Prove that every positive matrix is the square of a positive matrix.
Solution
1. Let \(A\) be a positive (positive-definite Hermitian) matrix.
hypothesis.
2. \(A\) is Hermitian, so it is unitarily diagonalizable: \(A = U D U^*\) with \(U\) unitary and \(D = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)\).
spectral theorem for Hermitian matrices.
3. Since \(A\) is positive, each \(\lambda_i > 0\).
the eigenvalues of a positive matrix are positive.
4. Define \(B = U \operatorname{diag}(\sqrt{\lambda_1}, \ldots, \sqrt{\lambda_n}) U^*\).
definition.
5. \(B\) is positive (its eigenvalues \(\sqrt{\lambda_i} > 0\)).
3 and 4.
6. \(B^2 = U \operatorname{diag}(\lambda_1, \ldots, \lambda_n) U^* = A\).
4 (squaring the diagonal matrix squares the entries).
7. Hence \(A\) is the square of the positive matrix \(B\).
5 and 6.
8. Q.E.D.
7.