Let $A \in \mathbb{R}^{n\times n}$ be a real symmetric matrix. Please help me clear up some confusion about the relationship between the singular value decomposition of $A$ and the eigen-decomposition of $A$.

Let $A = U\Sigma V^T$ be the SVD of $A$. Since $A = A^T$, we have $AA^T = A^TA = A^2$ and: $$A^2 = AA^T = U\Sigma V^T V \Sigma U^T = U\Sigma^2 U^T$$ $$A^2 = A^TA = V\Sigma U^T U\Sigma V^T = V\Sigma^2 V^T$$

Both of these are eigen-decompositions of $A^2$. Now consider some eigen-decomposition of $A$

$$A = W\Lambda W^T$$

Then

$$A^2 = W\Lambda W^T W\Lambda W^T = W\Lambda^2 W^T$$

So $W$ also can be used to perform an eigen-decomposition of $A^2$.

So now my confusion: It seems that $A = W\Lambda W^T$ is also a singular value decomposition of A. But singular values are always non-negative, and eigenvalues can be negative, so something must be wrong.

What is going on?