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Question: Compute the singular values of the \({\bf{4 \times 4}}\) matrix in Exercise 9 in Section 2.3, and compute the condition number \(\frac{{{\sigma _1}}}{{{\sigma _4}}}\).

Short Answer

Expert verified

The value of condition number is \(\frac{{{\sigma _1}}}{{{\sigma _4}}} = 23,683\).

Step by step solution

01

Step 1: Find the singular value decomposition

Consider the matrix\(A = \left( {\begin{array}{*{20}{c}}4&0&{ - 7}&{ - 7}\\{ - 6}&1&{11}&9\\7&{ - 5}&{10}&{19}\\{ - 1}&2&3&{ - 1}\end{array}} \right)\).

Enter the matrix\(A\)in MATLAB:

\( > > {\rm{ }}A = \left( {4,0, - 7, - 7; - 6,1,11,9;7, - 5,10,19; - 1,2,3, - 1} \right)\)

Find the singular value decomposition of\(A\).

\( > > {\rm{ }}B = svd\left( A \right);\)

\(B = \left( {\begin{array}{*{20}{c}}{27.386}\\{12.091}\\{2.6116}\\{0.001156}\end{array}} \right)\)

02

Compute the condition number \(\frac{{{\sigma _1}}}{{{\sigma _4}}}\)

Find the condition number of\(A\).

\( > > C = cond\left( A \right);\)

\(C = 2.3683 \times {10^4}\)

Find the condition number\(\frac{{{\sigma _1}}}{{{\sigma _4}}}\)using singular values:

\(\begin{array}{c}\frac{{{\sigma _1}}}{{{\sigma _4}}} = \frac{{27.386}}{{0.001156}}\\ = 23,683\end{array}\)

Therefore, \(\frac{{{\sigma _1}}}{{{\sigma _4}}} = 23,683\).

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Most popular questions from this chapter

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 \({y_1}\)?

In Exercises 25 and 26, mark each statement True or False. Justify each answer.

26.

  1. There are symmetric matrices that are not orthogonally diagonizable.
  2. b. If \(B = PD{P^T}\), where \({P^T} = {P^{ - {\bf{1}}}}\) and D is a diagonal matrix, then B is a symmetric matrix.
  3. c. An orthogonal matrix is orthogonally diagonizable.
  4. d. The dimension of an eigenspace of a symmetric matrix is sometimes less than the multiplicity of the corresponding eigenvalue.

In Exercises 17–24, \(A\) 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 \(m \times n\) “diagonal” matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

21. Justify the statement in Example 2 that the second singular value of a matrix \(A\) is the maximum of \(\left\| {A{\bf{x}}} \right\|\) as \({\bf{x}}\) varies over all unit vectors orthogonal to \({{\bf{v}}_{\bf{1}}}\), with \({{\bf{v}}_{\bf{1}}}\) a right singular vector corresponding to the first singular value of \(A\). (Hint: Use Theorem 7 in Section 7.3.)

Classify the quadratic forms in Exercises 9–18. Then make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct \(P\) using the methods of Section 7.1.

11. \({\bf{2}}x_{\bf{1}}^{\bf{2}} - {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}} - x_{\bf{2}}^{\bf{2}}\)

Question: Let \({\bf{X}}\) denote a vector that varies over the columns of a \(p \times N\) matrix of observations, and let \(P\) be a \(p \times p\) orthogonal matrix. Show that the change of variable \({\bf{X}} = P{\bf{Y}}\) does not change the total variance of the data. (Hint: By Exercise 11, it suffices to show that \(tr\left( {{P^T}SP} \right) = tr\left( S \right)\). Use a property of the trace mentioned in Exercise 25 in Section 5.4.)

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