Let $B\in\mathbb{R}^{n\times n}$. Recall that the operator norm of $B$ is defined by $$ \|B\| = \max_{\|x\|\leq 1}\|Bx\|,$$ where the norm denotes the standard Euclidean norm.
Now imagine we put another Euclidean norm on $\mathbb{R}^n$: $$ \|x\|_M = \sqrt{\langle x,Mx\rangle}, $$ where $M$ is (symmetric and) positive definite. This norm will give rise to a corresponding $M$-operator norm: $$ \|B\|_M = \max_{\|x\|_M\leq 1}\|Bx\|_M.$$
My question is this: How small can we make $\|B\|_M$, treating $M$ as a variable? I checked my Linear Algebra texts (Horn-Johnson, Meyer, etc.) but nothing springs to mind. A lower bound appears to be the spectral radius $\rho(B)$. Motivation: In some cases, one could hope for $\|B\|_M<1$ and then one deals with a Banach contraction and thus gets convergence of iterates. So how small can $\|B\|_M$ get, and how to find the matrix $M$ that does the job? (Interesting counterparts are making it large, or moving to a general Hilbert space, by the way.)
Any reference to a book or paper would be greatly appreciated. I suspect that this is already well known, but I cannot find it in my books.
We have $\|x\|_M=\|M^{1/2}x\|_2$. Therefore $$ \begin{aligned} \|B\|_M &=\max_{\|x\|_M\le1}\|Bx\|_M\\ &=\max_{\|M^{1/2}x\|_2\le1}\|M^{1/2}Bx\|_2\\ &=\max_{\|v\|_2\le1}\|M^{1/2}BM^{-1/2}v\|_2\\ &=\|M^{1/2}BM^{-1/2}\|_2. \end{aligned} $$ When $M$ is allowed to be complex Hermitian, it is well-known that $$ \inf_{M\succ0}\|M^{1/2}BM^{-1/2}\|_2=\rho(B). $$ This result is often used to prove that the spectral radius is the infimum of all submultiplicative matrix norms.
With slight modification, the proof of this result can be adopted to the case where $M$ is required to be (real) symmetric positive definite.
For any $M\succ0$, let $x$ be a (possibly complex) unit eigenvector corresponding to the dominant eigenvalue of $M^{1/2}BM^{-1/2}$ over $\mathbb C$. Then $$ \|M^{1/2}BM^{-1/2}\|_2\ge\|M^{1/2}BM^{-1/2}x\|_2=\|\rho(M^{1/2}BM^{-1/2})x\|_2=\rho(B). $$ It remains to show that there exists a sequence of positive definite matrices $\{M_t\}_{t\in\mathbb N}$ such that $\|M_t^{1/2}BM_t^{-1/2}\|_2\to\rho(B)$ as $i\to\infty$. Since the spectral norm is orthogonally invariant, we may assume without loss of generality that $B$ is in block-Schur form, i.e., we assume that $B$ is block-upper triangular, its diagonal sub-blocks are either $2\times2$ or $1\times1$ and these sub-blocks are in the form of $$ B_1,\ldots,B_r,\lambda_{r+1},\ldots,\lambda_{r+(n-2r)}, $$ where each $B_k$ is of the form $\pmatrix{a_k&-b_k\\ b_k&a_k}$ with $b_k\ne0$ (so that each conjugate pair of complex numbers $a_k\pm ib_k$ are non-real eigenvalues of $B$) and each $\lambda_k$ is a real eigenvalue of $B$. Now, for each $t\in\mathbb N$, let $$ M_t=\operatorname{diag}(I_2,\,tI_2,\,t^2I_2,\,\ldots,t^{r-1}I_2,\ t^r,\,t^{r+1},\ldots,t^{r+(n-2r)-1}). $$ Then $$ \begin{aligned} \lim_{t\to\infty}\|M_t^{1/2}BM_t^{-1/2}\|_2 &=\|\lim_{t\to\infty}M_t^{1/2}BM_t^{-1/2}\|_2\\ &=\|\operatorname{diag}(B_1,\ldots,B_r,\lambda_{r+1},\ldots,\lambda_{r+(n-2r)})\|_2\\ &=\rho(B). \end{aligned} $$