# Power sequence in matrices

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 $$\mathbb M_n(\mathbb C)$$.

• And your question is? May 15, 2021 at 7:08
• In this case I think it is easier to prove directly that $\lim_{m\to\infty}A^m\to0$. May 15, 2021 at 8:25
• @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. May 15, 2021 at 15:29
• @Sushil If you know that diagonalizable matrices are dense, you can simply use the fact that the norm is continuous. May 15, 2021 at 15:41
• @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. May 17, 2021 at 8:35

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 $$\|A^m\|\le\|A^M\|^q\max_{0\le r. When $$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.