Matrix norm and spectral radius

Question

It is well known that for every positive $\epsilon$ there is a matrix norm which is smaller than the spectral radius of the matrix plus $\epsilon$. Is there any improvement of this theorem for induced or consistent matrix norms? That is, that for every $\epsilon$ there is an induced norm (or at worst a matrix norm consistent with some vector norm) that is less than the spectral radius plus $\epsilon$.

Answer 1

Let $A=PJP^{-1}\in M_n(\mathbb C)$ where $J$ is the Jordan form of $A$. Let $D=\operatorname{diag}(1,t,t^2,\ldots,t^n)$ where $t>0$. The vector norm defined by $\|x\|_D=\|D^{-1}P^{-1}x\|_2$ then induces a matrix norm \begin{aligned} \|A\|_D &=\max_{x\ne0}\frac{\|Ax\|_D}{\|x\|_D}\\ &=\max_{x\ne0}\frac{\|D^{-1}P^{-1}Ax\|_2}{\|D^{-1}P^{-1}x\|_2}\\ &=\max_{y\ne0}\frac{\|D^{-1}P^{-1}APDy\|_2}{\|y\|_2}\\ &=\|D^{-1}P^{-1}APD\|_2\\ &=\|D^{-1}JD\|_2. \end{aligned} The bidiagonal matrix $D^{-1}JD$ has the diagonal entries as $J$, but its nonzero super-diagonal entries (if any) are equal to $t$ instead of $1$. Therefore $\lim_{t\to0}\|A\|_D=\rho(J)=\rho(A)$.