If A is a diagonalizable n × n matrix, prove that $A^{2020}$ is also diagonalizable.

Question

Would it be correct to prove this by referencing a theorem in our text that says that if x is an eigenvector of A corresponding to the eigenvalue t ,then x is an eigenvector of $A^m$ corresponding to the eigenvalue $t^m$? I said that since A is diagonalizable, we know there is a basis for $F^n$ consisting of eigenvectors of A. So then $A^{2020}$ has the same eigenvectors which also form a basis for $F^n$ and thus A is diagonalizable.

I guess I also used the theorem that states that a finite linear operator T on a vector space V is diagonalizable iff V has a basis of eigenvectors for T.

Answer 1

If $A$ is diagonalizable let's call $P$ the change of basis with brings $A$ in it's normal diagonal form, i.e. $PAP^{-1} = D$, with $D$ diagonal having the eigenvalues on the diagonal. Taking the $2020$- power we have $(PAP^{-1})^{2020} = D^{2020}$, but thanks to the conjugancy operator $(PAP^{-1})^{2020} = PA^{2020}P^{-1} = D^{2020}$ hence diagonalizable since $D^{2020}$ is a diagonal matrix with diagonal the eigenvalues of $A$ raised to the $2020$-th.

Answer 2

It's quite simple: let's prove by induction that, if $A=SBS^{-1}$, then $$ A^n=SB^nS^{-1} $$ for every nonnegative integer $n$. The case $n=0$ is obvious. If we assume the statement true for $n$, then $$ A^{n+1}=A^nA=SB^{n}S^{-1}SBS^{-1}=SB^nBS^{-1}=SB^{n+1}S^{-1} $$ hence the result.

If $B$ is diagonal, then $B^n$ is diagonal as well. No need to go to eigenvalues and multiplicity.