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Chapter 7: Q9E (page 395)

Determine which of the matrices in Exercises 7鈥�12 are orthogonal. If orthogonal, find the inverse.

9. \(\left[ {\begin{aligned}{{}}{ - 4/5}&{\,\,\,3/5}\\{3/5}&{\,\,4/5}\end{aligned}} \right]\)

Short Answer

Expert verified

\(P\) is an orthogonal matrix and\({P^{ - 1}} = \left[ {\begin{aligned}{{}}{ - 4/5}&{\,\,3/5}\\{3/5}&{\,\,4/5}\end{aligned}} \right]\).

Step by step solution

01

Find the characteristic equation

A matrix\(P\) with, \(n \times n\) dimension, is orthogonal if it satisfies the equation\({P^T}P = {I_n}\).

It is given that\(P = \left( {\begin{aligned}{{}}{ - 4/5}&{\,\,3/5}\\{3/5}&{\,\,4/5}\end{aligned}} \right)\). Find the matrix\({P^T}P\)as shown below:

\(\begin{aligned}{}{P^T}P &= \left( {\begin{aligned}{{}}{ - 4/5}&{\,\,3/5}\\{3/5}&{\,\,4/5}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 4/5}&{\,\,3/5}\\{3/5}&{\,\,4/5}\end{aligned}} \right)\\& = \left( {\begin{aligned}{{}}1&0\\0&1\end{aligned}} \right)\\ &= {I_2}\end{aligned}\)

02

Find the inverse

As\({P^T}P = {I_2}\), so it can be concluded that\(P\)is an orthogonal matrix. Also, the inverse of matrix\(P\)is\({P^T}\).

Find\({P^T}\)as follows:

\(\begin{aligned}{}{P^{ - 1}} &= {P^T}\\ &= \left[ {\begin{aligned}{{}}{ - 4/5}&{\,\,3/5}\\{3/5}&{\,\,4/5}\end{aligned}} \right]\end{aligned}\)

Thus, \(P\) is an orthogonal matrix and\({P^{ - 1}} = \left[ {\begin{aligned}{{}}{ - 4/5}&{\,\,3/5}\\{3/5}&{\,\,4/5}\end{aligned}} \right]\).

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Most popular questions from this chapter

Orthogonally diagonalize the matrices in Exercises 13鈥�22, giving an orthogonal matrix \(P\) and a diagonal matrix \(D\). To save you time, the eigenvalues in Exercises 17鈥�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.

15. \(\left( {\begin{aligned}{{}}{\,3}&4\\4&9\end{aligned}} \right)\)

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. \(Q\left( x \right) = x_1^2 + x_2^2 - 10x_1^{}x_2^{}\).

Question: 14. Exercises 12鈥�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 \(A\).

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

Question: [M] The covariance matrix below was obtained from a Landsat image of the Columbia River in Washington, using data from three spectral bands. Let \({x_1},{x_2},{x_3}\) denote the spectral components of each pixel in the image. Find a new variable of the form \({y_1} = {c_1}{x_1} + {c_2}{x_2} + {c_3}{x_3}\) that has maximum possible variance, subject to the constraint that \(c_1^2 + c_2^2 + c_3^2 = 1\). What percentage of the total variance in the data is explained by \({y_1}\)?

\[S = \left[ {\begin{array}{*{20}{c}}{29.64}&{18.38}&{5.00}\\{18.38}&{20.82}&{14.06}\\{5.00}&{14.06}&{29.21}\end{array}} \right]\]
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