Get eigenvector and eigenvalues from symmetrical Matrix

I used a Lapack library to compute Singular Value Decomposition of symmetrical Matrix in c# language like:

$\left( \begin{array}{ccc} 1 & 2 & 5 \\ 2 & 1 & 7 \\ 5 & 7 & 1 \end{array} \right) = \left( \begin{array}{ccc} -0.467 & -0.328 & -0.821 \\ -0.582 & -0.586 & 0.564 \\ -0.666 & 0.741 & 0.083 \end{array} \right) \left( \begin{array}{ccc} 10.62 & 0 & 0 \\ 0 & 6.74 & 0 \\ 0 & 0 & 0.88 \end{array} \right) \left( \begin{array}{ccc} -0.467 & -0.582 & -0.666 \\ 0.328 & 0.586 & -0.741 \\ 0.821 & -0.564 & -0.083 \end{array} \right)$

but How do I get eigenvectors and eigenvalues from $USV^T$, I was reading that $S$ are the eigenvalues but like the sign must be changed... also the $U$ gives the eigenvectors but there is an issue with the sign, is there a rule to get the correct sign?

The singular values of a matrix $A$ (not necessarily square) are the square roots of the eigenvalues of $A^TA$. Squaring $S$ will NOT give the eigenvalues of $A$, but of $A^TA$. Additionally, the singular vectors (the columns of $U$ and $V$) are the eigenvectors of $A^TA$ and $AA^T$. A priori, there is no direct connection between the eigenvalues of $A$ and the singular values of $A$, or between eigenvectors and singular vectors.

Of course, there are some relations. Given a diagonal matrix, it scales the length of a vector by at most the largest entry in the matrix. Because orthogonal matrices preserve length, the largest singular value is the largest amount by which the length of a vector can be scaled. If $A$ is symmetric, the same description holds of the largest eigenvalue. However, in general, if $\lambda$ is the largest eigenvalue of $A$, then in general $|\lambda|$ will be smaller than the largest singular value. For example, with the matrix

$$\pmatrix{1 & 2\\ 3 & 4}$$

the eigenvalues are $5.372$ and $-.372$, while the singular values are $5.465$ and $.366$.

If $A$ is symmetric, the eigen-decomposition is an SVD decomposition. Unfortunately, I know of no good way to go from one decomposition to another in general.

There is a nice relationship for normal matrices, that is matrices which satisfy $AA*=A*A$. (or in the case of real matrices, $AA^T=A^TA$). A matrix is normal if and only if it can be diagonalized by a unitary matrix, that is $A=USU^*$ where $UU^*=I$ and $S$ is diagonal. In this case, $AA^*=(USU^*)(US^*U^*)=U(SS^*)U^*$. In this case, $SS^*$ will be diagonal and will contain the squares of the lengths of the eigenvalues of $A$, and therefore the singular values of $A$ are in fact the absolute values of the eigenvalues of $A$.

Unfortunately, this makes it easy to go from an eigenvalue decomposition of $A$ to an SVD, but not the other way around.

• If I understood right, If I have a summetric matrix, I can use SVD to get eigenvalues and eigenvectors right?.... So, the only detail is the sign of $U$ to get eigenvectors and the diagonal of $S$ are eigenvalues right? Jan 23 '12 at 16:44
• @cMinor For a symmetric matrix, the SVD decomposition is the eigen-decomposition. No work is required to go from one to the other. You will have that $U=V$ to begin with (or rather, you can, as technically, one could change the signs of any of the column's of $U$ to produce a $V$ that works for SDV. So really, you can just ignore $V$ in entirety). I don't know what you mean by the sign of $U$, but yes, the diagonal entries of $S$ will be the eigenvalues. However, when you're normal, they are only the lengths of the eigenvalues, and when you're not normal, there is no clear relationship. Jan 23 '12 at 17:12