Relationship of spectral radius to matrix norm

Question

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?

Answer 1

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 $\|A\|=p$.

Answer 2

As a standalone statement, this can't be true, because the LHS of the identity is a funtion that satisfies $f(\lambda A)=\lvert\lambda\rvert f(A)$, whereas the RHS of the identity is a function that satisfies $g(\lambda A)=\lvert\lambda\rvert^n g(A)$.