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.

19. Show that the columns of\(V\)are eigenvectors of\({A^T}A\), the columns of\(U\)are eigenvectors of\(A{A^T}\), and the diagonal entries of\({\bf{\Sigma }}\)are the singular values of \(A\). (Hint: Use the SVD to compute \({A^T}A\) and \(A{A^T}\).)

Short answer

It is verified that, \(V\) and \(U\) are eigenvectors of \({A^T}A\) and \(A{A^T}\) respectively, and the columns of \(U\) re eigenvectors of \(A{A^T}\) and the diagonal elements of \(\Sigma \) are the singular values of \(A\).

Step-by-step solution

Find the product of \({A^T}\) and \(A\)

As singular value decomposition of\(A\)is \(A = U\Sigma {V^T}\), then

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

Therefore, from the Diagonalization theorem. Singular values are the diagonal entries in \(\Sigma \) matrix. The columns of matrix \(V\) are the eigenvectors of \({A^T}A\).

Find the product of \(A\) and \({A^T}\)

Now, find the product of \(A = U\Sigma {V^T}\) and its transpose.

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

Thus, it can be observed that\(U\)diagonalizes\(A{A^T}\)and the columns of\(U\)must be eigenvectors of\(A{A^T}\).