Find eigen values of $B = \begin{bmatrix} 0 & A^* \\ A &0 \end{bmatrix}$

Question

$$B = \begin{bmatrix} 0 & A^* \\ A&0 \end{bmatrix}$$

I think that $\det(B) = \det(A) * \det(A^*)$ and probably eigen values just get squared. What is the right answer?

EDIT: $\operatorname{rank}(B) = 2 * \operatorname{rank}(A)$, so there are twice more eigenvalue or their multiples. And the additional eigenvalues are just negatives of the eigenvalues of $A$ (Thanks to Simon).

Still can't figure out.

Answer 1

It is linked to the SVD deomposition, cf. http://en.wikipedia.org/wiki/Singular_value_decomposition

$B$ is a hermitian matrix. Since $I$ and $A$ commute, $\det(B-\lambda I)=\det(\lambda^2 I-AA^*)$ and the eigenvalues of $B$ are the square roots of the singular values of $A$. In particular $\det(B)=(-1)^n |\det(A)|^2$. For the eigenvectors, use the mookid's post or "wikipedia" above.

Answer 2

Hint: the eigenvalue equation is equivalent to $$ x = (x_1,x_2)\\ A^*x_2 = \lambda x_1\\ Ax_1 = \lambda x_2 $$