I'm trying to prove that a real matrix $A$ can be decomposed as $A = U \Sigma V^T$, where $V$ has orthogonal eigenvalues of $A^TA$ along its columns, $\Sigma$ contains the square roots of the corresponding eigenvalues along its diagonal, and $U = AV\Sigma^{-1}$.
I am trying to use the fact that since $A^TA$ is a square symmetric matrix with real values, it can be diagonalized as $$A^TA = V \Sigma^2 V^T.$$ I'm having trouble getting rid of the $A^T$ on the left hand side, in order to get an expression for $A$.
I know that the proof sort of just falls out by how I defined the matrices, since $U\Sigma V^T = (AV\Sigma^{-1}) \Sigma V^T = A$, but I'm trying to see how my above reasoning can be used.