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.

20. Show that if\(A\)is an orthogonal\(m \times m\)matrix, then \(PA\) has the same singular values as \(A\).

Short answer

It is verified that \(P\) is an orthogonal square matrix and, \(A\) and \(PA\) have the same singular values.

Step-by-step solution

Show that \(PU\) is orthogonal

Consider the equation as:

\(\begin{array}{c}PA = P\left( {U\Sigma {V^T}} \right)\\ = \left( {PU} \right)\Sigma {V^T}\end{array}\)

Since \(P\) and \(U\) are orthogonal, \(PU\) is also orthogonal.

Show that the matrices \(A\) and \(PA\) have the same singular values

Since \(PU\)and\(V\)are orthogonal and\(\Sigma \) is a diagonal matrix, so that \(PA = \left( {PU} \right)\Sigma {V^T}\) is the singular value decomposition of\(PA\).

Therefore, the diagonal entries of\(\Sigma \)are the singular values of\(PA\). Hence,

The matrices \(A\) and \(PA\) have the same singular values.