Suppose $A\in\mathbb{R}^{m\times n}$ and consider the matrix $2$-norm $$\|A\|_{2} = \max_{\|x\|_{2} = 1}\|Ax\|_{2}$$ Show that $\|A\|_{2} \geq \|A_1\|_{2}$ where $$A = \begin{pmatrix} A_1\\ A_2\\ \end{pmatrix}$$ $m = m_1 + m_2$, $A_1\in\mathbb{R}^{m_1\times n}$ and $A_2\in\mathbb{R}^{m_2\times n}$
Attempted solution - There exists $x_1\in\mathbb{R}^n$ such that $$\|A_1 x_1\|_{2} = \|A_1\|_{2}$$ Therefore, $$\|A\|_{2}^{2} \geq \|Ax_1\|_{2}^{2} = \|A_1 x_1\|_{2}^{2} + \|A_2 x_1\|_{2}^{2} \geq \|A_1 x_1\|_{2}^{2} = \|A_1\|_{2}^{2}$$
I feel like this proof is not that elegant enough. If anyone has any suggestions on this that would be great.
Here's the proof I like: $$ \|A\|^2 = \max_{\|x\| = 1} \|Ax\|^2 = \max_{\|x_1\|^2 + \|x_2\|^2 = 1} \|Ax_1\|^2 + \|Ax_2\|^2 \geq \\ \max_{\|x_1\| = 1, x_2 = 0} \|A_1x_1\|^2 + \|A_2x_2\|^2 = \max_{\|x_1\| = 1} \|A_1x_1\|^2= \|A_1\|^2 $$ The second max is necessarily smaller since $$ \{(x_1,x_2) : \|x_1\|^2 + \|x_2\|^2 =1 \} \supseteq \{(x_1,x_2): \|x_1\| = 1, x_2 = 0\} $$