Assume $A$ is a block matrix of the form:

$$A=\left[\begin{array}{cccc} A_{11}&A_{12}&\ldots&A_{1n}\\ A_{21}&A_{22}&\ldots&A_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ A_{n1}&A_{n2}&\ldots&A_{nn}\\ \end{array}\right]$$

and let $A^e$ denote an extended version of $A$ given by:

$$A^e=\left[\begin{array}{cccccccc} A_{11}&0&A_{12}&0&\ldots&A_{1n}&0\\ 0&(A_{11}^*)^\top&0&(A_{12}^*)^\top&\ldots&0&(A_{1n}^*)^\top\\ A_{21}&0&A_{22}&0&\ldots&A_{2n}&0\\ 0&(A_{21}^*)^\top&0&(A_{22}^*)^\top&\ldots&0&(A_{2n}^*)^\top\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ A_{n1}&0&A_{n2}&0&\ldots&A_{nn}&0\\ 0&(A_{n1}^*)^\top&0&(A_{n2}^*)^\top&\ldots&0&(A_{nn}^*)^\top\\ \end{array}\right]$$

where $A_{ij}^*$ is the conjugate transpose of the block $A_{ij}$ and $(\cdot)^\top$ is the transpose operation.

If $VJV^{-1}$ is the Jordan canonical decomposition of $A$ and $V_eJ_eV^{-1}_e$ is the Jordan canonical decomposition of $A^e$, can we find a relation between the Jordan blocks $J$ and $J_e$ (and $V$ and $V_e$)?

  • $\begingroup$ Looks complicated, Jordan form comes after eigenvalues, and the eigenvalues of $A^e$ don't seem to be clearly related to those of $A$. IMO you should look into this eigenvalue question first: if that does work, there's a hope for Jordan too. $\endgroup$ – Richard Apr 8 '19 at 19:40
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    $\begingroup$ I’m not sure what you mean by the transpose of the conjugate transpose. Do you just mean the elementwise conjugate? $\endgroup$ – Anthony Quas Apr 9 '19 at 6:14
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    $\begingroup$ You matrix is equivalent to a block diagonal matrix with A in the top left, some conjugated transpose version of A in the bottom right, and 0s elsewhere. So your question boils down to what the Jordan form of the conjugate transpose part looks like. $\endgroup$ – Anthony Quas Apr 9 '19 at 6:18
  • $\begingroup$ Thank you @AnthonyQuas. It is true! $\endgroup$ – user293017 Apr 10 '19 at 7:15

I assume that the $A_{ij}$ are square blocks of the same size (say $m \times m$), but I suspect that this will work whenever the blocks diagonal $A_{ii}$ are all square.

Note that we can find a permutation matrix $P$ such that $$ M = PAP^{-1} = \pmatrix{A & 0\\0 & \bar A} $$ where $\bar A = (A^*)^T$ is simply the conjugate of the matrix $A$. Thus, if $A = VJV^{-1}$, then we can take $M = V_mJ_e V_m^{-1}$ with $$ V_m = \pmatrix{V & 0\\0 & \bar V}, \quad J_e = \pmatrix{J & 0\\0 & \bar J} $$ To find the $V_e$ corresponding to this same Jordan form, take $V_e = V_m P$.

  • $\begingroup$ Am I missing something? Is it not true that $(A_{ij}^*)^T = \bar A_{ij}$? $\endgroup$ – Ben Grossmann Apr 17 '19 at 13:07
  • $\begingroup$ Oops yes you are correct, I was misled by the notation. Yes, that's simply elementwise conjugation it seems. I'll be self-destructing my first message soon. $\endgroup$ – Federico Poloni Apr 17 '19 at 13:21

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