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Exercises \({\bf{13}}\) and \(14\) apply to a \({\bf{3 \times 3}}\) matrix \(A\) whose eigenvalues are estimated to be \({\bf{4}}\), \({\bf{4}}\), and \({\bf{3}}\).

14. Suppose the eigenvalues close to \({\bf{4}}\) and \({\bf{ - 4}}\) are known to have exactly the same absolute value. Describe how one might obtain a sequence that estimates the eigenvalue close to \({\bf{4}}\).

Short Answer

Expert verified

The power method will not work however, the inverse power method can be used. If the initial estimate is chosen near the eigenvalue close to \(4\), then the inverse power method should produce a sequence that estimates the eigenvalue close to \(4\).

Step by step solution

01

Write the definition of the Power Method

The Power Method: An \(n \times n\) matrix \(A\) with a strictly dominant eigenvalue \({\lambda _1}\) that means \({\lambda _1}\) must be larger in absolute value than all the other eigenvalues.

02

Check whether the power method will work or not

If the eigenvalues are close to\(4\)and\( - 4\)have the same absolute values, then none of these is a strictly dominant eigenvalue, so the power method will not work.

However, the inverse power method can be used. If the initial estimate is chosen near the eigenvalue close to \(4\), then the inverse power method should produce a sequence that estimates the eigenvalue close to \(4\).

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Most popular questions from this chapter

Use Exercise 12 to find the eigenvalues of the matrices in Exercises 13 and 14.

14. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{5}}&{{\bf{ - 6}}}&{{\bf{ - 7}}}\\{\bf{2}}&{\bf{4}}&{\bf{5}}&{\bf{2}}\\{\bf{0}}&{\bf{0}}&{{\bf{ - 7}}}&{{\bf{ - 4}}}\\{\bf{0}}&{\bf{0}}&{\bf{3}}&{\bf{1}}\end{array}} \right)\)

Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

13. \(\left[ {\begin{array}{*{20}{c}}6&- 2&0\\- 2&9&0\\5&8&3\end{array}} \right]\)

Define \(T:{{\rm P}_2} \to {\mathbb{R}^3}\) by \(T\left( {\bf{p}} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 0 \right)}\\{{\bf{p}}\left( 1 \right)}\end{aligned}} \right)\).

  1. Find the image under\(T\)of\({\bf{p}}\left( t \right) = 5 + 3t\).
  2. Show that \(T\) is a linear transformation.
  3. Find the matrix for \(T\) relative to the basis \(\left\{ {1,t,{t^2}} \right\}\)for \({{\rm P}_2}\)and the standard basis for \({\mathbb{R}^3}\).

Use Exercise 12 to find the eigenvalues of the matrices in Exercises 13 and 14.

13. \(A = \left( {\begin{array}{*{20}{c}}3&{ - 2}&8\\0&5&{ - 2}\\0&{ - 4}&3\end{array}} \right)\)

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

7. \(\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{6}}&{{\bf{ - 1}}}\end{array}} \right)\)

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