Limit of matrix powers. Consider an arbitrary matrix $A$ with eigenvalues within the unit circle. Is there a nice formula for $A^\infty = \lim_{n \rightarrow \infty} A^n$?
In particular, maybe there is a formula which involves the three matrices from the SVD of A?
I am asking this from an algorithmic standpoint, i.e. is there a better way to compute $A^\infty$ than simply squaring the matrix $A$ many times?
 A: Edit: You may use eigendecomposition. Let $A=PJP^{-1}$ where $J=J_{r_1}(\lambda_1)\oplus\cdots\oplus J_{r_s}(\lambda_s)$ is the Jordan form of $A$ and each $J_{r_i}(\lambda_i)$ is a Jordan block of size $r_i$ corresponding to the eigenvalue $\lambda_i$. Clearly, $A^m$ converges if and only if $J_{r_i}(\lambda_i)^m$ converges.
Now, consider a Jordan block $B=J_r(\lambda)$.


*

*If $|\lambda|\ge1$ and $\lambda\neq1$, the diagonal entries of $B^m$, which are equal to $\lambda^m$, do not converge.

*If $\lambda=1$ and $B$ is a nontrivial Jordan block ($r>1$), the diagonal entries of the superdiagonal of $B^m$, which are equal to $m$, diverge.

*If $B$ is a $1\times1$ Jordan block corresponding to the eigenvalue $1$, clearly $B^m=1$ and $\lim_{m\to\infty}B^m=1$.

*If $|\lambda|<1$, consider $DBD^{-1}$, where $D$ is a diagonal matrix of the form $\operatorname{diag}(\varepsilon,\varepsilon^2,\ldots,\varepsilon^n)$ with $\varepsilon>0$. $B^m$ converges if and only if $(DBD^{-1})^m$ converges. However, the effect of the conjugation $B\mapsto DBD^{-1}$ is to scale the superdiagonal of $B$ by $\varepsilon$. Therefore, when $\varepsilon$ is sufficiently small, the maximum row sum norm of $DBD^{-1}$, $\|DBD^{-1}\|_\infty$, is strictly smaller than $1$. Hence $C^m$ and in turn $B^m$ converge to $0$.


Therefore, for any $n\times n$ complex matrix $A$,

$A^m$ converges if and only if the Jordan decomposition of $A$ has the form $P(J_{r_1}(\lambda_1)\oplus\cdots\oplus J_{r_t}(\lambda_t)\oplus I)P^{-1}$, where $|\lambda_1|,\ldots,|\lambda_t|<1$ (the identity block $I$ is void if $r_1+\cdots+r_t=n$). If this is the case, $\lim_{m\to\infty}A^m=P(0\oplus I)P^{-1}$. In particular, if all eigenvalues of $A$ lie inside the open unit disc, $\lim_{m\to\infty}A^m=0$.

If $A$ is real, since $\lim_{m\to\infty}A^m=X$ over $\mathbb{R}$ if and only if $\lim_{m\to\infty}A^m=X$ over $\mathbb{C}$, the above argument still applies and $P(0\oplus I)P^{-1}$ is real (given that $A^m$ converges) despite $P$ may be complex.
