# Expected value of the inverse of a random, truncated Haar matrix

Let $Q$ be a (say 4x4) unitary matrix, distributed according to the Haar distribution. Denote the upper left 2x2 submatrix of $Q$ as $Q_{1:2,1:2}$. I am interested in the following expectation:

$E(I - Q_{1:2,1:2}^HQ_{1:2,1:2})^{-1}$, where $H$ is Hermitian transpose and the expectation is across $Q_{1:2,1:2}$. Anyone that knows how to do it (or can point me to a reference)? Thanks!

• You can use left- and right-unitary invariance of the Haar measure to show that if this expectation exists, it must be a multiple of the identity matrix. However, numerical evidence seems to suggest that the given integral doesn't actually converge. – Nathaniel Johnston Oct 27 '14 at 17:22
• Thanks for your comment! I also noticed numerically that it does not seem to converge. I agree that if it converges, it should converge to a scaled identity. However, I found it strange that it doesn't converge, since it is related to the inverse of a Wishart matrix (where the eigenvalues of the 4 submatrices in the Wishart matrix are fixed). – DzeKap Oct 27 '14 at 18:14

I will first consider the case of a $4\times 4$ matrix $Q$, and generalize to a higher dimensional $Q$ at the end.

A. The four-by-four case.

It is helpful to start from the polar decomposition

$$Q= \left(\begin{array}{cc}U'&0\\ 0&V'\end{array}\right) \left(\begin{array}{cc}\sqrt{1-T}&\sqrt{T}\\ \sqrt{T}&-\sqrt{1-T} \end{array}\right) \left(\begin{array}{cc}U&0\\ 0&V\end{array}\right)$$ where $U,V,U',V'$ are four unitary matrices and $$T=\left(\begin{array}{cc}T_1&0\\ 0&T_2\end{array}\right)$$ is a diagonal matrix with diagonal elements $0\leq T_n\leq 1$.

You seek the expectation value of the matrix $$M=[I-U^{\rm H}(1-T)U]^{-1}=U^{\rm H}T^{-1}U$$ (I have used that $U^{\rm H}U=I$.)

The Haar distribution for $Q$ implies a Haar distribution for $U$, and moreover implies for $T_1,T_2$ the following distribution [*]

$$P(T_1,T_2)=6(T_1-T_2)^2,\;\;0\leq T_n\leq 1$$

The marginal distribution for $T_1$ is

$$P(T_1)=2-6T_1+6T_1^2$$

So you see that the expectation value of $M$ diverges: $E(M)=E(1/T_1)\,I=\infty$.

B. The higher-dimensional case.

Now let's consider an $N\times N$ dimensional unitary matrix $Q$, with $N\geq 4$. The $2\times 2$ upper-left block is still of the form $U'TU$, with unitary $U'$, $U$, so we still have $M=U^{\rm H}T^{-1}U$. The Haar distribution of $Q$ still implies a Haar distribution for $U$, what changes is the distribution of $T_1$ and $T_2$ [*]

$$P(T_1,T_2)=C(T_1-T_2)^2 (T_1 T_2)^{N-4}$$

with normalization constant $C=\frac{1}{2}(N-2)^2(N^2-4N+3)$. The marginal distribution of $T_1$ is

$$P(T_1)=CT_1^{N-4}\left(\frac{1}{N-1}-\frac{2T_1}{N-2}+\frac{T_1^2}{N-3}\right)$$

Now the expectation of $M$ converges,

$$E(M)=\frac{N-2}{N-4}I$$

[*] See Equation (2.10) of this review.

• That's a great derivation! Thanks a lot! – DzeKap Oct 27 '14 at 22:59