Square Root Matrix of a Diagonalizable Matrix is Diagonalizable

Question

I have a $2 \times 2$ matrix $A$ that is diagonalizable with two real, positive eigenvalues. I am asked to find all square roots of A

A is diagonalizable. Therefore, $A = SDS^{-1}$ where $D$ is diagonal and has entries as the two eigenvalues. Therefore, there are obviously at least 4 square root matrices: $SD'S^{-1}$ where $D'$ has $\pm \sqrt{\text{eigenvalue}}$ at each entry.

But now I want to show that these are the only ones. These are obviously the only diagonalizable square roots. But how do I prove that there are no other square root matrices that are not themselves diagonalizable.

In other words, how do I prove that if A is as above then sqrt(A) must be diagonalizable?

Answer 1

Prove the contrapositive. If $B$ is not diagonalizable, it is of the form $SJS^{-1}$, where $J$ is a size-2 Jordan block. So if $B$ also has positive eigenvalues, $B^2=SJ^2S^{-1}$ will not be diagonalizable.

However, the rest of your proof isn't totally valid. In particular, the identity matrix satisfies all your conditions, but has a lot more than 4 diagonalizable square roots...

EDIT: If $A$ has distinct eigenvalues, then $\sqrt{A}$ does as well, which means it's guaranteed to be diagonalizable.

Answer 2

Let the two eigenvalues of $A$ be $a,b>0$ and let $B$ be a square root of $A$.

If $a\ne b$, the polynomial $(x^2-a)(x^2-b)$ annihilates $B$; if $a=b$, the polynomial $x^2-a$ annihilates $B$. In either case, $B$ is annihilated by a polynomial with distinct real roots (because $a,b>0$). Hence the minimal polynomial of $B$ has distinct real roots and $B$ is diagonalisable over $\mathbb R$.