Power sequence in matrices I read the following result:
If $A$ is $n\times n$ matrix and $\rho(A)$ denotes the spectral radius of $A$. Then if $\rho(A)&lt;1$, 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 $|d_i|&lt;1$. 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 $\mathbb M_n(\mathbb C)$.
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 $\rho(A)&lt;1$ iff $\lim_{m\to\infty}A^m=0$. Suppose $\rho(A)&lt;1$. Pick any submultiplicative matrix norm. By Gelfand's formula, $\lim_{m\to\infty}\|A^m\|^{1/m}&lt;1$. It follows that $\|A^M\|&lt;1$ for some positive integer $M$. Now, for all $m\ge 0$, write $m=qM+r$ by Euclidean division. Then $\|A^m\|\le\|A^M\|^q\max_{0\le r&lt;M}\|A^r\|$. 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.
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 $\rho(A)&lt;1$ iff $\lim_{m\to\infty}A^m=0$. Suppose $\rho(A)&lt;1$. Pick any submultiplicative matrix norm. By Gelfand's formula, $\lim_{m\to\infty}\|A^m\|^{1/m}&lt;1$. It follows that $\|A^M\|&lt;1$ for some positive integer $M$. Now, for all $m\ge 0$, write $m=qM+r$ by Euclidean division. Then $\|A^m\|\le\|A^M\|^q\max_{0\le r&lt;M}\|A^r\|$. 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.