Chapter 7: Q33E (page 395)
Construct a spectral decomposition of A from Example 2.
Short Answer
The spectral decomposition of A is \(\left[ {\begin{aligned}{{}{}}6&{ - 2}&{ - 1}\\{ - 2}&6&{ - 1}\\{ - 1}&{ - 1}&5\end{aligned}} \right]\).
Chapter 7: Q33E (page 395)
Construct a spectral decomposition of A from Example 2.
The spectral decomposition of A is \(\left[ {\begin{aligned}{{}{}}6&{ - 2}&{ - 1}\\{ - 2}&6&{ - 1}\\{ - 1}&{ - 1}&5\end{aligned}} \right]\).
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Get started for freeFind the matrix of the quadratic form. Assume x is in \({\mathbb{R}^{\bf{3}}}\).
a. \(3x_1^2 - 2x_2^2 + 5x_3^2 + 4{x_1}{x_2} - 6{x_1}{x_3}\)
b. \(4x_3^2 - 2{x_1}{x_2} + 4{x_2}{x_3}\)
Make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form \(x_{\bf{1}}^{\bf{2}} + {\bf{10}}{x_{\bf{1}}}{x_{\bf{2}}} + x_{\bf{2}}^{\bf{2}}\) into a quadratic form with no cross-product term. Give P and the new quadratic form.
Question: 12. Exercises 12–14 concern an \(m \times n\) matrix \(A\) with a reduced singular value decomposition, \(A = {U_r}D{V_r}^T\), and the pseudoinverse \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).
Verify the properties of\({A^ + }\):
a. For each\({\rm{y}}\)in\({\mathbb{R}^m}\),\(A{A^ + }{\rm{y}}\)is the orthogonal projection of\({\rm{y}}\)onto\({\rm{Col}}\,A\).
b. For each\({\rm{x}}\)in\({\mathbb{R}^n}\),\({A^ + }A{\rm{x}}\)is the orthogonal projection of\({\rm{x}}\)onto\({\rm{Row}}\,A\).
c. \(A{A^ + }A = A\)and \({A^ + }A{A^ + } = {A^ + }\).
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.
13. \(\left( {\begin{aligned}{{}}3&1\\1&{\,\,3}\end{aligned}} \right)\)
Determine which of the matrices in Exercises 7–12 are orthogonal. If orthogonal, find the inverse.
12. \(P = \left( {\begin{aligned}{{}}{.5}&{.5}&{ - .5}&{ - .5}\\{.5}&{.5}&{.5}&{.5}\\{.5}&{ - .5}&{ - .5}&{.5}\\{.5}&{ - .5}&{.5}&{ - .5}\end{aligned}} \right)\)
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