block matriceseigenvalues-eigenvectorslinear algebramatrices
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!
I will answer your question just for the cases $N = 2$ and $N = 3$:
Let
$$ B_2 = \left( \begin{array}{cccc} 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{array} \right), \quad B_3 = \left( \begin{array}{cccccc} 0 & 1 & 1 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 1 & 1 & 0 \end{array} \right), $$
then, $\text{Spec}{(B_2)} = {(-2,2,0,0)} \, $ and $\text{Spec}{(B_3)} = (-2,-2,0,0,0,4). $
With the help of numerics, I've been able to show (at least for sufficiently large values of $N$) that the characteristic polynomial is given by:
$$\color{blue}{p(\lambda) =\det{(B_N - \lambda I_{2N})} = (\lambda - 2N +2)(\lambda+2)^{N-1} \lambda^N }$$
which tells you that the only eigenvalues of this kind of matrices are $-2,0,2N-2 \ $ with the corresponding multiplicities given by $p(\lambda)$.
Here is an animation showing the spectrum of the matrices $B_N$ for $N \in (2,30)$:
Here's the same approach in the case we have the $B_N$ matrices defined as:
$$B_N = \begin{bmatrix} C_N & A_N \\ A_N & C_N \end{bmatrix},$$
then:
$$\color{blue}{p(\lambda) =\det{(B_N - \lambda I_{2N})} = \left\{ \begin{array}{ll} \left(\lambda - 2N + 2\right) (\lambda-2)^{N/2}(\lambda+2)^{(N-2)/2} \lambda^{N} & N \text{ even} \\ \left(\lambda - 2N + 2\right) (\lambda-2)^{(N-1)/2}(\lambda+2)^{(N-1)/2} \lambda^{N} & N \text{ odd} \end{array}\right.}$$
which tells you that the only eigenvalues of this kind of matrices are $-2,0,2,2N-2 \ $ with the corresponding multiplicities given by $p(\lambda)$.
Here is another animation showing the spectrum of the matrices $B_N$ for $N \in (2,30)$:
pretty cool!
Hope somebody can shed some light on these results.
Cheers!
If $v$ is an eigenvector of $A_1$, then $\pmatrix{v\\ 0}$ is an eigenvector of $A$.
There might not be more eigenvectors: Consider the following example with $n=1$, $A_1=A_2=B=1$. then $$ A = \pmatrix{1 & 1 \\ 0 & 1} $$ only has a one-dimensional eigenspace to the eigenvector $1$.
Best Answer
$\det(\lambda I-C)=\det\pmatrix{\lambda I&-A\\ -B&\lambda I}$. 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.