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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: Find the principal components of the data for Exercise 2.

The principal components are \(\left[ {\begin{array}{*{20}{c}}{0.44013}\\{0.897934}\end{array}} \right]{\rm{ and }}\left[ {\begin{array}{*{20}{c}}{ - 0.897934}\\{0.44013}\end{array}} \right]\).

See the step by step solution

Step by Step Solution

Step 1: Mean Deviation form and Covariance Matrix

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

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

Whose \(p \times p\) covariance matrix is:

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

Step 2: The Principal Components

From exercise 2, the covariance matrix is:

\(S = \left[ {\begin{array}{*{20}{c}}{5.6}&8\\8&{18}\end{array}} \right]\)

The eigenvalues for this matrix will be:

\(\begin{array}{l}{\lambda _1} = 21.9213\\{\lambda _2} = 1.67874\end{array}\)

The respective eigenvectors are:

\(\begin{array}{l}{x_1} = \left[ {\begin{array}{*{20}{c}}{0.490158}\\1\end{array}} \right]\\{x_2} = \left[ {\begin{array}{*{20}{c}}{ - 2.04016}\\1\end{array}} \right]\end{array}\)

After normalization, we have:

\(\begin{array}{l}{u_1} = \left[ {\begin{array}{*{20}{c}}{0.44013}\\{0.897934}\end{array}} \right]\\{u_2} = \left[ {\begin{array}{*{20}{c}}{ - 0.897934}\\{0.44013}\end{array}} \right]\end{array}\)

Hence, these are the required components.


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