Chapter 7: Q29E (page 395)
Suppose A is invertible and orthogonally diagonalizable. Explain why \({A^{ - {\bf{1}}}}\) is also orthogonally diagonalizable.
The matrix \({A^{ - 1}}\) is diagonizable.
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Question 11: Prove that any \(n \times n\) matrix A admits a polar decomposition of the form \(A = PQ\), where P is a \(n \times n\) positive semidefinite matrix with the same rank as A and where Q is an \(n \times n\) orthogonal matrix. (Hint: Use a singular value decomposition, \(A = U\sum {V^T}\), and observe that \(A = \left( {U\sum {U^T}} \right)\left( {U{V^T}} \right)\).) This decomposition is used, for instance, in mechanical engineering to model the deformation of a material. The matrix P describe the stretching or compression of the material (in the directions of the eigenvectors of P), and Q describes the rotation of the material in space.
Orthogonally diagonalize the matrices in Exercises 13–22, giving an orthogonal matrix\(P\)and a diagonal matrix\(D\). To save you time, the eigenvalues in Exercises 17–22 are: (17)\( - {\bf{4}}\), 4, 7; (18)\( - {\bf{3}}\),\( - {\bf{6}}\), 9; (19)\( - {\bf{2}}\), 7; (20)\( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.
14. \(\left( {\begin{aligned}{{}}{\,1}&{ - 5}\\{ - 5}&{\,\,1}\end{aligned}} \right)\)
Determine which of the matrices in Exercises 1–6 are symmetric.
2. \(\left( {\begin{aligned}{{}}3&{\,\, - 5}\\{ - 5}&{ - 3}\end{aligned}} \right)\)
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 \(A{A^T}\).)
Compute the quadratic form \({{\bf{x}}^T}A{\bf{x}}\), when \(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\) and
a. \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)
b. \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)
c. \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)
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