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Q7.4-19E

Expert-verifiedFound in: Page 395

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**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 \(A{A^T}\).)**

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 \(A\)**.**

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 \({A^T}A\).

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 \(A{A^T}\)**.**

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