matrix square root

A square root of a matrix A is any B with B^2 = A. When does one exist, how many are there, and how do you build one? Jordan form plus functional calculus answers all three by reducing the question to one Jordan block at a time, where sqrt becomes a Taylor-series computation.

On a single Jordan block J_k(lambda) with lambda != 0, the function f(x) = sqrt(x) is analytic, so f(J) is computed by the function-of-a-block formula: sqrt(lambda) on the diagonal, and the derivatives of sqrt at lambda climbing the superdiagonals. Choosing a branch of sqrt(lambda) gives a square root of that block; doing this block by block (and conjugating back) gives a square root of A.

Existence and uniqueness hinge on the eigenvalue 0. If A is invertible (no zero eigenvalue), a square root always exists, and there is a distinguished 'principal' one (continuous branch with eigenvalues in the right half-plane). The number of square roots grows with the choices of branch per eigenvalue; for a diagonalizable invertible A with distinct eigenvalues there are 2^(number of eigenvalues) of them.

The subtle case is a singular A. A single nilpotent Jordan block of size >= 2 has NO square root, because sqrt is not analytic at 0 and the chain structure obstructs it (e.g. the 2x2 nilpotent [0,1;0,0] is not any B^2). So whether a singular matrix has a square root depends delicately on its zero-eigenvalue Jordan blocks — exactly the kind of question Jordan form is built to settle.