If a matrix $A$ satifies $A^TA=AA^T$, then its eigenvectors are
orthogonal.
I have not had a proof for the above statement yet. By the way, by the Singular Value Decomposition, $A=U\Sigma V^T$, and because $A^TA=AA^T$, then $U=V$ (following the constructions of $U$ and $V$). So $A=U\Sigma U^T$, thus $A$ is symmetric since $\Sigma$ is diagonal. In fact, the skew-symmetric or diagonal matrices also satisfy the condition $AA^T=A^TA$. So at which point do I misunderstand the SVD? And please also give me the proof of the statement.
Thank you very much.
Best Answer
Note that $\DeclareMathOperator{\im}{im}$ $$ \ker(A) = \ker(A^TA) = \ker(AA^T) = \ker(A^T) = \im(A)^\perp $$ Similarly, we have $\ker(A - \lambda I) = \im(A - \lambda I)^\perp$.
Suppose that $\lambda$ is an eigenvalue. Then any corresponding eigenvector lies in $\ker(A - \lambda I)$. Any eigenvector corresponding to a value other than $\lambda$ lies in $\im(A - \lambda I)$.
Thus, if two eigenvectors correspond to different eigenvalues, then they are orthogonal.