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Q7.5-11E

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Linear Algebra and its Applications
Found in: Page 395
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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 \({Y_k} = {P^T}{X_k}{\rm{ for }}k = 1,......,N\).

  1. Show that \({Y_1},................,{Y_N}\) are in mean-deviation form. (Hint: Let \(w\) be the vector in \({\mathbb{R}^N}\) with a 1 in each entry. Then \(\left( {{X_1},................,{X_N}} \right)w = 0\) (the zero vector in \({\mathbb{R}^p}\)).)
  2. Show that if the covariance matrix of \({X_1},................,{X_N}\) is \(S\), then the covariance matrix of \({Y_1},................,{Y_N}\) is \({P^T}SP\).

It is verified that:

  1. \(\left( {{{\bf{X}}_1},................,{{\bf{X}}_N}} \right)w = 0\)
  2. The covariance matrix is: \({S_Y} = {P^T}SP\)
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Mean Deviation form and Covariance Matrix

The Mean Deviation form of 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 matrix is:

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

Step 2: The Mean Deviation Form (a)

From the question, the \(w\)is a unit vector with all values equal to 1. Then, we have:

\(\begin{array}{c}\left( {{{\bf{X}}_1},................,{{\bf{X}}_N}} \right)w = \left( {{{\bf{X}}_1},{{\bf{X}}_2}, \ldots ,{{\bf{X}}_n}} \right)\left( {\begin{array}{*{20}{c}}1\\1\\ \vdots \\1\end{array}} \right)\\ = {{\bf{X}}_1} + ...... + {{\bf{X}}_N}\\ = 0\end{array}\)

The mean deviation form given is:

\(\begin{array}{c}\left( {{{\bf{Y}}_1},................,{{\bf{Y}}_N}} \right)w = \left( {{P^T}{{\bf{X}}_1},................,{P^T}{{\bf{X}}_N}} \right)w\\ = {P^T}\left( {{{\bf{X}}_1} + ...... + {{\bf{X}}_N}} \right)w\\ = {P^T}\left( {{{\bf{X}}_1} + ...... + {{\bf{X}}_N}} \right)\left( {\begin{array}{*{20}{c}}1\\1\\ \vdots \\1\end{array}} \right)\\ = {P^T}\left( {{{\bf{X}}_1} + ...... + {{\bf{X}}_N}} \right)\\ = 0\end{array}\)

Hence, this is the required proof.

Step 3: The Covariance Matrix (b)

From (a), the covariance matrix can be given as:

\(\begin{array}{c}{S_Y} = \frac{1}{{N - 1}}\left( {{{\bf{Y}}_1},................,{{\bf{Y}}_N}} \right){\left( {{{\bf{Y}}_1},................,{{\bf{Y}}_N}} \right)^T}\\ = \frac{1}{{N - 1}}\left( {{P^T}{{\bf{X}}_1},................,{P^T}{{\bf{X}}_N}} \right){\left( {{P^T}{{\bf{X}}_1},................,{P^T}{{\bf{X}}_N}} \right)^T}\\ = {P^T}\left\{ {\frac{1}{{N - 1}}\left( {{{\bf{X}}_1},......,{{\bf{X}}_N}} \right){{\left( {{{\bf{X}}_1},......,{{\bf{X}}_N}} \right)}^T}} \right\}P\\ = {P^T}SP\end{array}\)

Hence, this is the required proof.

Most popular questions for Math Textbooks

In Exercises 17–24, \(A\) 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 \(m \times n\) “diagonal” 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: In Exercises 15 and 16, construct the pseudo-inverse of \(A\). Begin by using a matrix program to produce the SVD of \(A\), or, if that is not available, begin with an orthogonal diagonalization of \({A^T}A\). Use the pseudo-inverse to solve \(A{\rm{x}} = {\rm{b}}\), for \({\rm{b}} = \left( {6, - 1, - 4,6} \right)\) and let \(\mathop {\rm{x}}\limits^\^ \)be the solution. Make a calculation to verify that \(\mathop {\rm{x}}\limits^\^ \) is in Row \(A\). Find a nonzero vector \({\rm{u}}\) in Nul\(A\), and verify that \(\left\| {\mathop {\rm{x}}\limits^\^ } \right\| < \left\| {\mathop {\rm{x}}\limits^\^ + {\rm{u}}} \right\|\), which must be true by Exercise 13(c).

15. \(A = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 3}&{ - 6}&6&{\,\,1}\\{ - 1}&{ - 1}&{ - 1}&1&{ - 2}\\{\,\,\,0}&{\,\,0}&{ - 1}&1&{ - 1}\\{\,\,\,0}&{\,\,0}&{ - 1}&1&{ - 1}\end{array}} \right]\)

Construct a spectral decomposition of A from Example 2.

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.

20. \(\left( {\begin{aligned}{{}}5&8&{ - 4}\\8&5&{ - 4}\\{ - 4}&{ - 4}&{ - 1}\end{aligned}} \right)\)

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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