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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

Classify the quadratic forms in Exercises 9鈥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.

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

Compute the quadratic form \({{\bf{x}}^T}A{\bf{x}}\), when \(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\) and

a. \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)

b. \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)

c. \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)

Question: 13. Exercises 12鈥14 concern an \(m \times n\) matrix \(A\) with a reduced singular value decomposition, \(A = {U_r}D{V_r}^T\), and the pseudoinverse \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).

Suppose the equation\(A{\rm{x}} = {\rm{b}}\)is consistent, and let\({{\rm{x}}^ + } = {A^ + }{\rm{b}}\). By Exercise 23 in Section 6.3, there is exactly one vector\({\rm{p}}\)in Row\(A\)such that\(A{\rm{p}} = {\rm{b}}\). The following steps prove that\({{\rm{x}}^ + } = {\rm{p}}\)and\({{\rm{x}}^ + }\)is the minimum length solution of\(A{\rm{x}} = {\rm{b}}\).

  1. Show that \({{\rm{x}}^ + }\) is in Row \(A\). (Hint: Write \({\rm{b}}\) as \(A{\rm{x}}\) for some \({\rm{x}}\), and use Exercise 12.)
  2. Show that\({{\rm{x}}^ + }\)is a solution of\(A{\rm{x}} = {\rm{b}}\).
  3. Show that if \({\rm{u}}\) is any solution of \(A{\rm{x}} = {\rm{b}}\), then \(\left\| {{{\rm{x}}^ + }} \right\| \le \left\| {\rm{u}} \right\|\), with equality only if \({\rm{u}} = {{\rm{x}}^ + }\).

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.

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.)

Determine which of the matrices in Exercises 1鈥6 are symmetric.

1. \(\left[ {\begin{aligned}{{}}3&{\,\,\,5}\\5&{ - 7}\end{aligned}} \right]\)

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