powers of a diagonal matrix to infinity

Question

Let $A$ be a square matrix that is diagonalizable. This means that it can be like this: $A = SDS^{-1}$, where $D$ is a diagonal matrix containing the eigenvalues of $A$. It follows $S$ contains the eigenvectors of $A$.

We also know the matrix powers $A^k = SD^kS^{-1}$.

Now, if we want to make $k \to \infty$, then the only way that the matrix will approach a 'stable' value is when $|\lambda|<1$. (And this is how most textbooks put it.)

My question is: Is there an additional condition imposed that $\lambda$ should be real only? Is it possible that when $\lambda$ is complex, then $A^k$ , as $k \to \infty$ still exists?

Your insights will be helpful.

Answer 1

No, if the eigenvalue is real and between 0 and 1 then the value of a $A^kv$, where v is the eigenvector, will exponentially decay to 0. If $\lambda$ is between -1 and 0 then the value will oscilate and decay to 0. If its complex and has an absolute value, or radius in the complex plane, between 0 and 1 then it will spiral, from the eigenvector, towards 0 either clockwise or anti-clockwise.

Answer 2

As you said $A^k=SD^kS^{-1}$ so the existence of the limit of $A^k$ reduces to that of $D^k$. Now, $D$ is a diagonal matrix, and so is $D^k$. If the element of the diagonal are $d_1,\dots, d_n$ then the elements of the diagonal of $D^k$ are $d_1^k,\dots,d_n^k$.

Therefore the question reduces to the following one :"given $d_i$, when has $d_i^k$ a limit for $k\to \infty$?"

Clearly if $|d_i|<1$ then $\lim_k d_i^k=0$. On the other hand, if $|d_i|=1$ then the limit exists only if $d_i=1$. Finally, if $|d_i|>1$ then the limit is infinite.