space spanned by $C_{2},...,C_{n}$.
Now $$d(C_{1},L)^2=||\Pi_{L^{\perp}}(C_{1})||^2,$$ where $\Pi_{L^{\perp}}(C_{1})$ denotes the orthogonal projection of $C_{1}$ to $L^{\perp}$, a subspace of $\mathbb{R}^{n}$ of dimension $m-n-1$ (because $L$ is of dimension $n-1$). By Cochran's theorem  we have that 
$$||\Pi_{L^{\perp}}(C_{1})||^{2} \sim \chi^{2}(m-n+1).$$
You need to apply the standard inner product on $\mathbb{R}^n$.

Edit: Do you know about $SO(3,1,\mathbb{F}_p )$ or $SO(3,1,\mathbb{C} )$?

Actul]ally the best way to solve this is via modular forms