Are all square roots of diagonalizable matrices diagonalizable?

If a matrix is normal/unitarily diagonalizable, then its square roots are readily computed by taking the square roots of its eigenvalues (in the complex plane if needed).

Any square root computed in this way is clearly going to also be normal.

However, are all square roots of a normal matrix $A$ also going to be normal? If this is the case, how can it be proven? How can the non-diagonal square roots be computed?

(The question could probably be generalized to refer to any inverse of a matrix function, but I am not sure about that).

Silly example: the zero matrix is clearly normal, but has $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ as a square root.

No. While $A:=\left(\begin{array}{cc} \cos x & k\sin x\\ k^{-1}\sin x & -\cos x \end{array}\right)$ is a square root of $I_2$,$$AA^T=\left(\begin{array}{cc} \cos^{2}x+k^{2}\sin^{2}x & \left(k^{-1}-k\right)\sin x\cos x\\ \left(k^{-1}-k\right)\sin x\cos x & k^{-2}\sin^{2}x+\cos^{2}x \end{array}\right).$$ Applying $k\mapsto k^{-1}$ obtains $A^T A$ as something different if $k\ne\pm 1$.

• is there a way to compute these non-normal square roots? Also, does this change if we consider general (not necessarily unitarily) diagonalizable matrices? – glS Jun 18 '18 at 17:19
• @glS I obtained all square roots of $I_2$ (which is diagonalisable by any definition) to see if a counterexample arose. The only real easy to square-root a general matrix is by solving simultaneous equations. – J.G. Jun 18 '18 at 17:22
• This does answer the question as it is stated in the question body. However, for the (current) title of the thread, Are all square roots of diagonalizable matrices diagonalizable? it does not provide an example of that, because while $A$ as constructed above is truly not unitarily diagonalizable for the reason you explain, it is still diagonalizable. The example $N=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ from the other answer is not diagonalizable at all. – Jeppe Stig Nielsen Jun 18 '18 at 21:26

The notion of normality has nothing to do here. Let $A\in M_n(\mathbb{C})$ s.t. $A$ is diagonalizable.

$\textbf{Poposition}$. If $dim(\ker(A))\leq 1$, then each of its square roots is diagonalizable. Otherwise, there is one of its square roots that is not diagonalizable.

$\textbf{Proof}$. Case 1. Suppose that $dim(\ker(A))=1$; we may assume that $A=diag(0,\lambda _1 I_{p_1},\cdots,\lambda_{n-1} I_{p_{n-1}})$ where the $(\lambda_i)$ are non-zero, distinct and $1+p_1+\cdots+p_{n-1}=n$. If $B$ is a square root of $A$, then $AB=BA$ and $B$ is in the form $diag(0,B_1,\cdots,B_{n-1})$, where $B_i^2=\lambda_i I_{p_i}$. Then each $B_i$ is diagonalizable and $B$ too. We solve the case $A$ invertible in the same way.

Case 2. If $dim(\ker(A))=k\geq 2$, then $A_{|\ker(A)}=0$ admits a square root in the form $diag(\begin{pmatrix}0&1\\0&0\end{pmatrix},0_{k-2})$, that is not diagonalizable. Therefore we obtain some $B$ that is not diagonalizable.

Remark. Of course, we can generalize the previous result. Let $f$ be an holomorphic function and consider the equation $f(B)=A$.Then the solutions $B$ are diagonalizable iff, for every $\lambda\in spectrum(A)$, the equation $f(x)=\lambda$ has only simple roots OR has multiple roots but, in this case, the eigenvalue $\lambda$ is simple..

• @glS , $x^2=\lambda_i$ has simple roots. – loup blanc Jun 19 '18 at 20:33
• does that mean that every square root of the identity is diagonalizable? – glS Jun 19 '18 at 20:41
• @gls . Yes it is a symmetry. – loup blanc Jun 19 '18 at 20:44
• sorry but I don't get it. In the first comment I guess you are referring to the fact that diagonalizability is equivalent to the characteristic equation having only simple roots. The roots here are the solution of $\det(B_i-\mu)=0$. How do you apply the fact that $B_i^2=\lambda_i 1$ in here? – glS Jun 19 '18 at 20:47
• @gls . You are wrong " the fact that diagonalizability is equivalent to the characteristic equation having only simple roots" is absolutely false. I show that the $(B_i)$ are diagonalizable (because their minimal polynomial has only simple roots), that implies that $B$ is diagonalizable. – loup blanc Jun 19 '18 at 20:53