The question states: If $|det(A)| > 1$, prove that the powers $A^n$ cannot stay bounded. But if $|det(A)| \leq 1$, show that some entries of $A^n$ might still grow large.

The answer provided by the solutions manual doesn't seem to be obvious (http://math.mit.edu/~gs/linearalgebra/ila5sol_05.pdf). I'd like to provide my undergraduate class with a proof this uses more concepts rather than some trick. So, I've told them to think about this in terms of the product of the eigenvalues.

I just wanted to see if there is something I am missing and you might add.

I have

Let $A$ be a $m\times m$ matrix. Since $det(A^n) = \lambda_1^n \times > \lambda_2^n \times \dots \times \lambda_m^n$, then we may consider what these values are when $|det(A)|>1$.

$$ |det(A)|>1\\ |det(A^n)|>1^n\\ |\lambda_1^n \times \lambda_2^n \times \dots \times \lambda_m^n|>1 $$

Similarly for $|det(A)| \leq 1$

$$ |det(A)|\leq 1\\ |det(A^n)|\leq 1^n\\ |\lambda_1^n \times \lambda_2^n \times \dots \times \lambda_m^n|\leq 1 $$

Before I begin analyzing the values of the $\lambda$'s that make these things true, I wanted to make sure this would work as a proof.

Thanks for your time and help!


1 Answer 1


For the first part it may help to consider diagonal matrices first. If $A$ is diagonal and $|\det A| > 1$ it has at least one entry satisfying $|a_{jj}| > 1$. Thus $A^k$ contains the entry $a_{jj}^k$ which satisfies $|a_{jj}^k| \to \infty$.

If $A$ is not diagonal and $|\det A| > 1$ you can write $A = P^{-1}DP$ where $D$ is diagonal. Then $D = PAP^{-1}$ and consequently $D^k = PA^kP^{-1}$ for all $k$. It is simple to see if the entries of $A^k$ remain bounded then so do the entries of $D^k$ -- but since $\det D = \det A$, the entries of $D^k$ do not remain bounded.

For the case $|\det A| \le 1$ look at the matrix $A$ consisting of all $1$'s. Then $\det A = 0$ but the entries of $A^k$ get arbitrarily large.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .