Construct a spectral decomposition of A from Example 2.

Short Answer

Expert verified

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

Step by step solution

01

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:

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

02

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 \(\left[ {\begin{aligned}{{}{}}6&{ - 2}&{ - 1}\\{ - 2}&6&{ - 1}\\{ - 1}&{ - 1}&5\end{aligned}} \right]\).

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Most popular questions from this chapter

Question: In Exercises 15 and 16, construct the pseudo-inverse of \(A\). Begin by using a matrix program to produce the SVD of \(A\), or, if that is not available, begin with an orthogonal diagonalization of \({A^T}A\). Use the pseudo-inverse to solve \(A{\rm{x}} = {\rm{b}}\), for \({\rm{b}} = \left( {6, - 1, - 4,6} \right)\) and let \(\mathop {\rm{x}}\limits^\^ \)be the solution. Make a calculation to verify that \(\mathop {\rm{x}}\limits^\^ \) is in Row \(A\). Find a nonzero vector \({\rm{u}}\) in Nul\(A\), and verify that \(\left\| {\mathop {\rm{x}}\limits^\^ } \right\| < \left\| {\mathop {\rm{x}}\limits^\^ + {\rm{u}}} \right\|\), which must be true by Exercise 13(c).

15. \(A = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 3}&{ - 6}&6&{\,\,1}\\{ - 1}&{ - 1}&{ - 1}&1&{ - 2}\\{\,\,\,0}&{\,\,0}&{ - 1}&1&{ - 1}\\{\,\,\,0}&{\,\,0}&{ - 1}&1&{ - 1}\end{array}} \right]\)

Let A be the matrix of the quadratic form

\({\bf{9}}x_{\bf{1}}^{\bf{2}} + {\bf{7}}x_{\bf{2}}^{\bf{2}} + {\bf{11}}x_{\bf{3}}^{\bf{2}} - {\bf{8}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{8}}{x_{\bf{1}}}{x_{\bf{3}}}\)

It can be shown that the eigenvalues of A are 3,9, and 15. Find an orthogonal matrix P such that the change of variable \({\bf{x}} = P{\bf{y}}\) transforms \({{\bf{x}}^T}A{\bf{x}}\) into a quadratic form which no cross-product term. Give P and the new quadratic form.

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.

22. Show that if \(A\) is an \(n \times n\) positive definite matrix, then an orthogonal diagonalization \(A = PD{P^T}\) is a singular value decomposition of \(A\).

Question: In Exercises 15 and 16, construct the pseudo-inverse of \(A\). Begin by using a matrix program to produce the SVD of \(A\), or, if that is not available, begin with an orthogonal diagonalization of \({A^T}A\). Use the pseudo-inverse to solve \(A{\rm{x}} = {\rm{b}}\), for \({\rm{b}} = \left( {6, - 1, - 4,6} \right)\) and let \(\mathop {\rm{x}}\limits^\^ \)be the solution. Make a calculation to verify that \(\mathop {\rm{x}}\limits^\^ \) is in Row \(A\). Find a nonzero vector \({\rm{u}}\) in Nul\(A\), and verify that \(\left\| {\mathop {\rm{x}}\limits^\^ } \right\| < \left\| {\mathop {\rm{x}}\limits^\^ + {\rm{u}}} \right\|\), which must be true by Exercise 13(c).

16. \(A = \left( {\begin{array}{*{20}{c}}4&0&{ - 1}&{ - 2}&0\\{ - 5}&0&3&5&0\\{\,\,\,2}&{\,\,0}&{ - 1}&{ - 2}&0\\{\,\,\,6}&{\,\,0}&{ - 3}&{ - 6}&0\end{array}} \right)\)

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.

24. Using the notation of Exercise 23, show that \({A^T}{u_j} = {\sigma _j}{v_j}\) for \({\bf{1}} \le {\bf{j}} \le {\bf{r}} = {\bf{rank}}\;{\bf{A}}\)

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