$||\textbf{A}\textbf{B}||_2=||\textbf{B}\textbf{A}||_2$?

Question

Let $\textbf{A}$ and $\textbf{B}$ be $n\times n$ matrices. Is it true that $||\textbf{A}\textbf{B}||_2=||\textbf{B}\textbf{A}||_2$?

I tried to prove this by the following argument: $$\text{det}(\textbf{A}\textbf{B})=\text{det}(\textbf{B}\textbf{A})=\text{det}(\textbf{A})\text{det}(\textbf{B})=\prod_{i=1}^n\lambda_i\prod_{i=1}^n\gamma_i$$ where $\lambda_i$ and $\gamma_i$ denote the $i^{\text{th}}$ eigenvalues of $\textbf{A}$ and $\textbf{B}$ respectively. Also $$\text{tr}\{\textbf{A}\textbf{B}\}=\text{tr}\{\textbf{B}\textbf{A}\}$$

but I am not sure how to proceed.

Edit: What about when $\textbf{A}$ and $\textbf{B}$ are Hermitian?

Answer 1

Suppose that$$A=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}\text{ and that }B=\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}.$$Then $AB\neq0$, but $BA=0$. Therefore, $\lVert AB\rVert\neq\lVert BA\rVert$, for any norm $\lVert\cdot\rVert$.