How to find SVD and eigenvalue decomposition of these matrices? Let be $A$ an $m\times m$ complex matrix with singular value decomposition $U\Sigma V^*$, then


*

*is the singular value decomposition of $A+I$, $U(\Sigma+I)V^*$?

*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
 A: 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
$$
\{1\pm\lambda^{1/2}:\ \lambda \ \text{ is an eigenvalue of } A^*A\}
$$
A: 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 $A$.
