Americas
Europe
Q7.5-4E
Expert-verifiedQuestion: Find the principal components of the data for Exercise 2.
The principal components are \(\left[ {\begin{array}{*{20}{c}}{0.44013}\\{0.897934}\end{array}} \right]{\rm{ and }}\left[ {\begin{array}{*{20}{c}}{ - 0.897934}\\{0.44013}\end{array}} \right]\).
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:
\(S = \frac{1}{{N - 1}}B{B^T}\)
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.
94% of StudySmarter users get better grades.
Sign up for free