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Q5.3-9E

Expert-verified
Linear Algebra and its Applications
Found in: Page 267
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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

9. \(\left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right)\)

The general solution is \({{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\). Since we cannot generate an eigenvector basis for \({\mathbb{R}^2}\), \(A\) is not diagonalizable.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the Diagonalization Theorem

The Diagonalization Theorem: An \(n \times n\) matrix \(A\) is diagonalizable if and only if \(A\) has \(n\) linearly independent eigenvectors. As \(A = PD{P^{ - 1}}\) which has \(D\) a diagonal matrix if and only if the columns of \(P\) are \(n\) linearly independent eigenvectors of \(A\).

Step 2: Find eigenvalues of the matrix

Consider the given matrix \(A = \left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right)\). We need to diagonalize the given matrix if possible.

Write the characteristic equation:

\[\begin{array}{c}\det \left( {A - \lambda I} \right) = \det \left( {\left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right) - \lambda \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\\ = \det \left( {\left( {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 1}\\1&{5 - \lambda }\end{array}} \right)} \right)\\ = \left( {3 - \lambda } \right)\left( {5 - \lambda } \right) - \left( { - 1} \right)\left( 1 \right)\\ = {\lambda ^2} - 8\lambda + 16\\ = {\left( {\lambda - 4} \right)^2}\end{array}\]

Therefore, \(\det \left( {A - \lambda I} \right) = {\left( {\lambda - 4} \right)^2}\).

Step 3: Find the eigenvalues

\[\begin{array}{c}\det \left( {A - \lambda I} \right) = 0\\{\left( {\lambda - 4} \right)^2} = 0\\\lambda = 4,4\end{array}\]

So, the eigenvalue of the matrix is \(4\) with a multiplicity of \(2\).

Step 4: Find the eigenvectors

\[\begin{array}{c}\left( {A - \lambda I} \right){\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 1}\\1&{5 - \lambda }\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{3 - 4}&1\\1&{5 - 4}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{ - 1}&{ - 1}\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{ - 1}&{ - 1}\\0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}\]

Therefore, the eigenvectors are shown below:

\[ - {x_1} - {x_2} = 0\]

\[\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = k\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\]

So, eigenvector is \({{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\).

Since we cannot generate an eigenvector basis for \({\mathbb{R}^2}\), \(A\) is not diagonalizable.

Thus, the matrix is not-Diagonalizable.

Most popular questions for Math Textbooks

For the matrix A, find real closed formulas for the trajectory x→(t+1)=Ax¯(t)where x→=[01]. Draw a rough sketch

A = [15-27]

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 \({\mathbb{R}^4}\).

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: Is \(\lambda = 2\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\)? Why or why not?

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

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

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