Gelfand's theorem

Question

Gelfand's theorem for matrices says that if $X$ is a complex square matrix, and $\|X\|$ is the operator 2-norm for $X$, then $\|X^n\|^{1/n}\to \rho(X)$, where $\rho(X)$ is the spectral radius of $X$.

This is easy to show using normal matrices, using the fact that $\|X^2\|=\|X\|^2$ for normal matrices. Can one deduce the general case from the case using normal matrices?

Answer 1

This is doable, but I'm not sure if the effort worths it.

Let $A$ be a complex square matrix. Since both spectral radius and operator norm are unitarily invariant, we may assume that $A$ is triangular. Denote the entrywise absolute value of any matrix/vector/scalar $X$ by $|X|$. Clearly, $|A|$ is still triangular and $\rho(A)=\rho(|A|)$.

For any given $\epsilon>0$, we can always choose a positive diagonal matrix $D$ whose diagonal entries of $D$ are smaller than $\epsilon$, such that $B:=D+|A|$ has distinct eigenvalues.

Entrywise absolute value of matrix/vector has the property that $|XY|\le|X||Y|$ entrywise. Since $|A|\le B$, we have $|A^nv|\le|A|^n|v|\le B^n|v|$ for every $n\ge0$ and every vector $v$. In particular, if $v$ is a unit singular vector for the largest singular value of $A^n$, we see that $$ \|A^n\|=\|A^nv\|=\||A^nv|\|\le\|B^n|v|\|\le\|B^n\|.\tag{1} $$ Since $B$ has distinct eigenvalues, we may diagonalise it as $B=P\Lambda P^{-1}$. It follows that \begin{align} \lim_{n\to\infty}\|B^n\|^{1/n} &=\lim_{n\to\infty}\|P\Lambda ^nP^{-1}\|^{1/n}\\ &\le\lim_{n\to\infty}(\|P^{-1}\|\|P\|)^{1/n} \|\Lambda ^n\|^{1/n}\\ &=\lim_{n\to\infty}\|\Lambda ^n\|^{1/n}\\ &=\rho(\Lambda)=\rho(B).\tag{2} \end{align} (Technically, normality of $\Lambda$ is used in last line.) Combining $(1)$ and $(2)$, we get $$ \lim_{n\to\infty}\|A^n\|^{1/n}\le\rho(B)<\rho(A)+\epsilon. $$ Let $\epsilon\to0$, we get $\lim_{n\to\infty}\|A^n\|^{1/n}\le\rho(A)$. Clearly, we also have $\rho(A)\le\lim_{n\to\infty}\|A^n\|^{1/n}$ (because $\rho(A)^n=\rho(A^n)\le\|A^n\|$). The proof is now complete.

Answer 2

Let $A\in M_n$, $u$ be a non-zero real. Then $\lim_{k\rightarrow\infty}(u||A^k||)^{1/k}=\rho(A)$ iff $\lim_{k\rightarrow\infty}||A^k||^{1/k}=\rho(A)$. Then we may choose any norm on $M_n$; there is an invertible $P$ s.t. $P^{-1}AP=diag(\lambda_1I_{n_1}+N_1,\cdots,\lambda_pI_{n_p}+N_p)$ is in Jordan form. We put $||U||=||P^{-1}UP||_{\infty}$.

Step 1. Consider a Jordan block $J_n=\lambda I+N$ where $N$ is nilpotent.

Case 1. $\lambda\not= 0$. If $k\geq n-1$, then $||J_n^k||_{\infty}=||\lambda^kI+\cdots+\binom{k}{j}\lambda^{k-j}N^j+\cdots+\binom{k}{n-1}\lambda^{k-n+1}N^{n-1}||_{\infty}$. Since $\binom{k}{j} \leq k^{j}$, there is $\tau>0$, that depends on $n$ but not on $k$, s.t. $||J_n^k||_{\infty}\leq \tau|\lambda|^kk^{n-1}$.

On the other hand, since we chose the $\infty$ norm, $||J_n^k||_{\infty}\geq ||\lambda^kI||_{\infty}=|\lambda|^k||I||_{\infty}$.

Case 2. $\lambda=0$. Then $J_n^k=0$ when $k\geq n$.

Finally $(||J_n^k||_{\infty})^{1/k}$ tends to $|\lambda|$.

Step 2. $||A^k||= \max_{j\leq p} ||(\lambda_jI_{n_j}+N_j)^k||_{\infty}$; according to Step 1, $(\max_{j\leq p} ||(\lambda_jI_{n_j}+N_j)^k||_{\infty})^{1/k}$ tends to $\rho(A)$.

Assume (for example) that $|\lambda_1|=\rho(A)$; then $||A^k||\geq |\lambda_1^k|=(\rho(A))^k$.

And we are done.