# Matrix norm and spectral radius

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 $$\epsilon$$.

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 $$\lim_{t\to0}\|A\|_D=\rho(J)=\rho(A)$$.
• Thank you for your answer, I accepted it since the way my question was posed makes it spot on. However, I was wondering if this result is optimal in some sense. What I mean is, your answer rephrases the traditional proof of the theorem I mentioned (I finally am able to check a resource about it). However the vector norm we get is thus forced to have memory of the matrix we are working with, according to this construction. Do you think it is possible to improve on that? (This would mean getting rid of the change of coordinates$P^{-1}$, the matrix $D$ doesnt bother me)
• @GiordanoGiambartolomei If you want a matrix norm (induced or not) such that $\|A\|<\rho(A)+\epsilon$ for every matrix $A$, there is no such norm. In fact, if $A=\pmatrix{0&1\\ 0&0}$, then $\|A\|\ne0$ because $A$ is nonzero. Hence $t\|A\|=\|tA\|$ will eventually be greater than any prespecified $\epsilon$ when $t>0$ is sufficiently large, but that makes $\|tA\|>\epsilon=0+\epsilon=\rho(tA)+\epsilon$. Commented Sep 29, 2020 at 15:01