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Q33E

Expert-verified
Linear Algebra and its Applications
Found in: Page 395
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Construct a spectral decomposition of A from Example 2.

The spectral decomposition of A is \(\left[ {\begin{aligned}{{}{}}6&{ - 2}&{ - 1}\\{ - 2}&6&{ - 1}\\{ - 1}&{ - 1}&5\end{aligned}} \right]\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write given values from example 2

The eigenvalues of the matrix \(A = \left[ {\begin{aligned}{{}{}}6&{ - 2}&{ - 1}\\{ - 2}&6&{ - 1}\\{ - 1}&{ - 1}&5\end{aligned}} \right]\) are 8, 6, and 3. The matrix P is defined as:

\(\begin{aligned}{}P &= \left[ {\begin{aligned}{{}{}}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}&{{{\bf{u}}_3}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}{ - \frac{1}{{\sqrt 2 }}}&{ - \frac{1}{{\sqrt 6 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 2 }}}&{ - \frac{1}{{\sqrt 6 }}}&{\frac{1}{{\sqrt 3 }}}\\0&{\frac{2}{{\sqrt 6 }}}&{\frac{1}{{\sqrt 3 }}}\end{aligned}} \right]\end{aligned}\)

Step 2: Find the spectral decomposition of A

The spectral decomposition of A can be calculated as:

\(\begin{aligned}{}A &= {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + {\lambda _2}{{\bf{u}}_2}{\bf{u}}_2^T + {\lambda _3}{{\bf{u}}_3}{\bf{u}}_3^T\\ &= 8{{\bf{u}}_1}{\bf{u}}_1^T + 6{{\bf{u}}_2}{\bf{u}}_2^T + 6{{\bf{u}}_3}{\bf{u}}_3^T\\ &= 8\left[ {\begin{aligned}{{}{}}{ - \frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\\0\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}{ - \frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}&0\end{aligned}} \right] + 6\left[ {\begin{aligned}{{}{}}{ - \frac{1}{{\sqrt 6 }}}\\{ - \frac{1}{{\sqrt 6 }}}\\{\frac{2}{{\sqrt 6 }}}\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}{ - \frac{1}{{\sqrt 6 }}}&{ - \frac{1}{{\sqrt 6 }}}&{\frac{2}{{\sqrt 6 }}}\end{aligned}} \right] + 3\left[ {\begin{aligned}{{}{}}{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\end{aligned}} \right]\end{aligned}\)

Solve further,

\(\begin{aligned}{}A &= 8\left[ {\begin{aligned}{{}{}}{\frac{1}{2}}&{ - \frac{1}{2}}&0\\{ - \frac{1}{2}}&{\frac{1}{2}}&0\\0&0&0\end{aligned}} \right] + 6\left[ {\begin{aligned}{{}{}}{\frac{1}{6}}&{\frac{1}{6}}&{ - \frac{2}{6}}\\{\frac{1}{6}}&{\frac{1}{6}}&{ - \frac{2}{6}}\\{ - \frac{2}{6}}&{ - \frac{2}{6}}&{\frac{4}{6}}\end{aligned}} \right] + 3\left[ {\begin{aligned}{{}{}}{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}4&{ - 4}&0\\{ - 4}&4&0\\0&0&0\end{aligned}} \right] + \left[ {\begin{aligned}{{}{}}1&1&{ - 2}\\1&1&{ - 2}\\{ - 2}&{ - 2}&4\end{aligned}} \right] + \left[ {\begin{aligned}{{}{}}1&1&1\\1&1&1\\1&1&1\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}6&{ - 2}&{ - 1}\\{ - 2}&6&{ - 1}\\{ - 1}&{ - 1}&5\end{aligned}} \right]\end{aligned}\)

Thus, the spectral matrix of A is \(\left[ {\begin{aligned}{{}{}}6&{ - 2}&{ - 1}\\{ - 2}&6&{ - 1}\\{ - 1}&{ - 1}&5\end{aligned}} \right]\).

Most popular questions for Math Textbooks

Orhtogonally diagonalize the matrices in Exercises 37-40. To practice the methods of this section, do not use an eigenvector routine from your matrix program. Instead, use the program to find the eigenvalues, and for each eigenvalue \(\lambda \), find an orthogonal basis for \({\bf{Nul}}\left( {A - \lambda I} \right)\), as in Examples 2 and 3.

37. \(\left( {\begin{aligned}{{}}{\bf{6}}&{\bf{2}}&{\bf{9}}&{ - {\bf{6}}}\\{\bf{2}}&{\bf{6}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{9}}&{ - {\bf{6}}}&{\bf{6}}&{\bf{2}}\\{\bf{6}}&{\bf{9}}&{\bf{2}}&{\bf{6}}\end{aligned}} \right)\)

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.

23. Let \(U = \left( {{u_1}...{u_m}} \right)\) and \(V = \left( {{v_1}...{v_n}} \right)\) where the \({{\bf{u}}_i}\) and \({{\bf{v}}_i}\) are in Theorem 10. Show that \(A = {\sigma _1}{u_1}v_1^T + {\sigma _2}{u_2}v_2^T + ... + {\sigma _r}{u_r}v_r^T\).

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.

22. \(\left( {\begin{aligned}{{}}4&0&1&0\\0&4&0&1\\1&0&4&0\\0&1&0&4\end{aligned}} \right)\)

Question: 13. The sample covariance matrix is a generalization of a formula for the variance of a sample of \(N\) scalar measurements, say \({t_1},................,{t_N}\). If \(m\) is the average of \({t_1},................,{t_N}\), then the sample variance is given by

\(\frac{1}{{N - 1}}\sum\limits_{k = 1}^n {{{\left( {{t_k} - m} \right)}^2}} \)

Show how the sample covariance matrix, \(S\), defined prior to Example 3, may be written in a form similar to (1). (Hint: Use partitioned matrix multiplication to write \(S\) as \(\frac{1}{{N - 1}}\) times the sum of \(N\) matrices of size \(p \times p\). For \(1 \le k \le N\), write \({X_k} - M\) in place of \({\hat X_k}\).)

Suppose A is invertible and orthogonally diagonalizable. Explain why \({A^{ - {\bf{1}}}}\) is also orthogonally diagonalizable.
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