
Re: why the sigma of symmetrical svd for a real symmetric matrix is negative?
Posted:
Dec 27, 2012 2:22 AM


"Rick" wrote in message <kbgbi5$g5n$1@newscl01ah.mathworks.com>... > Hi everyone, > as we know, for a real symmetric matrix A, A, there exists a singular value decomposition as A=USU', and S should be a rectangular diagonal matrix with nonnegative real numbers on the diagonal.
It is a wrong statement, as Roger has pointed it out.
Take an extreme case where A is 1x1. Any matrix is symmetric, the svd is: U = V = 1, and S = A(1,1). There is no reason for A(1,1) to be positive.
Bruno

