I just read that for a real symmetric matrix, the matrix $(A)$ norm equals the spectral radius $(p)$ to the $n^{th}$ power : $||A||=p^n$.
I don't think this is true, is it? If so, where does it come from, what is the theorem?
Relationship of spectral radius to matrix norm
matricesmatrix-normsspectral-radiussymmetric matrices
Best Answer
The general result (no requirements on $A$) is that $$\tag1 p=\lim_{n\to\infty}\|A^n\|^{1/n}. $$ The above limit always exists.
When $A$ is selfadjoint (in particular, real symmetric), it is easy to show that $\|A^{2^n}\|=\|A\|^{2^n}$, so $(1)$ implies that $p=\|A\|$.
This last result can also be obtained directly by noticing that $A$ selfadjoint admits an orthonormal basis of eigenvectors. From this is follows easily that $\|A\|=p$.