# Jordan decomposition of a block matrix

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 $$V_e$$)?

• 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. – Richard Apr 8 '19 at 19:40
• I’m not sure what you mean by the transpose of the conjugate transpose. Do you just mean the elementwise conjugate? – Anthony Quas Apr 9 '19 at 6:14
• 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. – Anthony Quas Apr 9 '19 at 6:18
• Thank you @AnthonyQuas. It is true! – 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 $$V_e = V_m P$$.
• Am I missing something? Is it not true that $(A_{ij}^*)^T = \bar A_{ij}$? – Ben Grossmann Apr 17 '19 at 13:07