Square root of a matrix

Remarks

Example: for every real $\theta$, $B = \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix}$ is a square root of $I_2$. Check: the entries of $B^2$ are $\cos^2\theta + \sin^2\theta = 1$ on the diagonal and $\cos\theta\sin\theta - \sin\theta\cos\theta = 0$ off it. Non-example: $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has no square root. If $B^2 = N$, then $B^4 = 0$, so $B$ is nilpotent; a nilpotent $2 \times 2$ matrix satisfies $B^2 = 0 \ne N$. A positive semidefinite matrix has a unique positive semidefinite square root, written $A^{1/2}$; see square root of a positive semidefinite matrix. A complex matrix with no eigenvalue in $(-\infty, 0]$ has a unique square root whose eigenvalues have positive real part, the principal square root.