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Q7.5-7E

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: 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 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.