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.

18. Suppose \(A\) is square and invertible. Find a singular value decomposition of \({A^{ - 1}}\)

Short answer

The required singular decomposition of \({A^{ - 1}}\) is \(V\mathop \Sigma \limits^{ - 1} {U^T}\).

Step-by-step solution

Find \(A\)

It is given that\(A\)is square and invertible. As the matrix\(V\)is an orthogonal matrix,\({V^T} = {V^{ - 1}}\), then we have,

\(\begin{array}{c}A = U\Sigma {V^T}\\ = U\Sigma {V^{ - 1}}\end{array}\)

Find the singular value decomposition of the matrix \({A^{ - 1}}\)

\(\begin{array}{c}{A^{ - 1}} = {\left( {U\Sigma {V^{ - 1}}} \right)^{ - 1}}\\ = V\mathop \Sigma \limits^{ - 1} {U^{ - 1}}\\ = V\mathop \Sigma \limits^{ - 1} {U^T}\end{array}\)

Therefore, \({A^{ - 1}} = V\mathop \Sigma \limits^{ - 1} {U^T}\).