What is the operator norm of this matrix $\left [\begin{smallmatrix}1&3\\2&0\end{smallmatrix}\right ]$? Let$$A=\begin{bmatrix}1&3\\2&0\end{bmatrix}.$$Question: What is $\|A\|$?
Here, $\|A\|$ is the operator norm of $A$ in a sense that$$\|A\|=\inf \left \{C\geq 0:\|Ax\|\leq C\|x\|,\quad x\in\mathbb{R}^2\right \}.$$Or, equivalently, in a senset that$$\|A\|=\sup \left \{\frac{\|Ax\|}{\|x\|}:x\neq 0,\quad x\in\mathbb{R}^2\right\}.$$
Trial 1)$$AA^T=\begin{bmatrix}10&2\\2&4\end{bmatrix}.$$And the characteristic equation becomes$$|AA^T-\lambda I|=(10-\lambda )(4-\lambda )-4=\lambda ^2-14\lambda +36,$$which has roots $\lambda _1=7+\sqrt{13}$ and $\lambda _2=7-\sqrt{13}$.
This leads to$$\|A\|=\sqrt{\|AA^T\|}=\sqrt{\max _i|\lambda _i|}=\sqrt{7+\sqrt{13}}.$$Trial 2) Although $A$ is not symmetric, it has full rank of eigenspace.
So I can proceeds as follows:$$|A-\lambda I|=(1-\lambda )(-\lambda )-6=\lambda ^2-\lambda -6=(\lambda -3)(\lambda +2).$$$\lambda _1'=-2$, $\lambda _2'=3$, $\max \limits _i\lambda _i'=3$.$$\|A\|=\max _i|\lambda _i'|=3.$$So, they don't coincide. Which answer is correct?
For a reference, I've read Conrad : Computing the Norm of a Matrix.
For a detailed explanation for the second trial, see my own answer below
 A: Your argument would work if $A$ were unitarily diagonalizable. But it isn't, it is diagonalizable but not unitarily diagonalizable. Your $x_2$ as written is not an eigenvector for $A$. The eigenspace for $\lambda=3$ runs along $(3,2)$, which is not orthogonal to $(1,1)$.
To add, even before calculating anything: a real matrix with real eigenvalues is unitarily diagonalizable if and only if it is symmetric.
A: A detailed explanation for the second trial.
The matrix
$$A=\begin{bmatrix}1&3\\2&0\end{bmatrix}$$
has characteristic polynomial $(1-\lambda)(-\lambda)-6=(\lambda+2)(\lambda-3)$.
Let $\lambda_1=-2$ and $\lambda_2=3$.
Then
\begin{align*}
A+2I&=\begin{bmatrix}3&3\\2&2\end{bmatrix}\\
A-3I&=\begin{bmatrix}-2&3\\2&-3\end{bmatrix}
\end{align*}
and the corresponding unit eigenvectors are
$$x_1=\begin{bmatrix}\frac1{\sqrt2}\\-\frac1{\sqrt2}\end{bmatrix}\text{ and }
x_2=\begin{bmatrix}\frac3{\sqrt{13}}\\\frac{-2}{\sqrt{13}}\end{bmatrix}.$$
They form an orthonormal basis for $\mathbb R^2$.
Now, for any $x\in\mathbb R^2$,
$$x=c_1x_1+c_2x_2$$
for some $c_1,c_2\in\mathbb R$.
It follows that
\begin{align*}
Ax
&=A(c_1x_1+c_2x_2)\\
&=c_1(Ax_1)+c_2(Ax_2)\\
&=\lambda_1c_1x_1+\lambda_2c_2x_2
\end{align*}
and that
$$\frac{||Ax||^2}{||x||^2}=\frac{{\lambda_1}^2{c_1}^2+{\lambda_2}^2{c_2}^2}{{c_1}^2+{c_2}^2}\le\left(\max\{|\lambda_1|,|\lambda_2|\}\right)^2.$$
Then
$$\frac{||Ax||}{||x||}\le\max\{|\lambda_1|,|\lambda_2|\},$$
and we can take the supremum on the left hand side, for all $x\in\mathbb R^2$, to conclude
$$||A||\le3.\tag{1}$$
For the converse inequality, we have
$$|\lambda_1|=\frac{||Ax_1||}{||x_1||}\le||A||.$$
This is similarly true for $i=2$, and thus,
$$3=\max\{|\lambda_1|,|\lambda_2|\}\le||A||.\tag{2}$$
By (1) and (2), we have $||A||=3$
My point is that, although $A$ is not symmetric, it can be orthogonally diagonalizable and we can get $||A||$ just by evaluating the eigenvalues of $A$, not by evaluating the eigenvalues of $AA^T$.
But, is the second trial right?
