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Chapter 5: Q5.3-35E (page 267)

Question: Diagonalize the matrices in Exercises 33-36. Use your matrix program’s eigenvalue command to find the eigenvalues, and then compute bases for the eigenspaces as in section 5.1.

35. \(\left[ {\begin{array}{*{20}{c}}{{\bf{11}}}&{ - {\bf{6}}}&{\bf{4}}&{ - {\bf{10}}}&{ - {\bf{4}}}\\{ - {\bf{3}}}&{\bf{5}}&{ - {\bf{2}}}&{\bf{4}}&{\bf{1}}\\{ - {\bf{8}}}&{{\bf{12}}}&{ - {\bf{3}}}&{{\bf{12}}}&{\bf{4}}\\{\bf{1}}&{\bf{6}}&{ - {\bf{2}}}&{\bf{3}}&{ - {\bf{1}}}\\{\bf{8}}&{ - {\bf{18}}}&{\bf{8}}&{ - {\bf{14}}}&{ - {\bf{1}}}\end{array}} \right]\)

Short Answer

Expert verified

The diagonal form of the given matrix is \(\left[ {\begin{array}{*{20}{c}}5&0&0&0&0\\0&5&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&3\end{array}} \right]\).

Step by step solution

01

Find the eigenvalues

Let,

\[A = \left[ {\begin{array}{*{20}{c}}{11}&{ - 6}&4&{ - 10}&{ - 4}\\{ - 3}&5&{ - 2}&4&1\\{ - 8}&{12}&{ - 3}&{12}&4\\1&6&{ - 2}&3&{ - 1}\\8&{ - 18}&8&{ - 14}&{ - 1}\end{array}} \right]\]

Use the following code in MATLAB to find the eigenvalues of the matrix.

\[\begin{array}{l} > > A = \left[ \begin{array}{l}\begin{array}{*{20}{c}}{11}&{ - 6}&4&{ - 10}&{ - 4;\,}\end{array}\,\begin{array}{*{20}{c}}{ - 3}&5&{ - 2}&4&{1;\,}\end{array}\,\begin{array}{*{20}{c}}{ - 8}&{12}&{ - 3}&{12}&{4;\,\,}\end{array}\,\\\begin{array}{*{20}{c}}1&6&{ - 2}&3&{ - 1;\,}\end{array}\begin{array}{*{20}{c}}8&{ - 18}&8&{ - 14}&{ - 1}\end{array}\end{array} \right];\\ > > {\rm{ev}} = eigs\left( A \right);\end{array}\]

So, the eigenvalues of A are:

\(ev = \left( {5,\,1,\,3,\,5,1} \right)\)

02

Find eigenvectors of A

Find the eigenvalues of A using the following MATLAB code:

\({\rm{ > > nulbasis}}\left( {A - {\rm{ev}}\left( 1 \right) * {\rm{eye}}\left( 5 \right)} \right);\)

So, the eigenvector for \(\lambda = 5\) is:

\({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}{2.0000}\\{ - 0.3333}\\{ - 1.0000}\\{1.0000}\\0\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{1.0000}\\{ - 0.3333}\\{ - 1.0000}\\0\\{1.0000}\end{array}} \right]\)

The basis of the eigenspace of \(\lambda = 5\) is \(\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\{ - 3}\\3\\0\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}3\\{ - 1}\\{ - 3}\\0\\3\end{array}} \right]\).

\({\rm{ > > nulbasis}}\left( {A - {\rm{ev}}\left( 2 \right) * {\rm{eye}}\left( 5 \right)} \right);\)

So, the eigenvector for \(\lambda = 1\) is:

The basis of the eigenspace of \(\lambda = 1\) is \(\left[ {\begin{array}{*{20}{c}}4\\{ - 3}\\{ - 2}\\5\\0\end{array}} \right],\,\left[ {\begin{array}{*{20}{c}}3\\{ - 1}\\{ - 4}\\0\\5\end{array}} \right]\).

