Suppose I have a SPD matrix, $A$, and a symmetric square matrix of appropriate size, $B$. I wish to show that $AB$ is diagonalizable, has all real eigenvalues (which are the same as $BA$), and that the eigenvectors of $AB$ are orthogonal.
My approach: It seems if I can prove that $AB = BA$ then I am done and I have proven everything at once. Is this correct? Assuming I am correct, this is where I am having trouble. My thought is since $A$ is SPD I can use its Cholesky factorization and try to get somewhere with that.
Since $A$ is a real symmetric matrix, there exists an orthogonal matrix $U$ and a real diagonal matrix $\Lambda$ such that $A = U\Lambda U^T$. Since $A$ is symmetric positive definite the diagonal values of $\Lambda$ are strictly positive. It follows that $A$ has a nonsingular symmetric square root $A^{\frac{1}{2}}$ given by $$ A^{\frac{1}{2}} = U \Lambda^{\frac{1}{2}} U^T. $$ The square root of a diagonal matrix is obtained by taking the square root of the individual diagonal entries.
The matrix $C = AB$ is similar to the matrix $D = A^{-\frac{1}{2}} C A^{\frac{1}{2}} = A^{\frac{1}{2}} B A^{\frac{1}{2}} $. It is clear that $D$ is symmetric. Hence there exists an orthogonal matrix $V$ and a real diagonal matrix $E$ such that $D = V E V^T$. It follows that $$ A^{\frac{1}{2}} B A^{\frac{1}{2}} = V E V^T$$ We conclude that $$ AB = A^{\frac{1}{2}} (V E V^T) A^{-\frac{1}{2}} = (A^{\frac{1}{2}} V) E (A^{\frac{1}{2}} V)^{-1}.$$ This exhibits the eigenvalues and the eigenvectors of $AB$. The eigenvalues of $AB$ are real, but the eigenvectors of $AB$ are not necessarily orthogonal.