Does every complex matrix \(A\) admit a complex matrix square root \(B\) with \[ B^2=A? \]
Solution
No.
Consider the nilpotent Jordan block \[ A=J_2(0)= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}. \] Suppose \(B^2=A\). Every eigenvalue \(\mu\) of \(B\) would satisfy \[ \mu^2\in\operatorname{Spec}(A)=\{0\}, \] so every eigenvalue of \(B\) is \(0\). Thus a \(2\times2\) complex matrix \(B\) would be nilpotent, and hence by Cayley–Hamilton \[ B^2=0. \] This contradicts \(B^2=A\ne0\).
Therefore not every complex matrix has a square root.
By contrast, every invertible complex matrix does have a square root: put it in Jordan form and choose a square root of each nonzero eigenvalue; on a block \(J_r(\lambda)=\lambda(I+N/\lambda)\), the finite binomial series for \((I+N/\lambda)^{1/2}\) gives a square root because \(N\) is nilpotent.