$$A= \begin{pmatrix} 0 & 1 & i \\ 1 & 1 & 2 \\ -i & 2 & 3 \end{pmatrix} $$ Calculate $\|A\|$ and demonstrate $\|A^+A\|=\|A\|^2 $.
Which is the fastest way to resolve this?
I can demonstrate that property using first a $\leq$ and then a $\geq$, and prove that if the two statements are valid then it's $=$ .
And $A^+A $ would mean that it is a Hermitian matrix.
But I'm not sure how to calculate $\|A\|$. Should it be the maximum of the eigenvalues of the matrix?
Your matrix is Hermitian, so $$\|A\|=\max\{|\lambda|:\ \lambda\in\sigma(A)\}.$$ And so is $A^*A$, so this problem is only about calculating eigenvalues. Which are a pain for the concrete matrix given.