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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

Classify the quadratic forms in Exercises 9–18. Then make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct \(P\) using the methods of Section 7.1.

11. \({\bf{2}}x_{\bf{1}}^{\bf{2}} - {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}} - x_{\bf{2}}^{\bf{2}}\)

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 \({\bf{1}} \le {\bf{j}} \le {\bf{r}} = {\bf{rank}}\;{\bf{A}}\)

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: Let \({\bf{X}}\) denote a vector that varies over the columns of a \(p \times N\) matrix of observations, and let \(P\) be a \(p \times p\) orthogonal matrix. Show that the change of variable \({\bf{X}} = P{\bf{Y}}\) does not change the total variance of the data. (Hint: By Exercise 11, it suffices to show that \(tr\left( {{P^T}SP} \right) = tr\left( S \right)\). Use a property of the trace mentioned in Exercise 25 in Section 5.4.)

Question: [M] A Landsat image with three spectral components was made of Homestead Air Force Base in Florida (after the base was hit by Hurricane Andrew in 1992). The covariance matrix of the data is shown below. Find the first principal component of the data, and compute the percentage of the total variance that is contained in this component.

\[S = \left[ {\begin{array}{*{20}{c}}{164.12}&{32.73}&{81.04}\\{32.73}&{539.44}&{249.13}\\{81.04}&{246.13}&{189.11}\end{array}} \right]\]

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