Singular value inequality for block matrices

Question

Suppose that $A \in \mathbb{C}^{m \times n}$, $m \geq n$, has the block form $$A = \begin{bmatrix} A_1 \\ A_2 \end{bmatrix} $$ where $A_1 \in \mathbb{C}^{n \times n}$ and $A_2 \in \mathbb{C}^{(m-n) \times n}$ is arbitrary. Let $\sigma_1(A_1)$, $\sigma_1(A_2)$, and $\sigma_1(A)$ be the largest singular values of $A_1$, $A_2$, and $A$ respectively. Similarly, let $\sigma_{min}(A_1)$, $\sigma_{\min}(A_2)$, and $\sigma_{\min}(A)$ be the smallest singular values of $A_1$, $A_2$, and $A$ respectively. Show that:

$$\sigma_{\min}(A_1)+\sigma_1(A_2) \geq \sigma_{\min}(A) \geq \max\{\sigma_{\min}(A_1),\sigma_{\min}(A_2)\} > 0$$

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So far, I have shown that $\max\{\sigma_{\min}(A_1),\sigma_{\min}(A_2)\} > 0$. I want to use the fact that the largest singular value of matrix is its 2-norm and $$\|B^{-1}\|_2 = \frac{1}{\|B\|_2}$$ on $A_1$ since it is invertible. However, I don't know how to deal with $A_2$ since I have no information about it.

Answer 1

It helps to note that $\sigma_{\min}(A) = \min_{\|x\| = 1} \|Ax\|$. For the upper bound, note that $$ \sigma_{\min}(A) = \min_{\|x\| = 1} \|Ax\| = \min_{\|x\| = 1} \left\|\pmatrix{A_1x\\A_2x}\right\| \leq\\ \min_{\|x\| = 1} \|A_1 x\| + \|A_2 x\| \leq\\ \min_{\|x\| = 1} \|A_1 x\| + \left(\max_{\|y\| = 1}\|A_2 y\|\right) =\\ \sigma_{\min}(A_1) + \sigma_{\max}(A_2) $$ The proof of the lower bound is similar, but this time we use the inequality $$ \left\|\pmatrix{A_1x\\ A_2x} \right\| \geq \max \{\|A_1x\|,\|A_2 x\|\} $$ We only need the invertibility of $A_1$ to say that $\sigma_{\min}(A_1) > 0$.