Finding eigenvalues and eigenvectors after SVD?

Question

The elements represent the SVD of a square matrix and its transpose. I'm confused on how this works for a matrix within a matrix. Computing the characteristic polynomial doesn't seem to get me anywhere, because I just end up with $\lambda^2=U\Sigma V^tV\Sigma U^t$.

Answer 1

Let us take the following notations

$$M=U\Sigma V^T\tag{1}$$ and

$$A=\begin{bmatrix}0 & M\\ M^T & 0\end{bmatrix}\tag{2}$$

We have the following decomposition:

$$\underbrace{\begin{bmatrix}0 & U\Sigma V^T \\ V\Sigma U^T & 0\end{bmatrix}}_A=\underbrace{\begin{bmatrix}U\Sigma & 0\\0& V\Sigma \end{bmatrix}}_P \underbrace{\begin{bmatrix}0&\Sigma\\ \Sigma&0\end{bmatrix}}_B\underbrace{\begin{bmatrix}\Sigma^{-1}U^{-1} & 0\\0& \Sigma^{-1}V^{-1}\end{bmatrix}}_{P^{-1}}\tag{3}$$

The eigenvalues of $B$ are known to be the eigenvalues of $\Sigma$ (i.e., the singular values) and their opposites (see for example here).

Therefore, $A$ has the same eigenvalues as $B$.

Dealing with eigenvectors, as (1) can be written in two ways,

$$MV=U\Sigma \ \ \text{and} \ \ M^TU=V\Sigma\tag{4}$$

which is equivalent to say that, for any $k$,

$$MV_k=\sigma_kU_k \ \ \text{and} \ \ M^TU_k=\sigma_kV_k\tag{5}$$

(we call $U_k,V_k$ the $k$th column of $U,V$ resp. and $\sigma_k$ the $k$-th singular value).

Therefore, using (5), it is easy to check that the eigenvectors of $A$ are:

$$\binom{U_k}{V_k}\ \ \text{and} \ \ \binom{\ \ \ U_k}{-V_k}$$

resp. associated with eigenvalues $\sigma_k$ and $-\sigma_k$.

Answer 2

Let $U\Sigma V^t$ and $\Sigma$ both be $n \times n$. Let $A$ be the block matrix.

Zero is an eigenvalue of $A$ if and only if zero is an eigenvalue of $U\Sigma V^t$. (Why?)

For nonzero eigenvalues, using the Schur complement, we can compute the characteristic polynomial $p(\lambda)$ for $\lambda \ne 0$ \begin{align} \det(A - \lambda I) &= \det(-\lambda I_n) \det(-\lambda I_n - V\Sigma U^t(-\lambda^{-1} I_n) U\Sigma V^t) \\ & = (-\lambda)^n \det (-\lambda I_n - (-\lambda^{-1}) V\Sigma^2 V^t) \\ &= (-\lambda)^n \prod_{i=1}^n (-\lambda + \sigma_i^2/\lambda) \end{align}