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Chapter 5: Q8SE (page 267)

a. Let \(A\) be a diagonalizable \(n \times n\) matrix. Show that if the multiplicity of an eigenvalue \(\lambda \) is \(n\), then \(A = \lambda I\).

b. Use part (a) to show that the matrix \(A =\left({\begin{aligned}{*{20}{l}}3&1\\0&3\end{aligned}}\right)\) is not diagonalizable.

Short Answer

Expert verified

a. It is proved that if the multiplicity of an eigenvalue \(\lambda \)is \(n\), then\(A = \lambda I\).

b. By part (a), if A would be diagonalizable then A would be \(3I\). Since \(A \ne 3I\)It is not diagonalizable.

Step by step solution

01

Showing that if the multiplicity of an eigenvalue

If\(A\)is diagonalizable, then\(A = PD{P^{ - 1}}\), for some invertible matrix\(P\)and diagonal matrix\(D\), where entries in\(D\)are eigenvalues of\(A\).

Since multiplicity of\(\lambda \)is\(n\)and \(\lambda \)appears\(n\)times in every diagonal entry in\(D\).

In other words,\(D = \lambda I\), so:

\(\begin{aligned}{c}A &= PD{P^{ - 1}}\\ &= P\lambda I{P^{ - 1}}\\ &= \lambda PI{P^{ - 1}}\\ &= \lambda I\end{aligned}\)

Hence \(A = \lambda I\).

It is proved that if the multiplicity of an eigenvalue \(\lambda \)is \(n\), then\(A = \lambda I\).

02

Showing that the matrix \(A\) not diagonalizable

The matrix is triangular, so its eigenvalues are on its diagonal.

Only eigenvalue is 3 with multiplicity 2.

By part a), if\(A\)would be diagonalizable then\(A\)would be\(3I\).

Since \(A \ne 3I\), it is not diagonalizable.

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Most popular questions from this chapter

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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

10. \(\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{3}}\\{\bf{4}}&{\bf{1}}\end{array}} \right)\)

Show that if \(A\) is diagonalizable, with all eigenvalues less than 1 in magnitude, then \({A^k}\) tends to the zero matrix as \(k \to \infty \). (Hint: Consider \({A^k}x\) where \(x\) represents any one of the columns of \(I\).)

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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

15. \(\left( {\begin{array}{*{20}{c}}{\bf{7}}&{\bf{4}}&{{\bf{16}}}\\{\bf{2}}&{\bf{5}}&{\bf{8}}\\{{\bf{ - 2}}}&{{\bf{ - 2}}}&{{\bf{ - 5}}}\end{array}} \right)\)

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

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

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.
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