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Q5.2-3E

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

3. \(\left[ {\begin{array}{*{20}{c}}3&-2\\1&-1\end{array}} \right]\)

Characteristic polynomial: \({\lambda ^2} - 2\lambda - 1\).

Eigenvalues: \(\lambda = 1 + \sqrt 2 \) and \(\lambda = 1 - \sqrt 2 \).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Find the characteristic polynomial

If is an \(n \times n\) matrix, then \(det\left( {A - \lambda I} \right)\), which is a polynomial of degree \(n\), is called the characteristic polynomial of \(A\).

It is given that \(A = \left[ {\begin{array}{*{20}{c}}3&- 2\\1&- 1\end{array}} \right]\) and \(I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\) is identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\) as shown below:

\[\begin{array} - \lambda I = \left[ {\begin{array}{*{20}{c}}3&- 2\\1&- 1\end{array}} \right] - \lambda \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 2}\\1&{ - 1 - \lambda }\end{array}} \right]\end{array}\]

Now, calculate the determinant of the matrix \(\left( {A - \lambda I} \right)\) as shown below:

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

So, the characteristic polynomial of is \({\lambda ^2} - 2\lambda - 1\).

Step 2: Describe the characteristic equation

To find the eigenvalues of the matrix, we must calculate all the scalars such that \(\left( {A - \lambda I} \right)x = 0\) has a non-trivial solution which is equivalent to finding all such that the matrix \(\left( {A - \lambda I} \right)\)is not invertible, that is, when determinant of \(\left( {A - \lambda I} \right)\)is zero.

Thus, the eigenvalues of \(A\) are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, find the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

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

Step 3: Find roots of characteristic equation

For the quadratic equation, \(a{x^2} + bx + c = 0\) , the general solution is given as\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} \;\;}}{{2a}}\).

Thus, the solution of the characteristic equation \[{\lambda ^2} - 2\lambda - 1 = 0\] is obtained as follows:

\[\begin{array}{\lambda ^2} - 2\lambda - 1 = 0\\\lambda = \frac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( { - 1} \right)} }}{2}\\ = \frac{{2 \pm 2\sqrt 2 }}{2}\\ = 1 \pm \sqrt 2 \end{array}\]

The eigenvalues of \(A\) are \(\lambda = 1 + \sqrt 2 \) and \(\lambda = 1 - \sqrt 2 \).

Most popular questions for Math Textbooks

Show that \(I - A\) is invertible when all the eigenvalues of \(A\) are less than 1 in magnitude. (Hint: What would be true if \(I - A\) were not invertible?)

Let\(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\) be a basis for a vector space\(V\). Find \(T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)\) when \(T\) isa linear transformation from \(V\) to \(V\) whose matrix relative to \(B\) is

\({\left( T \right)_B} = \left( {\begin{aligned}0&{}&{ - 6}&{}&1\\0&{}&5&{}&{ - 1}\\1&{}&{ - 2}&{}&7\end{aligned}} \right)\)

Let \(J\) be the \(n \times n\) matrix of all \({\bf{1}}\)’s and consider \(A = \left( {a - b} \right)I + bJ\) that is,

\(A = \left( {\begin{aligned}{*{20}{c}}a&b&b&{...}&b\\b&a&b&{...}&b\\b&b&a&{...}&b\\:&:&:&:&:\\b&b&b&{...}&a\end{aligned}} \right)\)

Use the results of Exercise \({\bf{16}}\) in the Supplementary Exercises for Chapter \({\bf{3}}\) to show that the eigenvalues of \(A\) are \(a - b\) and \(a + \left( {n - {\bf{1}}} \right)b\). What are the multiplicities of these eigenvalues?

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

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

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

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