Density of Gaussian Unitary Ensemble I'm trying to learn a bit about Gaussian matrix ensembles, and am having some trouble making the following connection. Sorry if I'm being a bit obtuse.
Take the Gaussian unitary ensemble (GUE) of $n \times n$ Hermitian matrices. A matrix $H$ from the GUE has diagonal entries that are real and independent $N(0,1)$, while off-diagonals are of form $X + iY$ with $X,Y$ independent $N(0,\frac{1}{2})$. All entries are independent subject to the matrix being Hermitian.
I have also seen the density of a $H$ described in terms of the Lebesgue measure on $\mathbb{R}^2$, i.e. the joint measure on the space of its $n^2$ entries:
$$ P(H) \;\propto\; \exp \Big(-\frac{n}{2} \mathrm{Tr}\, H^2 \Big) $$
I can kind of sense why this is just by the fact that the matrices are Hermitian and the entries are all independent (subject to the matrix being Hermitian) so you can express the joint as a product, but I'm not sure how one rigorously shows this...
Thanks so much,
 A: This result is indeed fairly intuitive.
To prove it rigorously though,
you must remember a few elementary facts in measure theory.
Given that I've actually never done this myself and that it is probably a good exercise,
I'll give it a try.

Let $\mathbb H$ be the space of Hermitian matrices in $\mathcal M_n(\mathbb C)$ (the space of $n\times n$ matrices with entries in $\mathbb C$).
Let $H=(H_{ij}:1\leq i,j\leq n)\in\mathbb H$ be arbitrary.
Then,
since $H$ is Hermitian,
it is clear that for any pair $i,j\leq n$, we have that $H_{ij}=\overline{H_{ji}}$.
Thus,
to specify $H$,
we don't really need $n^2$ points in $\mathbb C$,
but rather $n+\frac{n(n-1)}{2}$ points (i.e.,
the $n$ diagonal entries $H_{11},H_{22},\ldots,H_{nn}$,
as well as the entries above the diagonal $(H_{ij}:1\leq i<j\leq n)$).
Alternatively (and this is what we'll use in the computation of the density),
we can specify $H$ with $n+n(n-1)=n^2$ points in $\mathbb R$ (i.e., the $n$ diagonal entries $H_{11},H_{22},\ldots,H_{nn}$ which we know are real since $H$ is Hermitian,
and the real/imaginary parts of the entries above the diagonal $(\Re(H_{ij}):1\leq i<j\leq n)$ and $(\Im(H_{ij}):1\leq i<j\leq n)$).

Now,
let $H$ be a $n\times n$ GUE matrix.
Looking at how you define the GUE ensemble,
we then see that $(H_{ii}:1\leq i\leq n)$ are i.i.d. $N(0,1)$ random variables;
$(\Re(H_{ij}):1\leq i<j\leq n)$ and $(\Im(H_{ij}):1\leq i<j\leq n)$ are i.i.d. $N(0,1/2)$ r.v.;
and the $H_{ii},\Re(H_{ij})$ and $\Im(H_{ij})$ are all independent.
So,
we can henceforth see $H$ as a random vector in $\mathbb R^{n^2}$:
\begin{align*}
H=(H_{11},\ldots,H_{nn},\Re(H_{12}),\ldots,\Re(H_{(n-1)n}),\Im(H_{12}),\ldots,\Im(H_{(n-1)n}))\tag{1}
\end{align*}
whose entries are all independent.

Let
$$A=A_{11}\times\ldots\times A_{nn}\times A^{\Re}_{12}\times\ldots\times A^{\Re}_{(n-1)n}\times A_{12}^{\Im}\times\ldots\times A^{\Im}_{(n-1)n}$$
be a Borel measurable rectangle in $\mathbb R^{n^2}$.
(The indexation of the components of $A$ may seem a bit messy,
but I'm trying to make it match that of $H$ in equation $(1)$.)
Then,
we get by independence of the entries of $H$ that
\begin{align*}
\Pr[H\in A]=\prod_{i=1}^n\Pr[H_{ii}\in A_{ii}]\prod_{1\leq i<j\leq n}\Pr[\Re(H_{ij})\in A_{ij}^{\Re}]\Pr[\Im(H_{ij})\in A_{ij}^{\Im}].
\end{align*}
At this point,
using the density of the normal distribution,
we have that
\begin{align*}
\Pr[H\in A]
=&\prod_{i=1}^n\int_{A_{ii}}\frac{\exp(-H_{ii}^2/2)}{\sqrt{2\pi}}~\text{d} H_{ii}
\prod_{1\leq i<j\leq n}\int_{A_{ij}^{\Re}}\frac{\exp(-\Re(H_{ij})^2)}{\sqrt{\pi}}~\text{d} \Re(H_{ij})\\
&~~~\times\prod_{1\leq i<j\leq n}\int_{A_{ij}^{\Im}}\frac{\exp(-\Im(H_{ij})^2)}{\sqrt{\pi}}~\text{d} \Im(H_{ij}).
\end{align*}
Then,
Using Fubini's theorem (to be truly rigorous,
one would have to check that the hypotheses of Fubini's theorem are satisfied)
this yields
\begin{align*}
\Pr[H\in A]&=2^{-n/2}\pi^{-n^2/2}\int_{A}\exp\left(\frac{-1}2\left(\sum_{i=1}^nH_{ii}^2+2\sum_{1\leq i<j\leq n}(\Re(H_{ij})^2+\Im(H_{ij})^2)\right)\right)~\text{d}H\\
&=2^{-n/2}\pi^{-n^2/2}\int_{A}\exp\left(\frac{-1}2\left(\sum_{i=1}^nH_{ii}^2+2\sum_{1\leq i<j\leq n}|H_{ij}|^2\right)\right)~\text{d}H\\
&=2^{-n/2}\pi^{-n^2/2}\int_{A}\exp\left(\frac{-1}2\left(\sum_{i=1}^nH_{ii}^2+2\sum_{1\leq i<j\leq n}H_{ij}\overline{H_{ij}}\right)\right)~\text{d}H\\
&=2^{-n/2}\pi^{-n^2/2}\int_{A}\exp\left(\frac{-1}2\left(\sum_{i=1}^nH_{ii}^2+2\sum_{1\leq i<j\leq n}H_{ij}H_{ji}\right)\right)~\text{d}H.
\end{align*}
Using the definition of matrix product,
one easily checks that
$$\text{Tr}(H^2)=\sum_{i=1}^nH_{ii}^2+2\sum_{1\leq i<j\leq n}H_{ij}H_{ji},$$
which proves the result since the product Borel $\sigma$-algebra on $\mathbb R^{n^2}$ is generated by the Borel measurable rectangles (recall that if two measures agree on a $\pi$-system that generates a $\sigma$-algebra,
then the two measures agree on the whole $\sigma$-algebra).

A quick note: this is slightly different than what you wanted to prove by a factor of $n$ in the exponential $\exp\left(-\frac{n}{2}\text{Tr}(H^2)\right)$.
I believe this is the density for the GUE random matrices whose entries scaled by a factor of $1/\sqrt{n}$ or something like that.
