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Q7.5-4E

Expert-verified
Found in: Page 395

### 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

# 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 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.