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

Short Answer

Expert verified

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

Step by step solution

01

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:

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

02

The Principal Components

From exercise 2, thecovariance 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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Most popular questions from this chapter

In Exercises 17鈥24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) ,

\(\)

where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) 鈥渄iagonal鈥 matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

18. Suppose \(A\) is square and invertible. Find a singular value decomposition of \({A^{ - 1}}\)

(M) Compute an SVD of each matrix in Exercises 26 and 27. Report the final matrix entries accurate to two decimal places. Use the method of Examples 3 and 4.

26. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{ - {\bf{18}}}&{{\bf{13}}}&{ - {\bf{4}}}&{\bf{4}}\\{\bf{2}}&{{\bf{19}}}&{ - {\bf{4}}}&{{\bf{12}}}\\{ - {\bf{14}}}&{{\bf{11}}}&{ - {\bf{12}}}&{\bf{8}}\\{ - {\bf{2}}}&{{\bf{21}}}&{\bf{4}}&{\bf{8}}\end{array}} \right)\)

Let B be an \(n \times n\) symmetric matrix such that \({B^{\bf{2}}} = B\). Any such matrix is called a projection matrix (or an orthogonal projection matrix.) Given any y in \({\mathbb{R}^n}\), let \({\bf{\hat y}} = B{\bf{y}}\)and\({\bf{z}} = {\bf{y}} - {\bf{\hat y}}\).

a) Show that z is orthogonal to \({\bf{\hat y}}\).

b) Let W be the column space of B. Show that y is the sum of a vector in W and a vector in \({W^ \bot }\). Why does this prove that By is the orthogonal projection of y onto the column space of B?

Determine which of the matrices in Exercises 7鈥12 are orthogonal. If orthogonal, find the inverse.

9. \(\left[ {\begin{aligned}{{}}{ - 4/5}&{\,\,\,3/5}\\{3/5}&{\,\,4/5}\end{aligned}} \right]\)

Question: 13. The sample covariance matrix is a generalization of a formula for the variance of a sample of \(N\) scalar measurements, say \({t_1},................,{t_N}\). If \(m\) is the average of \({t_1},................,{t_N}\), then the sample variance is given by

\(\frac{1}{{N - 1}}\sum\limits_{k = 1}^n {{{\left( {{t_k} - m} \right)}^2}} \)

Show how the sample covariance matrix, \(S\), defined prior to Example 3, may be written in a form similar to (1). (Hint: Use partitioned matrix multiplication to write \(S\) as \(\frac{1}{{N - 1}}\) times the sum of \(N\) matrices of size \(p \times p\). For \(1 \le k \le N\), write \({X_k} - M\) in place of \({\hat X_k}\).)

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