Let $$B:= \begin{bmatrix} O_m & A \\ A^\ast & O_n \end{bmatrix}$$ where $A$ is an $m \times n$ matrix. Find the relationship between the two:
Singular values and singular vectors of $A$.
Eigenvalues and eigenvectors of $B$.
Try
Let $|\lambda_1| \ge \cdots \ge |\lambda_{m+n}|$ and $x_1, \cdots, x_{m+n}$ are eigenvalues and corresponding eigenvectors of $B$. (Note that eigenvalues are real, since $B$ is Hermitian.
Let $\eta_1 \ge \cdots \ge \eta_{k} \ge 0$ and $u_1, \cdots, u_m$ and $v_1, \cdots, v_n$ be singular values and corresponding left- and right- singular vectors of $A$, with $k := \min\{m,n\}$.
Note that $B^2 = B^\ast B= \begin{bmatrix} A A^\ast & 0 \\ 0 & A^\ast A \end{bmatrix}$ thus $B^2 x_i = B \lambda_i x_i = \lambda_i^2 x_i, \forall i$. Therefore,
$$
B^2x_i = \begin{bmatrix} AA^\ast x_i^{(1)} \\ A^\ast A x_i^{(2)} \end{bmatrix} = \begin{bmatrix} \lambda_i^2 x_i^{(1)} \\ \lambda_i^2 x_i^{(2)} \end{bmatrix}
$$
where $x_i := [\left(x_i^{(1)}\right)^T_m | \left(x_i^{(2)} \right)^T_n]^T_{m+n}$.
I have noticed that $x_i^{(2)}$, $\lambda_i^2$ are eigenpairs of $A^\ast A$, thus by the definition of singular value, $|\lambda_i|$ are the singular values of $A$.
By the way, from SVD of $A = U \Sigma V^\ast$, we have
$$
A^\ast A V = V\Sigma^2_{n \times n} \Leftrightarrow A^\ast A v_j = \eta_j^2 v_j \forall j=1,\cdots, k \\[7pt]
A A^\ast U = U\Sigma^2_{m \times m} \Leftrightarrow A A^\ast u_l = \eta_l^2 u_l \forall l=1,\cdots, k
$$
Question
Since $B$ has $(m+n)$ different(possibly same) $\lambda_i$ (i.e. $i = 1, \cdots, m+n)$, from
$$
B^2x_i = \begin{bmatrix} AA^\ast x_i^{(1)} \\ A^\ast A x_i^{(2)} \end{bmatrix} = \begin{bmatrix} \lambda_i^2 x_i^{(1)} \\ \lambda_i^2 x_i^{(2)} \end{bmatrix}
$$
we have $(m+n)$ formulas for $x_i^{(1)}$,$x_i^{(2)}$, i.e.
$$
AA^\ast x_i^{(1)} = \lambda_i^2 x_i^{(1)} \\
A^\ast A x_i^{(2)} = \lambda_i^2 x_i^{(2)} \\
$$
for $i = 1,\cdots, m+n$, but for
$$
A^\ast A V = V(\Sigma^T\Sigma)_{n \times n} \Leftrightarrow A^\ast A v_j = \eta_j^2 v_j \forall j=1,\cdots, k \\[7pt]
A A^\ast U = U(\Sigma\Sigma^T)_{m \times m} \Leftrightarrow A A^\ast u_l = \eta_l^2 u_l \forall l=1,\cdots, k
$$
we only have $k$ formulas. So I'm stuck at specifying the relationship.
Is there anyone to help solving the problem?
Best Answer
The eigenvectors/eigenvalues of matrix $B$ can be easily constructed out of singular vectors/values of matrix $A$. Here is how.
In order to have simple explanations, let us consider the case where $A$ is square, i.e. with $m=n$ (for the rectangular case, see below).
Let
$$A=U\Sigma V^T=\sum_{k=1}^n \sigma_k U_kV_k^T$$
be the SVD of $A$ where $U_k$ and $V_k$ $(k=1 \cdots n)$ denote the columns of $U$ and $V$ resp.
Recall that : $$AV_k = \sigma_k U_k \ \ \text{and} \ \ A^TU_k = \sigma_k V_k\tag{1}$$
Therefore,
$$\underbrace{\begin{pmatrix} O & A \\ A^T & O \end{pmatrix}}_B \underbrace{\begin{pmatrix} U_k \\ V_k \end{pmatrix}}_{X_k}=\sigma_k\underbrace{\begin{pmatrix} U_k \\ V_k \end{pmatrix}}_{X_k} \ \text{and} \ \underbrace{\begin{pmatrix} O & A \\ A^T & O \end{pmatrix}}_B \underbrace{\begin{pmatrix} \ \ \ U_k \\ -V_k \end{pmatrix}}_{Y_k}=-\sigma_k\underbrace{\begin{pmatrix} \ \ \ U_k \\ -V_k \end{pmatrix}}_{Y_k}\tag{2}$$
which means that each singular triple $(\sigma_k,U_k,V_k)$, generates two eigenpairs : $(\sigma_k, X_k)$ and $(-\sigma_k, Y_k)$.
From this remark, it is possible to retrieve in particular an eigendecomposition of matrix $B$.
If $A$ is $n \times p$ with, say, $n<p$, the only difference is that, to these eigenvalues $\pm \sigma_k$, one must add eigenvalue $0$ with an order of multiplicity $p-n$ ; up to you for exhibiting the corresponding eigenvectors (don't forget that the last $p-n$ elements of $V$ constitute a basis of $\ker A$).