How to prove that $\|A\|_2=5$?

Question

In the exercise textbook, I was asked to find a $4 \times 4$ matrix $A$ so $\|A\|_2 = 5$. I understand that $\|A\|_2=\sigma_{\max}(A)$ where $\sigma_{\max}(A)$ denotes the largest singular value of matrix $A$. So, I chose the following matrix:

$$A = \begin{pmatrix} 5 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$$

But, how to prove $\|A\|_2 = 5$?

Answer 1

You can read off the singular value decomposition of $A$. To wit $A=5 e_1 e_1^t+ e_2e_3^t$ where $e_i$ are the standard basis vectors. Since $e_1\perp e_2$ and $e_1\perp e_3$, it follows that this $5$ and $1$ are the singular values of $A$.

Answer 2

Since the $2-$norm is also given by $$\|A\|_2=\max_{\|\mathbf{x}\|=1}{\|A\mathbf{x}\|}.$$ So for $\mathbf{x}=\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$ with $a^2+b^2+c^2+d^2=1$, we have $$\|A\mathbf{x}\|=\underbrace{\sqrt{25a^2+c^2} \leq \sqrt{24a^2+1}}_{\because a^2+c^2 \leq 1}\leq\sqrt{25}=5.$$ The max is attained for $\mathbf{x}=(1,0,0,0)$, so the norm is $5$.