Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 5 down vote favorite
1

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?

share|improve this question
2  
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
2  
@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
1  
@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
show 1 more comment

1 Answer

up vote 2 down vote

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 $A-B$.

share|improve this answer
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 7 '12 at 11:44
Maybe I misunderstood the problem but in the comments you assumed $A$ and $B$ Hermitian. – Davide Giraudo Aug 7 '12 at 11:50
Yes, $A$ and $B$ are Hermitian. – Martin Aug 7 '12 at 11:56
1  
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 7 '12 at 11:59
1  
No, it's me. Reverse the order of multiplication in my last comment. – Davide Giraudo Aug 7 '12 at 12:18
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.