# Eigenvalues, diagonalization and convergence of matrices

I am trying to wrap my head around some basic results in Linear Algebra.

I am trying to avoid more abstract concepts like Rank-Nullity, and stay in simple properties at an introductory level. (I've taken more advanced Linear Algebra before, but it's been a long while, just trying to brush up on some basic facts).

Suppose I have a $$n \times n$$ matrix $$A$$.

Which of these facts are true?

1. $$\lim A^k$$ exists if and only if all eigenvalues are strictly less than 1 in absolute value.

2. If there are at least 2 eigenvalues that are equal to $$1$$, then the limit above does not exist.

3. If a matrix has all its eigenvalues less than or equal to one in absolute value (not necessarily unique), then the limit above exists.

4. If a matrix is diagonalizable and all its eigenvalues are less than or equal to one, then the limit above exists.

I think that statement 4 is the only one that is correct ("easy" to show in an informal way by the fact that $$A^k = PJ^kP^{-1}$$), but I've come across notes in Linear Algebra that state 1) and 2) as facts.

• Both 1 and 2 fail for the identity matrix. Also, when you say less than 1, do you mean they have absolute value less than 1? – eyeballfrog Mar 8 at 23:32
• yeah, in absolute value, i've updated it, thanks – Marcos Lee Mar 8 at 23:34
• Number 3 fails for the scalar ($1 \times 1$ matrix $A$) equal to $-1$. – Koen Tiels Mar 8 at 23:43
• Number 4 as well, for the same matrix as in 3. – enedil Mar 8 at 23:45
• Sorry, a more accurate one would be "If a PSD matrix is diagonalizable, and all its eigenvalues are $\leq 1$, with at least one $< 1$ inequality, then the limit above exists." Is this correct? – Marcos Lee Mar 9 at 1:44

To summarize the comments, none of these statements is true. An easy sufficient condition for this limit to exist is that $$A$$ is diagonalizable and all of its eigenvalues lie in the half-open interval $$(-1, 1]$$, and a necessary condition is that the eigenvalues of $$A$$ are either equal to $$1$$ or strictly less than $$1$$ in absolute value. It's a bit tricky to say what happens if $$A$$ isn't diagonalizable.