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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: Let \({x_1}\,,{x_2}\) denote the variables for the two-dimensional data in Exercise 1. Find a new variable \({y_1}\) of the form \({y_1} = {c_1}{x_1} + {c_2}{x_2}\), with\(c_1^2 + c_2^2 = 1\), such that \({y_1}\) has maximum possible variance over the given data. How much of the variance in the data is explained by \({y_1}\)?

The variance of the data by \({y_1}\) obtained as \(93.3374\% \).

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 Change in Variance

From exercise 1, the maximum eigenvalue is:

\({\lambda _1} = 95.2041\)

The respective unit vector is:

\({u_1} = \left( {\begin{array}{*{20}{c}}{0.946515}\\{ - 0.322659}\end{array}} \right)\)

The new variable will be:

\({y_1} = 0.946515{x_1} - 0.322659{x_2}\)

Now, the percentage of change in variance can be obtained as:

\(\begin{array}{c}\Delta = \frac{{{\lambda _1}}}{{tr\left( S \right)}} \times 100\\ = \frac{{95.2041}}{{86 + 16}} \times 100\\ = 93.3374\% \end{array}\)

Hence, this is the required answer.

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