I read the following result:

If $A$ is $n\times n$ matrix and $\rho(A)$ denotes the spectral radius of $A$. Then if $\rho(A)<1$, then $\lim\limits_{m\rightarrow \infty} A^m\rightarrow \boldsymbol{0}$, where $\boldsymbol{0}$ is zero matrix.

Now result is easily proved to be true for diagonalizable matrices because if $PAP^{-1}$ is diagonal matrix $D$, then $D=diag(d_1, d_2, \dots, d_n)$ where $|d_i|<1$. Then $PA^mP^{-1}=diag(d_1^m, d_2^m, \dots, d_n^m)\rightarrow \boldsymbol{0}$. So $\lim\limits_{m\rightarrow \infty} A^m\rightarrow \boldsymbol{0}$.

Now we know that $\lim\limits_{m\rightarrow \infty} A^m\rightarrow \boldsymbol{0}$ is same as $\lim\limits_{m\rightarrow \infty} \|A^m\|\rightarrow \boldsymbol{0}$ for any matrix norm $\|\cdot\|$ on $\mathbb M_n(\mathbb C)$.

And in any matrix norm, diagonalizable matrices are dense in $\mathbb M_n(\mathbb C)$. Can we use some continuity argument to prove the result for any matrix? (we also know eigenvalues are continuous function on $\mathbb M_n(\mathbb C)$.

  • 1
    $\begingroup$ And your question is? $\endgroup$
    – egreg
    May 15, 2021 at 7:08
  • $\begingroup$ In this case I think it is easier to prove directly that $\lim_{m\to\infty}A^m\to0$. $\endgroup$
    – user1551
    May 15, 2021 at 8:25
  • $\begingroup$ @egreg To prove $\lim\limits_{m\rightarrow\infty} A^m=\boldsymbol{0}$ for any matrix $A$. I only got it for diagonalizable matrices till now. And want to use the case of diagonalizable matrices to prove the general case, if possible using denseness of diagonalizable matrices. $\endgroup$
    – Sushil
    May 15, 2021 at 15:29
  • $\begingroup$ @Sushil If you know that diagonalizable matrices are dense, you can simply use the fact that the norm is continuous. $\endgroup$
    – egreg
    May 15, 2021 at 15:41
  • $\begingroup$ @egreg Yes I was trying in same way. I got given matrix $A$ with $\ho(A)<1$, I can find diagonalizable matrix $B$ such that $\rho(B)<1$ and I can choose $B$ as close as to $A$. But problem is $A^m$ and $B^m$ may not be close enough. $\endgroup$
    – Sushil
    May 17, 2021 at 8:35

1 Answer 1


Can we use some continuity argument to prove the result for any matrix?

Yes, I have such an argument, albeit a roundabout one.

Note that Gelfand's formula $\rho(A)=\lim_{m\to\infty}\|A^m\|^{1/m}$ can be proved without using the statement that $\rho(A)<1$ iff $\lim_{m\to\infty}A^m=0$. Suppose $\rho(A)<1$. Pick any submultiplicative matrix norm. By Gelfand's formula, $\lim_{m\to\infty}\|A^m\|^{1/m}<1$. It follows that $\|A^M\|<1$ for some positive integer $M$. Now, for all $m\ge 0$, write $m=qM+r$ by Euclidean division. Then $\|A^m\|\le\|A^M\|^q\max_{0\le r<M}\|A^r\|$. When $m\to\infty$, we have $q\to\infty$ and hence $\|A^m\|\to0$. Since all norms are equivalent on a finite-dimensional vector space, we conclude that $\lim_{m\to\infty}A^m=0$ with respect to every (submultiplicative or non-submultiplicative) matrix norm.

But where is continuity used? It's in the proof of Gelfand's formula.

  • $\begingroup$ Thanks for your answer. But my question was something else. And I basically wanted the question to prove Gelfand's formula itself. Still thanks a lot for your effort. $\endgroup$
    – Sushil
    May 15, 2021 at 15:28

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