This question is related to the general theory of Bogoliubov transformations that arises in physics but I've distilled it to a linear algebra question.

Suppose I have a $2M \times 2M$ unitary matrix, $U$, with complex entries. Now suppose I write this as

$$ U = \left(\begin{array}{cc} U_1 & U_2 \\ U_3 & U_4 \end{array}\right) $$ where $U_1, U_2$ are $n\times M$ sized matrices and $U_3,U_4$ are $M-n \times M$ sized matrices.

I now have the unitary constraints: $U U^\dagger = \mathbb{I}_{2M\times2M} = U^\dagger U$ which translate to constraints on the matrices $U_i$. E.g.,

$$ U_1 U_1^\dagger + U_2 U_2^\dagger = \mathbb{I}_{n\times n}\\ U_1^\dagger U_1 + U_3^\dagger U_3 = \mathbb{I}_{M\times M} $$ and 4 others.

Here, $U_2^\dagger U_2$ is an $n\times n$ Hermitian matrix and so I can write down its spectral decomposition, $$(U_2^\dagger U_2) X_2^j = \lambda^j_2 X_2^j$$ (j= 1,2,$\cdots n$) and similarly for $U_2 U_2^\dagger$, $U_3^\dagger U_3$, and $U_3 U_3^\dagger$.

My question is the following: why must $U_2^\dagger U_2$ and $U_3^\dagger U_3$ have some eigenvalues = 1 if $n\neq M$ and no eigenvalues = 1 if $n = M$?

I've played with this by creating random unitary matrices and partitioning them as described and this seems to be a general phenomenon. I am trying to reproduce the results of https://projecteuclid.org/euclid.cmp/1103859661 and right above Eq. 3.3 the author claims that these eigenvalues = 1 exist, but I don't see why or how to show it. I tried thinking in terms of singular value decomposition for $U_2, U_3$ but that didn't help either.


1 Answer 1


So I thought about this and it turns out to be a general property of unitary matrices. Given this matrix $U$, partitioned as described above, we have constraints on the $U_i$ stemming from unitarity.

Let's take the following two constraints into account - $$ U_1 U_1^\dagger + U_2 U_2^\dagger = \mathbb{I}_{n\times n} \\ U_1^\dagger U_1 + U_3^\dagger U_3 = \mathbb{I}_{M\times M} $$

Assume without loss of generality that $n<M$. Next, $$ (U_2 U_2^\dagger) v_j = \lambda_j v_j \quad j = 1,\cdots,n $$ The first constraint implies $$ (U_1 U_1^\dagger) v_j = (1-\lambda_j) v_j \quad j = 1,\cdots,n $$ We now need to invoke a fact about singular value decomposition. Namely, the non-zero singular values of $U_1$ equal the square root of the eigenvalues of $U_1^\dagger U_1$ and $U_1 U_1^\dagger$.

This immediately implies that the eigenvalues of $U_1^\dagger U_1$ are $$ \xi_j = \begin{cases}\begin{align} 1- \lambda_j, &\quad 1\leq j \leq n \\ 0, &\quad n< j \leq M \end{align}\end{cases} $$ since if $U_1^\dagger U_1$ had non-zero excess eigenvalues, they would be present in the singular values of $U_1$, but then must also be eigenvalues of $U_1 U_1^\dagger$, which would be a contradiction. So the eigenvalues of the two Hermitian matrices match modulo these extra zero eigenvalues present in the larger of the two matrices.

Using the second constraint, we can see that the eigenvalues of $U_3^\dagger U_3$ are $$ 1 - \xi_j = \begin{cases}\begin{align} \lambda_j, &\quad 1\leq j \leq n \\ 1, &\quad n< j \leq M \end{align}\end{cases} $$ which shows that there must be unity eigenvalues for this matrix. This argument can be used to then relate the eigenvalues of all of the hermitian matrices constructed from the partitioned unitary matrix, showing why unity and zero eigenvalues follow from unitarity and the singular value decomposition.


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