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

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.

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    $\begingroup$ What about $A = (1)$, so that $B$ is a $2 \times 2$ matrix? $\endgroup$ – lokodiz Oct 10 '14 at 7:52
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    $\begingroup$ $\TeX$ tip: when writing a matrix, separate each element on the same row with a & (rather than a backslash). It looks much nicer. $\endgroup$ – beep-boop Oct 10 '14 at 9:21
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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.

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  • $\begingroup$ Sorry, don't understand this step - $\det(B-\lambda I)=\det(\lambda^2 I-AA^*)$ $\endgroup$ – Yola Oct 10 '14 at 9:15
  • $\begingroup$ @ yola , see en.wikipedia.org/wiki/Determinant#Block_matrices Section 3.3 block matrices. $\endgroup$ – user91684 Oct 10 '14 at 9:21
  • $\begingroup$ But i still don't understand how step i ask about work. Though i understand that $\det(B) = \det(-A A^*).$ For what we need $I$ commute with $A?$ $\endgroup$ – Yola Oct 10 '14 at 9:55
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    $\begingroup$ @ Yola , if $CD=DC$, then $\det(\begin{pmatrix}A&B\\C&D\end{pmatrix}=\det(AD-BC)$. $\endgroup$ – user91684 Oct 10 '14 at 10:01
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Hint: the eigenvalue equation is equivalent to $$ x = (x_1,x_2)\\ A^*x_2 = \lambda x_1\\ Ax_1 = \lambda x_2 $$

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