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Chapter 5: Q21E (page 267)

A common misconception is that if \(A\) has a strictly dominant eigenvalue, then, for any sufficiently large value of \(k\), the vector \({A^k}{\bf{x}}\) is approximately equal to an eigenvector of \(A\). For the three matrices below, study what happens to \({A^k}{\bf{x}}\) when \({\bf{x = }}\left( {{\bf{.5,}}{\bf{.5}}} \right)\), and try to draw general conclusions (for a \({\bf{2 \times 2}}\) matrix).

a. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{{\bf{.8}}}&{\bf{0}}\\{\bf{0}}&{{\bf{.2}}}\end{aligned}} \right)\) b. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{{\bf{.8}}}\end{aligned}} \right)\) c. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{8}}&{\bf{0}}\\{\bf{0}}&{\bf{2}}\end{aligned}} \right)\)

Short Answer

Expert verified
  1. Conclusion: If\({\bf{x}} \ne 0\)and all the eigenvalues of the matrix\(A\)are less than\(1\)in magnitude then\({A^k}{\bf{x}}\)is approximately an eigenvector for the larger \(k\).
  2. Conclusion:If the strictly dominant eigenvalue of \(A\) is \(1\), and if \(x\) has a component in the direction of the corresponding eigenvector, then \({A^k}{\bf{x}}\) will converge to a multiple of that eigenvector.
  3. Conclusion: If the eigenvalues of \(A\) are all greater than \(1\) in magnitude, and if \(x\) is not an eigenvector, then the distance from \({A^k}{\bf{x}}\) to the nearest eigenvector will increase as \(k \to \infty \).

Step by step solution

01

Write about what happened to \({A^k}{\bf{x}}\)

Consider \(A = \left( {\begin{aligned}{ {20}{c}}{.8}&0\\0&{.2}\end{aligned}} \right)\), \({\bf{x}} = \left( {\begin{aligned}{ {20}{c}}{.5}\\{.5}\end{aligned}} \right)\)

Now the sequence for \({A^k}{\bf{x}}\), \(k = 1,...5\).

Therefore,

\(\left( {\begin{aligned}{ {20}{c}}{.4}\\{.1}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.32}\\{.02}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.256}\\{.004}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.2048}\\{.0008}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.16384}\\{.00016}\end{aligned}} \right)\)

From the above conclusion, \({A^k}{\bf{x}}\) is approximate \(.8\).

Conclusion: If \({\bf{x}} \ne 0\) and all the eigenvalues of the matrix \(A\) are less than \(1\) in magnitude then \({A^k}{\bf{x}}\) is approximately an eigenvector for the larger \(k\).

02

Write about what happened to \({A^k}{\bf{x}}\)

Consider \(A = \left( {\begin{aligned}{ {20}{c}}1&0\\0&{.8}\end{aligned}} \right)\), \({\bf{x}} = \left( {\begin{aligned}{ {20}{c}}{.5}\\{.5}\end{aligned}} \right)\)

Now the sequence for \({A^k}{\bf{x}}\), \(k = 1,...5\).

Therefore,

\(\left( {\begin{aligned}{ {20}{c}}{.5}\\{.4}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.32}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.256}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.2048}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.16384}\end{aligned}} \right)\)

From the above conclusion \({A^k}{\bf{x}}\) is approximate \(\left( {\begin{aligned}{ {20}{c}}{.5}\\0\end{aligned}} \right)\).

Conclusion: If the strictly dominant eigenvalue of \(A\) is \(1\), and if \(x\) has a component in the direction of the corresponding eigenvector, then \({A^k}{\bf{x}}\) will converge to a multiple of that eigenvector.

03

Write about what happened to \({A^k}{\bf{x}}\)

Consider \(A = \left( {\begin{aligned}{ {20}{c}}8&0\\0&2\end{aligned}} \right)\), \({\bf{x}} = \left( {\begin{aligned}{ {20}{c}}{.5}\\{.5}\end{aligned}} \right)\)

Now the sequence for \({A^k}{\bf{x}}\), \(k = 1,...5\).

Therefore,

\(\left( {\begin{aligned}{ {20}{c}}4\\1\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{32}\\2\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{256}\\4\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{2048}\\8\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{16384}\\{16}\end{aligned}} \right)\)

From the above conclusion distance of \({A^k}{\bf{x}}\) distance of from either eigenvector of \(A\) is increasing rapidly as \(k\) increases

Conclusion: If the eigenvalues of \(A\) are all greater than \(1\) in magnitude, and if \(x\) is not an eigenvector, then the distance from \({A^k}{\bf{x}}\) to the nearest eigenvector will increase as \(k \to \infty \).

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Most popular questions from this chapter

Let\(D = \left\{ {{{\bf{d}}_1},{{\bf{d}}_2}} \right\}\) and \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2}} \right\}\) be bases for vector space \(V\) and \(W\), respectively. Let \(T:V \to W\) be a linear transformation with the property that

\(T\left( {{{\bf{d}}_1}} \right) = 2{{\bf{b}}_1} - 3{{\bf{b}}_2}\), \(T\left( {{{\bf{d}}_2}} \right) = - 4{{\bf{b}}_1} + 5{{\bf{b}}_2}\)

Find the matrix for \(T\) relative to \(D\), and\(B\).

Question: For the matrices in Exercises 15-17, list the eigenvalues, repeated according to their multiplicities.

16. \(\left[ {\begin{array}{*{20}{c}}5&0&0&0\\8&- 4&0&0\\0&7&1&0\\1&{ - 5}&2&1\end{array}} \right]\)

M] In Exercises 19 and 20, find (a) the largest eigenvalue and (b) the eigenvalue closest to zero. In each case, set \[{{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0,0}}} \right)\] and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.

19.\[A{\bf{=}}\left[{\begin{array}{*{20}{c}}{{\bf{10}}}&{\bf{7}}&{\bf{8}}&{\bf{7}}\\{\bf{7}}&{\bf{5}}&{\bf{6}}&{\bf{5}}\\{\bf{8}}&{\bf{6}}&{{\bf{10}}}&{\bf{9}}\\{\bf{7}}&{\bf{5}}&{\bf{9}}&{{\bf{10}}}\end{array}} \right]\]

Question: In Exercises 21 and 22, \(A\) and \(B\) are \(n \times n\) matrices. Mark each statement True or False. Justify each answer.

  1. If \(A\) is \(3 \times 3\), with columns \({{\rm{a}}_1}\), \({{\rm{a}}_2}\), and \({{\rm{a}}_3}\), then \(\det A\) equals the volume of the parallelepiped determined by \({{\rm{a}}_1}\), \({{\rm{a}}_2}\), and \({{\rm{a}}_3}\).
  2. \(\det {A^T} = \left( { - 1} \right)\det A\).
  3. The multiplicity of a root \(r\) of the characteristic equation of \(A\) is called the algebraic multiplicity of \(r\) as an eigenvalue of \(A\).
  4. A row replacement operation on \(A\) does not change the eigenvalues.

Define\(T:{{\rm P}_3} \to {\mathbb{R}^4}\)by\(T\left( {\bf{p}} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 3} \right)}\\{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 1 \right)}\\{{\bf{p}}\left( 3 \right)}\end{aligned}} \right)\).

  1. Show that \(T\) is a linear transformation.
  2. Find the matrix for \(T\) relative to the basis \(\left\{ {1,t,{t^2},{t^3}} \right\}\)for \({{\rm P}_3}\)and the standard basis for \({\mathbb{R}^4}\).
See all solutions

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