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Question: Let \({\bf{X}}\) denote a vector that varies over the columns of a \(p \times N\) matrix of observations, and let \(P\) be a \(p \times p\) orthogonal matrix. Show that the change of variable \({\bf{X}} = P{\bf{Y}}\) does not change the total variance of the data. (Hint: By Exercise 11, it suffices to show that \(tr\left( {{P^T}SP} \right) = tr\left( S \right)\). Use a property of the trace mentioned in Exercise 25 in Section 5.4.)

Short Answer

Expert verified

It is verified that the total variance would not change when variables change as \({\bf{X}} = P{\bf{Y}}\).

Step by step solution

01

Mean Deviation form and Covariance Matrix.

The Mean Deviation formof any \(p \times N\)is given by:

\(B = \left( {\begin{array}{*{20}{c}}{{{{\bf{\hat X}}}_1}}&{{{{\bf{\hat X}}}_2}}&{........}&{{{{\bf{\hat X}}}_N}}\end{array}} \right)\)

Whose \(p \times p\) covariance matrixis:

\(S = \frac{1}{{N - 1}}B{B^T}\)

02

The Variance

From exercise 11,we have:

\({S_Y} = {P^T}SP\)

When variable changes as:\({\bf{X}} = P{\bf{Y}}\)

The traces of the covariance matrices\({S_Y}{\rm{ and }}S\)will be the same.

The total variance of the data is given by\({\bf{Y}}\)is\({\rm{tr}}\left( {{P^T}SP} \right)\).

For two similar matrices\(A,B\)are such that,\({\bf{B}} = P{\bf{A}}{P^{ - 1}}\)which implies\({\rm{tr}}\left( {\bf{B}} \right) = {\rm{tr}}\left( {P{\bf{A}}{P^{ - 1}}} \right)\).

In the obtained equation, if\(P\)is an orthogonal matrix, then\({P^T} = {P^{ - 1}}\).

Apply trace on both sides of\({P^T} = {P^{ - 1}}\)and simplify.

\(\begin{array}{c}{\rm{tr}}\left( {{P^T}SP} \right) = {\rm{tr}}\left( {{P^{ - 1}}SP} \right)\\ = {\rm{tr}}\left( {{P^{ - 1}}PS} \right)\\ = {\rm{tr}}\left( {\left( {{P^{ - 1}}P} \right)S} \right)\\ = {\rm{tr}}\left( {IS} \right)\\ = {\rm{tr}}\left( S \right)\end{array}\)

Thus, the total variance would not changewhen variables change as:\({\bf{X}} = P{\bf{Y}}\).

Hence, this is the required proof.

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Most popular questions from this chapter

Question: In Exercises 17-22, determine which sets of vectors are orthonormal. If a set is only orthogonal, normalize the vectors to produce an orthonormal set.

18. \(\left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right),{\rm{ }}\left( {\begin{array}{*{20}{c}}0\\{ - 1}\\0\end{array}} \right)\)

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 \({y_1}\)?

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

Suppose Aand B are orthogonally diagonalizable and \(AB = BA\). Explain why \(AB\) is also orthogonally diagonalizable.

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.

19. \(\left( {\begin{aligned}{{}}3&{ - 2}&4\\{ - 2}&6&2\\4&2&3\end{aligned}} \right)\)

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

a. \(3x_1^2 - 4{x_1}{x_2} + 5x_2^2\) b. \(3x_1^2 + 2{x_1}{x_2}\)

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