Show that if an $n\times n$ matrix $A$ is diagonalizable, so is $A^k$ for some integer $k$.

Question

We know that if $A$ is diagonalizable that means it is similar to a diagonal matrix $D$ whose columns are the eigenvectors of the linear transformation $T$ that corresponds to $A$.

I think that this fact could somehow be used. We know that if raise a diagonal matrix to a power, the only thing that changes are the diagonal entries (because every other entry is $0$). Therefore, connecting that to the eigenvalues of $A$ I think one might be able to make some good observations and possibly prove this.

Any help would be much appreciated.

Answer 1

In fact it is true that $A^k$ is diagonalizable for every $k \in \Bbb N$, because $A$ diagonalizable means that there is an invertible $n \times n$ matrix $S$ with

$SAS^{-1} = D, \tag 1$

where $D$ is a diagonal matrix; then

$SA^kS^{-1} = (SAS^{-1})^ k = D^k, \tag 2$

and $D^k$ is diagonal since $D$ is diagonal.

In the above we used the fact that

$SA^kS^{-1} = (SAS^{-1})^ k, \tag 3$

which may easily be proved by induction on $k$; I leave this to the members of my vast audience.

Answer 2

If $A$ is diagonalizable, it means it can be written as: $$A=PDP^{-1}$$ Now multiply both sides by $A$ $$AA=PDP^{-1}A$$ $$A^2=PDP^{-1}PDP^{-1}$$ $$A^2=PD(P^{-1}P)DP^{-1}$$ $$A^2=PDI_nDP^{-1}$$ $$A^2=PDDP^{-1}$$ $$A^2=PD^2P^{-1}$$ Where $I_n$ is the identity matrix of size n.

Therefore the statement has been proven for $k=2$. You can also prove it for an arbitrary positive integer $k$ by induction.