Relationship between eigendecomposition and singular value decomposition

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?

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If $A = U \Sigma V^T$ and $A$ is symmetric, then $V$ is almost $U$ except for the signs of columns of $V$ and $U$.

$$A = W \Lambda W^T = \displaystyle \sum_{i=1}^n w_i \lambda_i w_i^T = \sum_{i=1}^n w_i \left| \lambda_i \right| \text{sign}(\lambda_i) w_i^T$$ where $w_i$ are the columns of the matrix $W$.

The left singular vectors $u_i$ are $w_i$ and the right singular vectors $v_i$ are $\text{sign}(\lambda_i) w_i$. (You can of course put the sign term with the left singular vectors as well.)The singular values $\sigma_i$ are the magnitude of the eigen values $\lambda_i$.

Hence, $A = U \Sigma V^T = W \Lambda W^T$

and $$A^2 = U \Sigma^2 U^T = V \Sigma^2 V^T = W \Lambda^2 W^T$$

Note that the eigenvalues of $A^2$ are positive.

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Perfect -- thanks a lot! – user8478 Mar 20 '11 at 2:17