Given block matrices $\mathbf{A}$ and $\mathbf{B}$ which are block transposes of each other, i.e.,

$$\mathbf{A} = \left[ \begin{array}{cccc} \mathbf{M}_{11} & \mathbf{M}_{21} & \dots & \mathbf{M}_{J1}\\ \mathbf{M}_{12} & \mathbf{M}_{22} & \dots & \mathbf{M}_{J2}\\ \vdots & \vdots & \ddots & \vdots\\ \mathbf{M}_{1K} & \mathbf{M}_{2K} & \dots & \mathbf{M}_{JK} \end{array} \right]$$

$$\mathbf{B} = \left[\begin{array}{cccc} \mathbf{M}_{11} & \mathbf{M}_{12} & \dots & \mathbf{M}_{1K}\\ \mathbf{M}_{21} & \mathbf{M}_{22} & \dots & \mathbf{M}_{2K}\\ \vdots & \vdots & \ddots & \vdots\\ \mathbf{M}_{J1} & \mathbf{M}_{J2} & \dots & \mathbf{M}_{JK} \end{array} \right]$$

where $\mathbf{M}_{jk} \in \mathbb{C}^{n \times m}$, I'd like show that the non-zero eigenvalues of $\mathbf{A}' \mathbf{A}$ and $\mathbf{B}' \mathbf{B}$ are equivalent, where $A' = conj(A^T)$ is the conjugate transpose.

This is clearly true if $\mathbf{A}=\mathbf{B}'$ (e.g. see here), but can this result be extended to the block-transpose case?

If it matters, $\mathbf{M}_{jk}$ is a non-square lower-triangular toeplitz matrix for my particular problem, though I'm also interested if the result holds in general as well. For what its worth, numerical analysis suggests that this is true, at least for the special lower-triangular toeplitz case.

Shoot - I had computed the spectrum in MATLAB for an initial data set which initially seemed to suggest that this conjecture was true (at least to the limit of numerical noise). However further investigation of some smaller examples built from random matrices show that this is false... Nevertheless, the eigenvalue distribution does seem to be remarkably similar, at least for the cases I looked at. And it is true that trace(A'A) = trace(B'B). If there is anything else interesting that can be said about the relationship between the spectrum of A'A and B'B, I am still very interested. Thanks!


This is the eigenvalues of the actual dataset I am interested in, computed in MATLAB.

  • $\begingroup$ Have you done any numerical experiment to see if the statement is plausible or not? $\endgroup$
    – user1551
    Sep 3 '17 at 2:06
  • $\begingroup$ Yes, a numerical calculation suggests this is true. $\endgroup$
    – Jon
    Sep 3 '17 at 2:09
  • $\begingroup$ What is $A'$? The transpose? $\endgroup$
    – user357980
    Sep 3 '17 at 2:23
  • $\begingroup$ Sorry, A' is hermitian transpose $\endgroup$
    – Jon
    Sep 3 '17 at 2:24
  • $\begingroup$ So, the adjoint, the conjugate transpose? $\endgroup$
    – user357980
    Sep 3 '17 at 2:26

I found a counter-example with no restriction on the blocks, then also with Toeplitz matrices. Here is an example that is not a counter-example:



I do not remember the first example that I tried, but in both cases with those Toeplitz matrices, their eigenvalues were really close. Two things to note, since they have the same trace the first two terms of their characteristic polynomials are the same. Also, a lot of the matrix is the same, just with the block transpose some of the same products will be taken (recall the permutation form of the determinant). I tried to have the terms be prime and large and so little repetition since my earlier example where the eigenvalues were similar had a lot of $1$'s.

It is still kinda striking that the eigenvalues are so close.


(Edit: I misread the question. This answer does not apply, but I'll leave it here for reference.)

The answer is "no" in general (it is easy to general random counterexamples), but "yes" if all sub-blocks are $n\times m$ lower triangular Toeplitz.

The eigenvalues of $A^\ast A$ are just the squared singular values of $A$. Each sub-block of $A$ is lower triangular Toeplitz. So, when $m\ne n$, $A$ is guaranteed to have $|m-n|$ zero rows/columns in every $\max(m,n)$ rows/columns (and similarly for $B$). As zero rows/columns do not contribute to nonzero singular values, you may just remove them and assume that each sub-block is square (i.e. $n=m$).

Let $P=P^\top$ denotes the $n\times n$ reversal matrix (i.e. the anti-diagonal matrix obtained by flipping the identity matrix from left to right). Then $M=PM^\top P$. It follows that $B=(P\oplus\cdots\oplus P)A^\top(P\oplus\cdots\oplus P)$. Therefore $B$ is unitarily similar to $A^\top$ and they have identical singular values. But $A$ and $A^\top$ have identical singular values too. Hence the result follows.

  • $\begingroup$ Ah! Very nice. Perhaps this is obvious, but just because the extra rows/cols in each block do not affect the block's nonzero SVs, how can I prove that they do not affect the SVs of the overall matrix? -edit - just realised I can simply insert dummy zeros rows/cols as needed to make all blocks square, which would not change the eigenvalues. $\endgroup$
    – Jon
    Sep 3 '17 at 4:06
  • $\begingroup$ @Jon Yes, insertion of (i.e. padding with) zero rows/columns will do the trick, but the usual practice is not insertion but removal --- that's the spirit of thin/economic SVD. $\endgroup$
    – user1551
    Sep 3 '17 at 4:33
  • $\begingroup$ Actually, I realized that padding doesnt work because it breaks the toeplitz symmetry that is required. Removing the extra rows/cols makes intuitive sense given that they don't affect the SVD of each block, but I'm still not certain why it doesn't affect the SVD of the total matrix. Is there some theorem that states that the SVD of a block matrix depends only on the singular values of the individual blocks? $\endgroup$
    – Jon
    Sep 3 '17 at 4:50
  • $\begingroup$ @Jon (1) For padding, when there are more columns than rows, you don't pad at the bottom, but at the top. (2) Removal of zero columns can affect the SVD (because one can pick other sets of singular vectors). It just doesn't affect those nonzero singular values. E.g. consider $G=[C|0]$. Then $G^\ast G=\pmatrix{C^\ast C&0\\ 0&0}$ and $C^\ast C$ have identical nonzero eigenvalues. $\endgroup$
    – user1551
    Sep 3 '17 at 5:11
  • $\begingroup$ In my case, the blocks have more rows than columns, so there are no zero columns, but there are rank deficient rows. Simply truncating the bottom rows doesn't work, and its not clear to me how to transform into to a square block in a way that doesn't affect the singular values of the overall matrix. (Thanks for all your help BTW!) $\endgroup$
    – Jon
    Sep 3 '17 at 5:19

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