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Chapter 7: Q33E (page 395)

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: 11. Given multivariate data \({X_1},................,{X_N}\) (in \({\mathbb{R}^p}\)) in mean deviation form, let \(P\) be a \(p \times p\) matrix, and define \({Y_k} = {P^T}{X_k}{\rm{ for }}k = 1,......,N\).

  1. Show that \({Y_1},................,{Y_N}\) are in mean-deviation form. (Hint: Let \(w\) be the vector in \({\mathbb{R}^N}\) with a 1 in each entry. Then \(\left( {{X_1},................,{X_N}} \right)w = 0\) (the zero vector in \({\mathbb{R}^p}\)).)
  2. Show that if the covariance matrix of \({X_1},................,{X_N}\) is \(S\), then the covariance matrix of \({Y_1},................,{Y_N}\) is \({P^T}SP\).

(M) 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.

39. \(\left( {\begin{aligned}{{}}{.{\bf{31}}}&{.{\bf{58}}}&{.{\bf{08}}}&{.{\bf{44}}}\\{.{\bf{58}}}&{ - .{\bf{56}}}&{.{\bf{44}}}&{ - .{\bf{58}}}\\{.{\bf{08}}}&{.{\bf{44}}}&{.{\bf{19}}}&{ - .{\bf{08}}}\\{ - .{\bf{44}}}&{ - .{\bf{58}}}&{ - .{\bf{08}}}&{.{\bf{31}}}\end{aligned}} \right)\)

Determine which of the matrices in Exercises 1鈥�6 are symmetric.

\(\left( {\begin{array}{{}{}}1&2&2&1\\2&2&2&1\\2&2&1&2\end{array}} \right)\)

Compute the quadratic form \({x^T}Ax\), when \(A = \left( {\begin{aligned}{{}}3&2&0\\2&2&1\\0&1&0\end{aligned}} \right)\) and

a. \(x = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\)

b. \(x = \left( {\begin{aligned}{{}}{ - 2}\\{ - 1}\\5\end{aligned}} \right)\)

c. \(x = \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right)\)

Classify the quadratic forms in Exercises 9鈥�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. \({\bf{2}}x_{\bf{1}}^{\bf{2}} - {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}} - x_{\bf{2}}^{\bf{2}}\)
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