91Ó°ÊÓ

We shall try to draw some conclusions about the matrix A when U=V in the singular value decomposition of A .

i. We can write$A=U\mathrm{Î\pounds }{V}^T=U\mathrm{Î\pounds }{U}^T.$This means that A is similar to$\mathrm{Î\pounds }.$ Therefore, the eigenvalues of A and $\mathrm{Î\pounds }.$are same.$\mathrm{Î\pounds }$ is a diagonal matrix with non-negative diagonal entries. Therefore, its eigenvalues are non-negative. This implies that the eigenvalues of A are non-negative.

ii. $A^T={\left(U\mathrm{Î\pounds }{U}^T\right)}^T=U{\mathrm{Î\pounds }}^T{U}^T=U\mathrm{Î\pounds }{U}^T=A$which implies that A is a symmetric matrix.

That A is a symmetric matrix and that it has all non-negative eigenvalues implies that A is a symmetric positive semi-definite matrix.

On the other hand, let A be a symmetric positive semi-definite matrix.

Since A is symmetric, it is orthogonally diagonalizable. Thus, there exists an orthogonal matrix U such that $A=UD{U}^T$where D is a diagonal matrix. Since A is positive semi-definite, all its eigenvalues are non-negative. That A is symmetric and that its eigenvalues are non-negative together imply that the singular values of A are its eigenvalues. Thus, if we form D with its diagonal entries appearing in descending order, then $UD{U}^T$becomes the singular value decomposition of A

Hence, if A is symmetric positive semi-definite, then we have U=V in the singular value decomposition of A .

Together it implies that A is symmetric positive semi-definite if and only if we have U=V in the singular value decomposition of A.

A is symmetric positive semi-definite if and only if we have U=V in the singular value decomposition of A

A is symmetric positive semi-definite if and only if we have U=V in the singular value decomposition of A .