$\|A\|^2 =\|A^2\|$ implies $A$ is normal 
Assume $A$ is a complex $2 \times 2$ matrix with$\|A\|^2 =\|A^2\|$. Prove that $A$ is normal.

I have found this solution but it is not clear. Any help would be appreciated.
In the question it's not mentioned which norm to use but I am assuming it is operator norm i.e. $$\|A\|_2 := \sup_{x \neq 0}\frac{\|Ax\|_2}{\|x\|_2}$$
 A: I'll start from where you said that you were confused in the comments. $$\|SV^*US\|=1$$ as a consequence of $$\|USV^*USV^*\|=1$$ (this is because $\|ABC\|=\|B\|$ when $A$ and $C$ are unitary, when $\|\cdot\|$ is the two-norm, as in their post). Given that $1=\|A\|^2=\lambda_\text{max}(A^*A),$ we know that there exists a singular value $\sigma_1$ so that $\sigma_1=1$. Then, there must be some unit singular vector $x$ which corresponds to it. In the composition, their $f$ is $S$ and their $g$ is $V^*U$. The reason that $f$ shrinks norms is because $f$ is the diagonal matrix comprised of the singular values, and we know that they're, at most, $1$. The exception is, of course, when $x=ce_1$ (with $|c|=1$), since the singular values less than one will not affect any of the entries of $x$. Since $V^*U$ is unitary, it automatically preserves norms. Since $f\circ g\circ f (x)$ will shrink the norm of $x$ unless $x=ce_1$ and $V^*U (Sx)=de_1$ (with $|c|,|d|=1$), and we know that $\|f\circ g\circ fx\|=1=\|x\|,$ it follows that $x=ce_1$ and $V^*U(Sx)=de_1$, or $V^*Ue_1=\frac{d}{c}e_1$ (again, with $|c|,|d|=1$). Since $V^*U$ is unitary, this implies that it is a diagonal matrix, which they call $D$, and hence $$A=USV^*=USV^*UU^*=U(SD)U^*,$$ which is normal.
