For $2 \times 2$-matrix, $\|A^2\|=\|A\|^2$ implies that A is normal I am reading book "A Hilbert space problem book", written by Halmos and getting some trouble.
In problem 205, he claim that

For two-by-two matrices an unpleasant computation proves a strong converse: if $\|A^2\|=\|A\|^2$, then $A$ is normal.

I tried to use the determinant as the norm of $A$, but it seems to be useless.
I also found a similar question on Mathstackexchange. Someone said that we should only consider matrix $\begin{pmatrix}
\lambda & 1\\ 
0 & \lambda
\end{pmatrix}$. I, however, do not understand why.
Could you please give me some hint to solve this claim?
Thank you for your time.
 A: He probably meant to use the induced $2$-norm. The statement does not always hold if another norm is used. E.g. suppose the induced $\infty$-norm is used, where $\|A\|=\max_i\sum_j|a_{ij}|$. Then $\|A^2\|=\|A\|^2$ for every $n\times n$  row stochastic matrix (unless $n=1$), but obviously some row stochastic matrices are not normal.
The statement is correct when the induced $2$-norm $\|A\|=\sqrt{\lambda_\max(A^\ast A)}$, is used. The "unpleasant computation" probably means to calculate and compare $\|A^2\|=\sqrt{\lambda_\max\left((A^2)^\ast A^2\right)}$ and $\|A\|^2=\lambda_\max(A^\ast A)$ directly, and that really is unpleasant.
An easier way to prove the statement is to employ singular value decomposition. Let $A=USV^\ast$ be a singular value decomposition, where the two singular values of $A$ or $S$ are $\sigma_1\ge\sigma_2$. When $\sigma_1=\sigma_2$, we have nothing to prove because $A=UV^\ast$ is a unitary matrix, which is normal. So, suppose $\sigma_1>\sigma_2$. By scaling $A$ appropriately, we may also assume that $\|A\|=1$.
Now, from $\|A^2\|=\|A\|^2=1$, we get $\|USV^\ast USV^\ast\|=1$ and in turn $\|SV^\ast US\|=1$. Let $x$ be a unit singular vector of $SV^\ast US$ corresponding to the singular value $\sigma_1=1$, so that $\|x\|=\|SV^\ast USx\|=1$. The mapping $x\mapsto SV^\ast USx$ is a function in the form of $f\circ g\circ f$, where $f:x\mapsto Sx$ always shrinks the norm of a unit vector $x$ unless $x$ is a unit multiple of $e_1=(1,0,\ldots,0)^T$, and $g:x\mapsto V^\ast Ux$ preserves the norm of a vector $x$. It follows that $\|f\circ g\circ f(x)\|=1$ only if both $x$ and $V^\ast Ue_1$ are unit multiples of $e_1$. Hence $D:=V^\ast U$ is a diagonal matrix (because $V^\ast U$ is unitary $2\times2$). Thus $A=USV^\ast=U(SD)U^\ast$ is normal.
