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How do I prove that $$\rho(A)=\inf\limits_{\text{operator norms}}\|A\|,$$ $\rho$ being the spectral radius, $A$ being a complex $n\times n$ matrix and operator norms being induced from vector norms by defining $\|A\|=\min\limits_{\|x\|=1}\frac{\|Ax\|}{\|x\|}$? I have proved that $\rho(A)\leq\|A\|$ for any induced norm $\|\cdot\|$, so $\leq$ holds in the above equality. How do I prove $\geq$?

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Let's be precise about what "proving $\geq$" entails. What you need to show is that for any $\epsilon > 0$, there exists a matrix norm $\|\cdot\|$ such that $$ \|A\| < \rho(A) + \epsilon $$ The trick then, is to tailor a suitable example $\|\cdot \|$ to your matrix $A$ and choice of $\epsilon$.

I suggest the following: given the vector norm $|\cdot|$ of your choice, we can define the new vector norm $$ \|x\| = |Sx| $$ for any choice of invertible matrix $S$. Consider the resulting induced norm, under a clever choice of $S$.


Further hint:

Note that the induced norm can be written as $\|A\| = |SAS^{-1}|$, since $$ \max_{\|x\| = 1} \|Ax\| = \max_{|Sx| = 1}|SAx| = \max_{|y| = 1}|SAS^{-1}y| = |SAS^{-1}| $$


Further hint:

Note the following similarity: $$ \pmatrix{ \lambda&1\\ &\lambda&\ddots\\ &&\ddots&1\\ &&&\lambda} \sim \pmatrix{ \lambda&1/n\\ &\lambda&\ddots\\ &&\ddots&1/n\\ &&&\lambda} $$ Consider using the $\|\cdot \|_1$ norm; any should work here, though.

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  • $\begingroup$ Did this answer your question? Is something I said unclear? Is there something in my answer that you're unsure of? If so, let me know. If, on the other hand, you've found this answer satisfactory, accept it by clicking the check mark next to my question. $\endgroup$ – Omnomnomnom Aug 25 '14 at 20:46
  • $\begingroup$ OK. That answer wasn't really an answer, it was the starting point for one. I would suggest not using $|\cdot|$ for a norm as I normally interpret that as absolute value or determinant, but that's a detail. Basically what we are saying is that for all $\epsilon$ and all induced norms there exists at least an invertible matrix $S$ such that $\|S\|\|S^{-1}\|<\frac{1}{\|A\|}(\rho(A)+\epsilon)$, since a property of matrix norms is $\|AB\|\leq\|A\|\|B||$ and therefore the new induced norm of $A$ is not greater than $\|S\|\|A\|\|S^{-1}\|$,… $\endgroup$ – MickG Aug 28 '14 at 8:19
  • $\begingroup$ …where in inequality 1 I indicate a generic norm by $\|\cdot\|$, just like in 2, whereas in inequality three (the one with the inequality sign given in words) I use $\|\cdot\|$ to mean what you indicated by $|\cdot|$. In fact, it is enough to find them for one induced norm. That is the condition number of $A$ in the norm $|\cdot|$. Now I know that, in any norm, condition numbers are not less than 1. I don't have proofs that there are invertible matrixes with all possible values of the condition number, however. $\endgroup$ – MickG Aug 28 '14 at 8:29
  • $\begingroup$ Proving that should more or less complete the answer, viewing as by changing the norm I can avoid $\frac{1}{\|A\|}(\rho(A)+\epsilon)<1$ (right?). So let's say I'll accept the answer once that is proved, so the answer is complete. By the way, is there any relationship between $\|S\|$ and $\|S^{-1}\|$ in general? And what's the hurry to see your answer accepted? I only just saw it anyway. $\endgroup$ – MickG Aug 28 '14 at 8:34
  • $\begingroup$ A lot of things to answer here; I'll address them in the order they appear. "Basically what we are saying is that for all ϵ and all induced norms there exists at least an invertible matrix S such that $\|S\|\|S^{-1}\|<\frac{1}{\|A\|}(\rho(A)+\epsilon)$". No, that is not what we are saying. For operator norms, the best we can say for $S$ and $S^{-1}$ is that $1 = \|I\| = \|SS^{-1}\| \leq \|S\| \|S^{-1}\|$. No "hurry", it's just that a lot of askers give absolutely no response, question, comment, or otherwise. Note also that though you've just seen it, it's been up for 2 days. $\endgroup$ – Omnomnomnom Aug 28 '14 at 11:09
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I came here looking for an answer to this question for compact operators on Banach spaces. Here is a contruction that works in finite or infinite dimensions, assuming the spectral radius formula: $\rho = \lim_{n\rightarrow \infty} \|A^n\|^{1/n} $.

Define a norm in the underlying space as $ N(x) = \sum_{n \ge 0} \frac{\|A^nx\|}{(\rho +\epsilon)^n} .$ Trivially, $\|x\| \le N(x).$ From the spectral radius formula, given $\epsilon > 0$, there is a constant $C>0$ such that $\|A^n\| \le C(\rho + \epsilon/2)^n $ for all $n$. Then,

$$ \sum_{n \ge 0} \frac{\|A^nx\|}{(\rho +\epsilon)^n} \le \sum_{n \ge 0} \frac{\|A^n\|\|x\|}{(\rho +\epsilon)^n} \le \sum_{n \ge 0} \frac{C(\rho + \epsilon/2)^n}{(\rho +\epsilon)^n} \|x\| = M \|x\|$$ for some constant $M$. Therefore, $N$ is a norm in the underlying space, equivalent to the norm $\| \|$.

Note that $N(Ax) = \sum_{n \ge 0} \frac{\|A^{n+1}x\|}{(\rho +\epsilon)^n} = (\rho +\epsilon) \ \sum_{n \ge 0} \frac{\|A^{n+1}x\|}{(\rho +\epsilon)^{n+1}} \le (\rho +\epsilon) \ \sum_{n \ge 0} \frac{\|A^{n}x\|}{(\rho +\epsilon)^{n}} = (\rho +\epsilon) \ N(x)$, i.e.: the norm of $A$ in the operator norm derived from $N$ satisfies $ N(A) \le \rho + \epsilon $.

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