# Matrix norm and diagonalization of a matrix

I wonder if anyone knows a reference for this question:

For a complex matrix $$A$$ with only one eigenvalue $$\lambda$$ such that $$||A^k || \leq C|\lambda|^k$$ (for some constant $$C$$), can we say $$A$$ is diagonalizable?

The answer is yes. While the idea of the proof can usually be found in the development of Gelfand's formula, I don't think the result is stated as a theorem in many books.

In general, suppose $$\lambda$$ is one of the largest-sized eigenvalue of a complex square matrix $$A$$ (i.e. $$|\lambda|=\rho(A)$$). If there is a positive constant $$C$$ such that $$\|A^k\|\le C|\lambda|^k$$ for every positive integer $$k$$, then $$\lambda$$ must be semi-simple. (In your case, since $$\lambda$$ is the only eigenvalue, that $$\lambda$$ is semi-simple implies that $$A$$ is diagonalisable.)

Suppose the contrary. Then $$Av=\lambda v$$ and $$Au=\lambda u+v$$ for some eigenvector $$v$$ and generalised eigenvector $$u$$. It follows from mathematical induction that $$A^ku=\lambda^ku+k\lambda^{k-1}v$$. Now pick any vector norm $$\|\cdot\|$$ such that $$\|au+bv\|=\max(\|au\|,\|bv\|)$$ (e.g. take $$\|x\|=\|P^{-1}x\|_\infty$$ where $$P$$ is any invertible matrix whose first two columns are $$u$$ and $$v$$). Then it induces a matrix norm $$\|M\|=\sup_{x\ne0}\frac{\|Mx\|}{\|x\|}$$. Since all norms are equivalent on a finite-dimensional vector space, we may assume that this induced matrix norm is the one in the condition that $$\|A^k\|\le C|\lambda|^k$$. Thus $$k|\lambda|^{k-1}\|v\|\le\|\lambda^ku+k\lambda^{k-1}v\|=\|A^ku\|\le\|A^k\|\|u\|\le C|\lambda|^k\|u\|\tag{1}$$ for every $$k\ge1$$. But this is impossible: if $$\lambda=0$$, $$(1)$$ is violated when $$k=1$$; if $$\lambda\ne0$$, $$(1)$$ is violated when $$k$$ is large. Thus $$\lambda$$ must be semi-simple.

$$\newcommand{nrm}{\left\lVert {#1}\right\rVert}\newcommand{abs}{\left\lvert {#1}\right\rvert}$$ Let $$A$$ be a triangulable matrix with exactly one eigenvalue $$\lambda$$, and let there be some $$C>0$$ such that $$\nrm {A^k}\le C\abs\lambda^k$$ for all $$k\ge 1$$.

If $$\lambda=0$$, then $$A=0$$ by considering $$k=1$$.

If $$\lambda\ne0$$, then, by hypothesis, $$A-\lambda I$$ is nilpotent, therefore $$\nrm{A^k}=\nrm{(\lambda I+(A-\lambda I))^k}=\nrm{\sum_{h=0}^s \lambda^{k-h}\binom kh (A-\lambda I)^h}$$ where $$s$$ is the least natural number such that $$(A-\lambda I)^{s+1}=0$$. \begin{align}\nrm{A^k}&=\nrm{\sum_{h=0}^s \lambda^{k-h}\binom kh (A-\lambda I)^h}\ge\\&\ge\nrm{\lambda^{k-s}\binom ks(A-\lambda I)^s}-\nrm{\sum_{h=0}^{s-1}\lambda^{k-h}\binom kh(A-\lambda I)^h}\ge\\&\ge\abs\lambda^k\left(\nrm{\lambda^{-s}(A-\lambda I)^s}\abs{\binom ks}-\sum_{k=0}^{s-1}\abs{\binom kh}\nrm{\lambda^{-h}(A-\lambda I)^h}\right)\end{align}

Now, $$\nrm{\lambda^{-h}(A-\lambda I)}$$ for $$h=0,1,\cdots, s$$ are non-zero constants that depend on $$A$$ and $$\lambda$$. On the other hand, all the $$\binom kj$$ are polynomials (exactly) of degree $$j$$ in $$k$$. Therefore, $$\frac1{\abs\lambda^k}\nrm{A^k}$$ grows asymptotically at least as fast as some polynomial of degree $$s$$. Therefore the hypothesis demands $$s=0$$, which implies $$(A-\lambda I)^{0+1}=0$$.