Are all square roots of diagonalizable matrices diagonalizable?

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 $k\ne\pm 1$.

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Matrices

Eigenvalues Eigenvectors

Inverse Function

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