# Relationship between singular values of $A$ and eigenvalues of $B:= \begin{bmatrix} 0 & A \\ A^\ast & 0 \end{bmatrix}$

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?

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, $$n, the only difference is that, to these eigenvalues $$\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$$).