Symmetric complex matrices

Question

Prove that if A is real symmetric matrix, there is a real symmetric B such that $B^3$=A. I've tried using explicit assumptions, but can not prove.

Answer 1

Since $A$ is real symmetric, it is diagonalizable with orthogonal matrices $O$ in the form $OJO^T$, where $J$ is diagonal. Since $OO^T=1$ by definition of orthogonal, it suffices to write $B=OJ^{1/3}O^T$, where $J^{1/3}$ is the matrix given by taking the cube root of each diagonal entry (eigenvalue) of $J$. Notice that an odd power like $3$ is necessary, as some eigenvalues might be negative. This ensures $B$ is real.