Note, the post here answered the question that SVD's exists for all matrices. The post did not answer what such one might be for this particular question.
We know that $A$ (in our case) will have positive and negative eigenvalues. In addition, we know that all symmetric matrices can be decomposed by the eigenvector decomposition theorem that is
$$A=Q\Lambda Q^T$$
Where $\Lambda$ is the diagonal matrix matrix containing the eigenvalues for $A$ and $Q$ contains the linearly independent eigenvectors of $A$. Now using this equation, I am unsure how to manipulate it so that
$$A = (\text{"orthogonal matrix"})\Sigma(\text{"orthogonal matrix"}).$$
Note that the "orthogonal matrix" are of compatible size. I was given a hint to find an equation that satifsifes
$$A = (QD^{-1})(D\Lambda)Q^T$$
Where $\Sigma = D\Lambda$? I am not sure though. Any advice is much helpful!
Best Answer
Choose $D$ to be a diagonal matrix whose diagonal elements are either $1$ or $-1$, so that $D\Lambda$ becomes diagonal with nonnegative diagonal entries. $D^{-1}$ will similarly negate the columns of $Q$ corresponding to the negative eigenvalues. Check that $QD^{-1}$ is still orthogonal.