# Antidiagonal block matrix (eigenvalues and eigenvectors)

I have $2$ square matrices $A_m$ and $B_m$ which are symmetric and of size $m\times m$. And the 3rd matrix is, $C=\begin{bmatrix}0 & A \\ B & 0\end{bmatrix}$.

Now, I would like to calculate the eigenvalues and eigenvectors of matrix $C$. How can I get it? Or how does it related to the eigenvalues and eigenvectors of $A$ and $B$? Thank you very much in advance!

• What have you tried so far? Hint: detC = -det(A) det(B) because the zero matrix commutes. – Almentoe Sep 29 '15 at 5:29
• Consider $C^2$. – A.Γ. Sep 29 '15 at 5:29
• @ A.G. Clever... – Almentoe Sep 29 '15 at 5:33
• Sorry, I don't get. Could you plz explain some more details. Thanks. – Mike22LFC Sep 29 '15 at 5:36
• @A.G.: Very astute observation. – copper.hat Sep 29 '15 at 6:00

## 2 Answers

$\det(\lambda I-C)=\det\pmatrix{\lambda I&-A\\ -B&\lambda I}$. Since all square subblocks have the same sizes and the two subblocks at bottom commute, the determinant is equal to $\det(\lambda^2 I - AB)$. Therefore, the eigenvalues of $C$ are the square roots of eigenvalues of $AB$. That is, for each eigenvalue $t$ of $AB$, the two roots of $\lambda^2-t=0$ are eigenvalues of $C$.

As pointed out in a comment, we have $\det(C)=\det(-AB)$ and hence there is some relation between the product of the eigenvalues of $C$ and the products of the eigenvalues of $A$ and $B$, but besides that, very few about the spectrum or the eigenvectors of $AB$ can be said even if the spectra and eigenvectors of $A$ and $B$ are fully known. When both $A$ and $B$ are positive definite, we do have some bounds for the eigenvalues of $AB$. See "Evaluating eigenvalues of a product of two positive definite matrices" on this site or "Eigenvalues of product of two symmetric matrices" on MO.

@A.G. idea is a good start:

If we assume that $A,B$ commute, then we get an easy result.

$C^2$ is symmetric, so it is diagonalisable. Let $P$ be a matrix which is the change to the diagonal basis.

$(PC^2 P^{-1}) = (PCP^{-1})^2 = D$, where $D$ some diagonal matrix, so $PCP^{-1} = D^{1/2}$, which you can compute by taking the square roots of the entries of $D$.

• I am wondering whether there is any relation between the eigen-values of two matrices and their product matrices. – Rajat Sep 29 '15 at 5:52
• Thanks to A.G. and Almentoe.Yes, @Rajada is correct. In my case, $m$ is very large, I can calculate eigenvalues and eigenvectors of $A$ and $B$ easily, while it's time-costing for $C$. Hence I'm thinking if the eigenvalues and eigenvectors of $C$ has any relation with the ones of $A$ and of $B$? – Mike22LFC Sep 29 '15 at 5:59
• @user1551 If $A$ and $B$ are symmetric, then so are $AB$ and $BA$, that's what you get in the top left and bottom right respectively. So $C^2$ is symmetric. So long as you pick the same branch of the square root on $\mathbb{C}$ when you take the root, then you get a consistent answer; if you pick the other, then you can multiply by $-I$. – Almentoe Sep 29 '15 at 7:32
• @user1551 Ah, I should definitely be more careful. I will edit my answer to assume that. We must always be careful with spectra! – Almentoe Sep 29 '15 at 7:44
• I'm afraid some justification is still needed, if this approach ever works at all. E.g. consider $A=B=\operatorname{diag}(1,-1)$. Then $AB=BA=I_2$ and $C^2=I_4$. So, a square root of $C^2$ can assume five possible spectra: $\{1,1,1,1\},\{1,1,1,-1\},\{1,1,-1,-1\},\{1,-1,-1,-1\}$ and $\{-1,-1,-1,-1\}$. In this particular example, as $C$ is traceless, we know that its spectrum must be $\{1,1,-1,-1\}$, but how do you know which spectrum is correct in general? – user1551 Sep 29 '15 at 8:23