Chapter 7: Q30E (page 395)
Suppose Aand B are orthogonally diagonalizable and \(AB = BA\). Explain why \(AB\) is also orthogonally diagonalizable.
Short Answer
The matrix AB is orthogonally diagonizable.
Chapter 7: Q30E (page 395)
Suppose Aand B are orthogonally diagonalizable and \(AB = BA\). Explain why \(AB\) is also orthogonally diagonalizable.
The matrix AB is orthogonally diagonizable.
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Get started for freeClassify 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.
15. \( - {\bf{3}}x_{\bf{1}}^{\bf{2}} - {\bf{7}}x_{\bf{2}}^{\bf{2}} - {\bf{10}}x_{\bf{3}}^{\bf{2}} - {\bf{10}}x_{\bf{4}}^{\bf{2}} + {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{4}}{x_{\bf{1}}}{x_{\bf{3}}} + {x_{\bf{1}}}{x_{\bf{4}}} + {\bf{6}}{x_{\bf{3}}}{x_{\bf{4}}}\)
Question: 14. 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 \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).
Given any \({\rm{b}}\) in \({\mathbb{R}^m}\), adapt Exercise 13 to show that \({A^ + }{\rm{b}}\) is the least-squares solution of minimum length. [Hint: Consider the equation \(A{\rm{x}} = {\rm{b}}\), where \(\mathop {\rm{b}}\limits^\^ \) is the orthogonal projection of \({\rm{b}}\) onto Col \(A\).
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 \({Y_k} = {P^T}{X_k}{\rm{ for }}k = 1,......,N\).
Compute the quadratic form \({{\bf{x}}^T}A{\bf{x}}\), when \(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\) and
a. \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)
b. \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)
c. \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)
In Exercises 3-6, find (a) the maximum value of \(Q\left( {\rm{x}} \right)\) subject to the constraint \({{\rm{x}}^T}{\rm{x}} = 1\), (b) a unit vector \({\rm{u}}\) where this maximum is attained, and (c) the maximum of \(Q\left( {\rm{x}} \right)\) subject to the constraints \({{\rm{x}}^T}{\rm{x}} = 1{\rm{ and }}{{\rm{x}}^T}{\rm{u}} = 0\).
5. \(Q\left( x \right) = x_1^2 + x_2^2 - 10x_1^{}x_2^{}\).
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