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Suppose I have a special block, Hermitian matrix

$$H = \begin{bmatrix} A & B \\ B^* & A^* \end{bmatrix}$$

where $*$ denotes conjugate transpose. The blocks $A$ and $B$ are themself Hermitian in this case. Are there any theorems considering the eigenvalues and eigenvectors for this special matrix?

Rodrigo de Azevedo
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Martin
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    If they are Hermitian, then why do you write $A^,B^$, and not just $A,B$? – tomasz Jun 25 '12 at 13:20
  • If $H$ is hermitian, it is clear that off-diagonal blocks must be conjugate transposed, but there is an additional property for the diagonal blocks – Martin Jun 25 '12 at 13:27
  • It is not true that the eigenvalues of $A$ are always eigenvalues of $H$. Consider $A=B=1$. Then $A$ has eigenvalue $1$, while $H=\begin{bmatrix}1&1\1&1\end{bmatrix}$ has eigenvalues $0$ and $2$. – Martin Argerami Jun 25 '12 at 15:07
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    @Martin Based on your above comment and what you typed, you should see now that $B$ need not be Hermitian. For example $H=\begin{bmatrix}1&i\-i&1\end{bmatrix}$ with $B=[i]$. If you intended the blocks themselves to be Hermitian, please make that clearer in the statement, and possibly eliminate the asterisks. – rschwieb Jun 25 '12 at 15:30
  • Generally B need not to be Hermitian, but in case of my matrix it is. So does A. If I would write the matrix as $H = \begin{bmatrix} A & B \ B & A \end{bmatrix}$ then H might not be Hermitian. Sorry for the confusion. – Martin Jun 25 '12 at 16:29
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    @Martin: Hi, and welcome to math.SE! Users with any number of "reputation points" can comment on their own questions and answers (once you obtain 50 points, you gain the ability to comment anywhere), but you were not able to comment because you were not signed into the account that asked the question. I've now merged your duplicate account into the original. – Zev Chonoles Jun 25 '12 at 22:04

1 Answers1

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Since in the comment, you assumed $A$ and $B$ hermitian, we can compute the characteristic polynomial $\det(H-X_{2n})$. Add to the column $k$ the column $n+k$ for $1\leq k\leq n$ to see that $$\det(H-XI_{2n})=\det(A+B-XI_n)\det\pmatrix{I_n&B\\ I_n&A-XI_n}.$$ Then do $R_{n+k}\leftarrow R_{n+k}-R_k$, $1\leq k\leq n$, which gives $$\det(H-XI_{2n})=\det(A+B-XI_n)\det(A-B-XI_n).$$ So the spectrum of $H$ is the union of the spectra of $A+B$ and $A-B$.

Davide Giraudo
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  • Thank you for your answer. I have trouble understanding the first step. Adding to the column $k$ the column $n+k$ for $1≤k≤n$ gives $$det(H-XI_{2n}) = det \begin{bmatrix} A + B - XI_{n } & B \ A^* + B^* - XI_{n} & A^* - XI_{n} \end{bmatrix}$$. I cannot see how this implies that $$\det(H-XI_{2n})=\det(A+B-XI_n)\det\pmatrix{I_n & B\ I_n & A-XI_n}.$$ – Martin Aug 07 '12 at 11:44
  • Maybe I misunderstood the problem but in the comments you assumed $A$ and $B$ Hermitian. – Davide Giraudo Aug 07 '12 at 11:50
  • Yes, $A$ and $B$ are Hermitian. – Martin Aug 07 '12 at 11:56
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    Write $\pmatrix{A+B-XI_n&B\ A+B-XI_n&A-XI_n}=\pmatrix{A+B-XI_n&0\ 0&I_n}\cdot\pmatrix{I&B\ I&A-XIn}$. – Davide Giraudo Aug 07 '12 at 11:59
  • $\pmatrix{A+B-XI_n & 0\ 0 & I_n}\cdot\pmatrix{I & B\ I & A-XIn} = \pmatrix{A+B-XI_n & AB + BB - XB\ I & A - XI}$ or am I missing something? – Martin Aug 07 '12 at 12:17
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    No, it's me. Reverse the order of multiplication in my last comment. – Davide Giraudo Aug 07 '12 at 12:18
  • The multiplication is correct in reversed order. How do you obtain $$ det \pmatrix{A+B-XI_n & B\ A+B-XI_n&A-XI_n} = det \pmatrix{A+B-XI_n&B\ A^+B^-XI_n&A^*-XI_n}$$ ? – Martin Aug 07 '12 at 12:32