How to find SVD and eigenvalue decomposition of these matrices?

Question

Let be $A$ an $m\times m$ complex matrix with singular value decomposition $U\Sigma V^*$, then

1. is the singular value decomposition of $A+I$, $U(\Sigma+I)V^*$?
2. Find an eigenvalue decomposition of $$M = \begin{bmatrix} I & A^* \\ A & I \end{bmatrix}.$$

I think that the first one should be no, as \begin{align*} U(\Sigma+I)V^* &= U\Sigma V^* + UV^*\\ &= A + UV^* \end{align*} and $UV^*$ is not always the identity. Is this true? I do not think that $U$ and $V^*$ are each other's inverse.

I am more so confused with the second part. I know the fact that, if $A$ is non-defective and has eigenvalue decomposition $X\Lambda X^{-1}$, then $A+I$ has eigenvalue decomposition $X(\Lambda+I)X^{-1}$. I would think that it could be helpful if 1. were true, to combine these to get 2. . I also know that $M$ is self-adjoint, as $M=M^*$, which means that it has an orthonormal basis of eigenvectors.

Could I get some help?

Thanks

Answer 1

Your observations for the first question are correct.

In the second question the eigenvalue decomposition can be computed by using Schur' s determinantal identity: for $D$ invertible, $\det \begin{pmatrix} A & B \\ C & D\end{pmatrix}=\det{D}\det{(A-B D^{-1} C)}$. Using this formula it is possible to compute the characteristic polynomial of $M$ by substituting the block matrices from $M-\lambda I$. This allows one to express the eigenvalues of matrix $M$ in terms of the entries of $\Sigma$ and the eigenvectors are given by the SVD of $A$.

Answer 2

You are right about 1.

For the second part, your matrix $M$ is $I_{2n}+\begin{bmatrix} 0&A^*\\ A&0\end{bmatrix}$. If $\lambda$ is an eigenvalue of this last matrix, then there exist $x,y$ with $$ \begin{bmatrix} \lambda x\\ \lambda y\end{bmatrix} = \begin{bmatrix} 0&A^*\\ A&0\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix} =\begin{bmatrix} A^*y\\ Ax\end{bmatrix} . $$ Then $$ A^*Ax=A^*(\lambda y)=\lambda A^*y=\lambda ^2 x. $$ Similarly, $AA^*y=\lambda ^2 y$. As either $x$ or $y$ is nonzero (otherwise there is no eigenvector), we get that $\lambda^2$ is an eigenvalue of $A^*A$ (note that $A^*A$ and $AA^*$ have the same eigenvalues). This implies that $\lambda$ is real.

Conversely, if $A^*Ax=\lambda^2 x$ with $\lambda\ne0$, then $$ \begin{bmatrix} 0&A^*\\ A&0\end{bmatrix}\begin{bmatrix} x\\ \lambda^{-1}Ax\end{bmatrix}=\begin{bmatrix} (\lambda^{-1}A^*A x)\\ Ax\end{bmatrix}=\lambda\,\begin{bmatrix} x \\ \lambda^{-1}Ax\end{bmatrix}. $$ So the eigenvalues of $\begin{bmatrix} 0&A^*\\ A&0\end{bmatrix}$ are precisely $$ \{\pm\lambda^{1/2}:\ \lambda \ \text{ is an eigenvalue of } A^*A\} $$ and thus the eigenvalues of $M$ are precisely $$ \{1\pm\lambda^{1/2}:\ \lambda \ \text{ is an eigenvalue of } A^*A\} $$