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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 $\geq$?

Rodrigo de Azevedo
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MickG
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2 Answers2

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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.

Ben Grossmann
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  • 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. – Ben Grossmann Aug 25 '14 at 20:46
  • 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 $|S||A||S^{-1}|$,… – MickG Aug 28 '14 at 08:19
  • …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. – MickG Aug 28 '14 at 08:29
  • 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. – MickG Aug 28 '14 at 08:34
  • 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. – Ben Grossmann Aug 28 '14 at 11:09
  • Here's what I'm really suggesting. For your favorite operator norm $|\cdot|$, we can select a sequence of similar matrices ${S_n A S_n^{-1}}_{n=1}^\infty$ whose norm approaches the spectral radius. Importantly, note that you don't have to do this for every operator norm; you just have to choose a single one and show that this works. As a hint towards the choice of similar matrix, note that two matrices are similar if and only if they have the same Jordan canonical form. – Ben Grossmann Aug 28 '14 at 11:14
  • And I hadn't connected to the web for about 2 days :). Anyway I'll try to think about this and find the sequence. – MickG Aug 29 '14 at 08:40
  • Uh-huh. You have to make sure at least one of the $\lambda$ that are equal in absolute value to $\rho$ has a $\frac1n$ beside or above. How do you find that similarity? – MickG Aug 29 '14 at 17:15
  • In fact, no you don't need to worry about that, because if there isn't, the norm will be lesser, and since we need to major we are OK even with a lesser norm. The norm 1 of that matrix with the $\frac1n$'s is at most $|\rho(A)|+\frac1n$. So the last thing is how do I see that similarity, I mean, for which invertible $S_n$ is $S_nAS_n^{-1}=A_n$, $A$ being the matrix with ones and $A_n$ the one with $\frac1n$? – MickG Aug 29 '14 at 17:30
  • In fact, there is no need to find the similarity-matrix; it suffices to show that the matrices are similar by considering $\ker(A - \lambda I)$. However, if you prefer, you could use $$ S_n = \pmatrix{ n^{n-1}&&&\ &n^{n-2}&&\ &&\ddots&\ &&& 1 } $$ – Ben Grossmann Aug 29 '14 at 17:38
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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 $.

VictorZurkowski
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