\({\rm{ > > nulbasis}}\left( {A - {\rm{ev}}\left( 3 \right) * {\rm{eye}}\left( 5 \right)} \right);\)

So, the eigenvector for \(\lambda = 3\) is:

\({{\bf{v}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{0.5000}\\{ - 0.2500}\\{ - 1.0000}\\{ - 0.2500}\\{1.0000}\end{array}} \right]\)

The basis of the eigenspace of \(\lambda = 3\) is \(\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\{ - 4}\\{ - 1}\\4\end{array}} \right]\).

03

Write the matrix invertible matrix P and its diagonalized form

The invertible matrix P can be written as,

\(P = \left[ {\begin{array}{*{20}{c}}6&3&4&3&2\\{ - 1}&{ - 1}&{ - 3}&{ - 1}&{ - 1}\\{ - 3}&{ - 3}&{ - 2}&{ - 4}&{ - 4}\\3&0&5&0&{ - 1}\\0&3&0&5&4\end{array}} \right]\)

The diagonalized form can be written as,

\(D = \left[ {\begin{array}{*{20}{c}}5&0&0&0&0\\0&5&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&3\end{array}} \right]\)

So, the diagonal form of the given matrix is \(\left[ {\begin{array}{*{20}{c}}5&0&0&0&0\\0&5&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&3\end{array}} \right]\).

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Most popular questions from this chapter

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

15. \(\left[ {\begin{array}{*{20}{c}}4&- 7&0&2\\0&3&- 4&6\\0&0&3&{ - 8}\\0&0&0&1\end{array}} \right]\)

Exercises 19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

20. Let \(p\left( t \right){\bf{ = }}\left( {t{\bf{ - 2}}} \right)\left( {t{\bf{ - 3}}} \right)\left( {t{\bf{ - 4}}} \right){\bf{ = - 24 + 26}}t{\bf{ - 9}}{t^{\bf{2}}}{\bf{ + }}{t^{\bf{3}}}\). Write the companion matrix for \(p\left( t \right)\), and use techniques from chapter \({\bf{3}}\) to find the characteristic polynomial.

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.6}&{.3}\\{.4}&{.7}\end{array}} \right)\), \({v_1} = \left( {\begin{array}{*{20}{c}}{3/7}\\{4/7}\end{array}} \right)\), \({x_0} = \left( {\begin{array}{*{20}{c}}{.5}\\{.5}\end{array}} \right)\). (Note: \(A\) is the stochastic matrix studied in Example 5 of Section 4.9.)

  1. Find a basic for \({\mathbb{R}^2}\) consisting of \({{\rm{v}}_1}\) and anther eigenvector \({{\rm{v}}_2}\) of \(A\).
  2. Verify that \({{\rm{x}}_0}\) may be written in the form \({{\rm{x}}_0} = {{\rm{v}}_1} + c{{\rm{v}}_2}\).
  3. For \(k = 1,2, \ldots \), define \({x_k} = {A^k}{x_0}\). Compute \({x_1}\) and \({x_2}\), and write a formula for \({x_k}\). Then show that \({{\bf{x}}_k} \to {{\bf{v}}_1}\) as \(k\) increases.

Let \({\bf{u}}\) be an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \), and let \(H\) be the line in \({\mathbb{R}^{\bf{n}}}\) through \({\bf{u}}\) and the origin.

  1. Explain why \(H\) is invariant under \(A\) in the sense that \(A{\bf{x}}\) is in \(H\) whenever \({\bf{x}}\) is in \(H\).
  2. Let \(K\) be a one-dimensional subspace of \({\mathbb{R}^{\bf{n}}}\) that is invariant under \(A\). Explain why \(K\) contains an eigenvector of \(A\).

Question 18: It can be shown that the algebraic multiplicity of an eigenvalue \(\lambda \) is always greater than or equal to the dimension of the eigenspace corresponding to \(\lambda \). Find \(h\) in the matrix \(A\) below such that the eigenspace for \(\lambda = 5\) is two-dimensional:

\[A = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&h&0\\0&0&5&4\\0&0&0&1\end{array}} \right]\]
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