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?

  • 2
    $\begingroup$ If they are Hermitian, then why do you write $A^*,B^*$, and not just $A,B$? $\endgroup$
    – tomasz
    Jun 25, 2012 at 13:20
  • $\begingroup$ 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 $\endgroup$
    – Martin
    Jun 25, 2012 at 13:27
  • $\begingroup$ 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 $2$. $\endgroup$ Jun 25, 2012 at 15:07
  • 2
    $\begingroup$ @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. $\endgroup$
    – rschwieb
    Jun 25, 2012 at 15:30
  • 1
    $\begingroup$ @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. $\endgroup$ Jun 25, 2012 at 22:04

1 Answer 1


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 $A-B$.

  • $\begingroup$ 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 $$\det(H-XI_{2n})=\det(A+B-XI_n)\det\pmatrix{I_n & B\\ I_n & A-XI_n}.$$ $\endgroup$
    – Martin
    Aug 7, 2012 at 11:44
  • $\begingroup$ Maybe I misunderstood the problem but in the comments you assumed $A$ and $B$ Hermitian. $\endgroup$ Aug 7, 2012 at 11:50
  • $\begingroup$ Yes, $A$ and $B$ are Hermitian. $\endgroup$
    – Martin
    Aug 7, 2012 at 11:56
  • 1
    $\begingroup$ Write $\pmatrix{A+B-XI_n&B\\ A+B-XI_n&A-XI_n}=\pmatrix{A+B-XI_n&0\\ 0&I_n}\cdot\pmatrix{I&B\\ I&A-XIn}$. $\endgroup$ Aug 7, 2012 at 11:59
  • 2
    $\begingroup$ No, it's me. Reverse the order of multiplication in my last comment. $\endgroup$ Aug 7, 2012 at 12:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .