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Chapter 7: Q7.4-20E (page 395)

In Exercises 17鈥�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 \(m \times n\) 鈥渄iagonal鈥� matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

20. Show that if\(A\)is an orthogonal\(m \times m\)matrix, then \(PA\) has the same singular values as \(A\).

Short Answer

Expert verified

It is verified that \(P\) is an orthogonal square matrix and, \(A\) and \(PA\) have the same singular values.

Step by step solution

01

Show that \(PU\) is orthogonal

Consider the equation as:

\(\begin{array}{c}PA = P\left( {U\Sigma {V^T}} \right)\\ = \left( {PU} \right)\Sigma {V^T}\end{array}\)

Since \(P\) and \(U\) are orthogonal, \(PU\) is also orthogonal.

02

Show that the matrices \(A\) and \(PA\) have the same singular values

Since \(PU\)and\(V\)are orthogonal and\(\Sigma \) is a diagonal matrix, so that \(PA = \left( {PU} \right)\Sigma {V^T}\) is the singular value decomposition of\(PA\).

Therefore, the diagonal entries of\(\Sigma \)are the singular values of\(PA\). Hence,

The matrices \(A\) and \(PA\) have the same singular values.

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Most popular questions from this chapter

Find the matrix of the quadratic form. Assume x is in \({\mathbb{R}^2}\) .

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In Exercises 17鈥�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 \(m \times n\) 鈥渄iagonal鈥� matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

23. Let \(U = \left( {{u_1}...{u_m}} \right)\) and \(V = \left( {{v_1}...{v_n}} \right)\) where the \({{\bf{u}}_i}\) and \({{\bf{v}}_i}\) are in Theorem 10. Show that \(A = {\sigma _1}{u_1}v_1^T + {\sigma _2}{u_2}v_2^T + ... + {\sigma _r}{u_r}v_r^T\).

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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