Diagonalizing the matrix $A$, when $A^2$ is diagonalizable

Question

If the matrix $A^2$ is diagonalizable and $A$ is invertible, is $A$ diagonalizable? I know it is not true if we leave out the invertibility. For example if $ A= \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} $ (which is not diagonalizable), then $A^2= \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix} $ (and that is trivially diagonalizable).

But what if we require $A$ to be invertible? I believe in that case, answer to my original question is yes, but I'm not able to prove that. Can someone please help me?

Answer 1

I'm assuming this is over the complex numbers.

Put $A$ in Jordan canonical form. The square of a nontrivial Jordan block (with non-zero diagonal) is an upper triangular matrix with nonzero diagonal and above-diagonal elements, and the eigenvalue has geometric multiplicity $1$. So if $A$ is invertible and not diagonalizable, neither is $A^2$.

Answer 2

Notation: $E_{L}^{B}$ is the set of eigenvectors associated to $L$ for the matrix $B$.

As $A^2$ is diagonalizable, $$ E = \bigoplus_{l_i\neq 0} E_{l_i}^{A^2} $$

On each $E_{l_i}^{A^2}$, $A^2 -l_i I =0 $ and then, as $l_i\neq 0$,
$$E_{l_i}^{A^2} = E_{-\sqrt{l_i}}^{A} \oplus E_{\sqrt{l_i}}^{A}$$ Hence $$ E = \bigoplus_{l_i\neq 0} \left[E_{-\sqrt{l_i}}^{A} \oplus E_{\sqrt{l_i}}^{A}\right] $$and $A$ is diagonalizable.

Answer 3

Let me formulate what I believe is the precise condition to make this conclusion. No assumptions are made about the base field $K$ over which one wants to diagonalise

Proposition. Let $A$ be a square matrix with entries in a field $K$. Assume that

• $A^2$ is diagonalisable over$~K$, with eigenvalues in a set $S\subseteq K$, and
• every element of $S$ has two distinct square roots in $K$.

Then $A$ is diagonalisable over$~K$, with eigenvalues in $\sqrt S\overset{\rm def}=\{\,r\in K\mid r^2\in S\,\}$.

Note that $S$ need not be just the set of eigenvalues of $A^2$, and certainly $\sqrt S$ can be larger than the set of eigenvalues of$~A$. But one may, and we will, assume $S$ to be finite.

Proof. Let $P_1=\prod_{s\in S}(X-s)$, then by the first assumption $P_1[A^2]=0$. By the second assumption there exists for each $s\in S$ an element $r\in K$ such that $X^2-s=(X-r)(X+r)$, and $r\neq -r$. Choosing one such$~r$ and calling it $\sqrt s$, one has $$ P_2\overset{\rm def}=P_1[X^2]=\prod_{s\in S}(X-\sqrt s)(X+\sqrt s)=\prod_{r\in\sqrt S}(X-r). $$ Then $P_2[A]=P_1[A^2]=0$, and since $P_2$ is split with simple roots (because of $r\neq-r$) in $\sqrt S\subseteq K$, it follows that $A$ is diagonalisable over$~K$, with eigenvalues in $\sqrt S$. $\square$

Note that in fields not of characteristic$~2$ the second condition excludes $0$ and any non-square from the set $S$ of allowed eigenvalues for$~A$, while for fields of characteristic$~2$ it excludes everything (i.e., the proposition will never apply). These are precisely the cases where the conclusion fails miserably.