Norm of matrix and property $$A=  \begin{pmatrix}
  0 & 1  & i \\
  1 & 1 & 2 \\
  -i & 2  & 3 
 \end{pmatrix}
$$
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. 
