According to Wikpedia a positive semidefinite matrix has a unique positive semidefinite square root. If the first PSD matrix is $\sum_i \lambda_iv_iv_i^T$, the PSD square root must be $\sum_i \sqrt{\lambda_i}v_iv_i^T$.
If we drop the requirement that the root is positive semidefinite, other square roots can be obtained by changing the signs of the eigenvalues, like $\sum_i -\sqrt{\lambda_i}v_iv_i^T$. Can all the symmetric (not necessarily PSD) square roots of a PSD matrix be obtained in this way?
Best Answer
In a sense. Suppose $P \in M(n, \mathbb{R})$ is positive semifdefinite. Suppose $Q \in M(n, \mathbb{R})$ is a symmetric square root of $P$, meaning $Q^* = Q$, $Q^2 = P$. We have $QP = QQ^2 = Q^2Q = PQ$, so $Q$ commutes with $P$. Thus the eigenspaces of $P$ are invariant under $Q$, so since self adjoint operators are orthonormally diagonalizable, there is an orthonormal basis $\{v_1, \dots, v_n\}$ of $\mathbb{R}^n$ such that each $v_j$ is both an eigenvector of $P$ and of $Q$. Say $Qv_j = \mu_jv_j$, $Pv_j = \lambda_jv_j$. Then $\mu_j^2v_j = \lambda_jv_j$, so $\mu_j = \pm\lambda_j$. If you want all the eigenvectors of $P$ to be eigenvectors of $Q$, you have to ensure that if $\lambda_j = \lambda_k$, then $\mu_j = \mu_k$. If you ensure that, then (with appropriate relabeling) $$Q = \sum_{j = 1}^{K}\mu_jP_j,$$ where $\{\lambda_1, \dots, \lambda_K\}$ is the set of eigenvalues of $P$, and $P_j$ is the orthogonal projection onto the $\lambda_j$-eigenspace of $P$.