Prove that a real symmetric matrix has a real symmetric cube root; i.e., if \(A\) is real symmetric, there is a real symmetric \(B\) such that \(B^3 = A\).
Solution
Let \(A\) be a real symmetric \(n\times n\) matrix. By the real spectral theorem, there is a real orthogonal matrix \(P\) such that \[ P^tAP=D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n), \] where every \(\lambda_i\in\mathbb R\).
For each \(i\), let \[ \mu_i=\sqrt[3]{\lambda_i} \] be the real cube root, and put \[ C=\operatorname{diag}(\mu_1,\ldots,\mu_n). \] Then \(C\) is real symmetric and \[ C^3=D. \] Define \[ B=PCP^t. \] Since \(P\) and \(C\) are real, \(B\) is real. Also \[ B^t=(PCP^t)^t=PC^tP^t=PCP^t=B, \] so \(B\) is symmetric. Finally, using \(P^tP=I\), \[ B^3 =PCP^tPCP^tPCP^t =PC^3P^t =PDP^t =A. \] Thus every real symmetric matrix has a real symmetric cube root.