Give a reasonable definition of a non-negative \(n \times n\) matrix, and then prove that such a matrix has a unique non-negative square root.
Solution
Call a complex matrix \(A\) non-negative if \[ A=A^* \qquad\text{and}\qquad X^*AX\ge0 \] for every column vector \(X\). Equivalently, \(A\) is self-adjoint and all of its characteristic values are non-negative.
1. Every non-negative matrix has a non-negative square root.
By the spectral theorem there is a unitary matrix \(U\) and non-negative real numbers \(\lambda_1,\ldots,\lambda_n\) such that \[ A=U\operatorname{diag}(\lambda_1,\ldots,\lambda_n)U^*. \] Define \[ B =U\operatorname{diag}(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n})U^*. \] Then \(B=B^*\) and its characteristic values are the non-negative numbers \(\sqrt{\lambda_i}\), so \(B\) is non-negative. Moreover \[ B^2 =U\operatorname{diag}(\lambda_1,\ldots,\lambda_n)U^* =A. \] Thus a non-negative square root exists.
2. Any non-negative square root of \(A\) commutes with \(A\) and preserves every eigenspace of \(A\).
Suppose \(C\) is non-negative and \(C^2=A\). Then \[ CA=C^3=AC. \] If \(Av=\lambda v\), then \[ A(Cv)=C(Av)=\lambda Cv, \] so each eigenspace \[ E_\lambda=\ker(A-\lambda I) \] is invariant under \(C\).
3. The restriction of a non-negative square root to \(E_\lambda\) is \(\sqrt\lambda I\).
On \(E_\lambda\) one has \[ C^2=\lambda I. \] The restriction \(C|_{E_\lambda}\) is self-adjoint and non-negative, hence is unitarily diagonalizable with non-negative characteristic values \(\mu\). Every such \(\mu\) satisfies \[ \mu^2=\lambda. \] Because \(\mu\ge0\), necessarily \[ \mu=\sqrt\lambda. \] Therefore all characteristic values of \(C|_{E_\lambda}\) are equal to \(\sqrt\lambda\), and since the restriction is diagonalizable, \[ C|_{E_\lambda}=\sqrt\lambda I. \]
4. The non-negative square root is unique.
1.Therefore\[ \boxed{A\text{ has a unique non-negative square root}.} \]