Question

In Exercises 17–24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) , where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) “diagonal” matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

22. Show that if \(A\) is an \(n \times n\) positive definite matrix, then an orthogonal diagonalization \(A = PD{P^T}\) is a singular value decomposition of \(A\).

Short answer

The given statement is verified.

Step-by-step solution

Show that vectors orthogonal to \({{\bf{v}}_{\bf{1}}}\)

Since the matrix\(P\)is a square and orthogonal matrix then we have,

\(\begin{array}{c}P{P^T} = I\\{P^T} = {P^{ - 1}}\\{({P^T})^{ - 1}} = {({P^{ - 1}})^{ - 1}}\\ = P\end{array}\)

Show that \(A = PD{P^T}\) is a singular value decomposition of \(A\)

Simplify\(P{P^T}\).

\(\begin{array}{c}P{P^T} = I\\{P^T} = {P^{ - 1}}\\{({P^T})^T} = {({P^{ - 1}})^T}\\ = {\left( {{P^T}} \right)^{ - 1}}\\ = P\end{array}\)

Therefore,\({P^T}\)is an orthogonal matrix, and the diagonal matrix becomes the\(\sum \)matrix.

Thus, the factorization \(A = PD{P^{ - 1}}\) satisfies the properties which make it singular value decomposition.