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Q10E

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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

In Exercises 9–16, find a basis for the eigenspace corresponding to each listed eigenvalue.

10. \(A = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right)\), \(\lambda = 4\)

\(\left( {\begin{array}{*{20}{c}}{\frac{3}{2}}\\1\end{array}} \right)\) is the basis for the eigenspace for \(\lambda = 4\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definitions

Eigenvalue: Let \(\lambda \) is a scaler, \(A\) is an \(n \times n\) matrix and \({\bf{x}}\) is an eigenvector corresponding to \(\lambda \), \(\lambda \) is said to be an eigenvalue of the matrix \(A\) if there exists a nontrivial solution \({\bf{x}}\) of \(A{\bf{x}} = \lambda {\bf{x}}\).

Eigenvector: For a \(n \times n\) matrix \(A\), whose eigenvalue is \(\lambda \), the set of a subspace of \({\mathbb{R}^n}\) is known as an eigenspace, where the set of the subspace of is the set of all the solutions of \(\left( {A - \lambda I} \right){\bf{x}} = 0\).

Step 2: Find a basis of eigenspace for \(\lambda  = 4\)

The given matrix is \(A = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right)\), where \(\lambda = 4\).

As, \(\lambda = 4\) is the eigenvalue of the matrix \(A\), so it satisfies the equation \(A{\bf{x}} = \lambda {\bf{x}}\).

For \(\lambda = 4\), solve \(\left( {A - \lambda I} \right){\bf{x}} = 0\), for which first evaluate \(\left( {A - \lambda I} \right)\).

\(\begin{array}{c}\left( {A - 4I} \right) = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right) - 4\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}4&0\\0&4\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}6&{ - 9}\\4&{ - 6}\end{array}} \right)\end{array}\)

Write the obtained matrix in the form of an augmented matrix, where for \(A{\bf{x}} = 0\), the augmented matrix given by \(\left( {\begin{array}{*{20}{c}}A&0\end{array}} \right)\).

\(\left( {\begin{array}{*{20}{c}}6&{ - 9}&0\\4&{ - 6}&0\end{array}} \right)\)

The obtained matrix is not in a reduced form, so reduce it in row echelon form by applying row operations.

\(\begin{gathered} \hfill \left( {\begin{array}{*{20}{c}} 6&{ - 9}&0 \\ 4&{ - 6}&0 \end{array}} \right)\xrightarrow{{{R_1} \to \frac{{{R_1}}}{6}}}\left( {\begin{array}{*{20}{c}} 1&{ - \frac{3}{2}}&0 \\ 4&{ - 6}&0 \end{array}} \right) \\ \hfill \xrightarrow{{{R_2} \to {R_2} - 4{R_1}}}\left( {\begin{array}{*{20}{c}} 1&{ - \frac{3}{2}}&0 \\ 0&0&0 \end{array}} \right) \\ \end{gathered} \)

Write a system of equations corresponding to the obtained matrix.

\(\begin{array}{c}{x_1} - \frac{3}{2}{x_2} = 0\\{x_2},{\rm{ free variable}}\end{array}\)

As \({x_2}\) is a free variable, let \({x_2} = 1\). Then;

\(\begin{array}{c}{x_1} = \frac{3}{2}\\{x_2} = 1\end{array}\)

So, the general solution is given as:

\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}{\frac{3}{2}}\\1\end{array}} \right)\)

So, \(\left( {\begin{array}{*{20}{c}}{\frac{3}{2}}\\1\end{array}} \right)\) is the basis for the eigenspace for \(\lambda = 4\).

Most popular questions for Math Textbooks

Question 19: Let \(A\) be an \(n \times n\) matrix, and suppose A has \(n\) real eigenvalues, \({\lambda _1},...,{\lambda _n}\), repeated according to multiplicities, so that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) . Explain why \(\det A\) is the product of the n eigenvalues of A. (This result is true for any square matrix when complex eigenvalues are considered.)

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

Consider the growth of a lilac bush. The state of this lilac bush for several years (at year’s end) is shown in the accompanying sketch. Let n(t) be the number of new branches (grown in the year t) and a(t) the number of old branches. In the sketch, the new branches are represented by shorter lines. Each old branch will grow two new branches in the following year. We assume that no branches ever die.

(a) Find the matrix A such that [nt+1at+1]=A[ntat]

(b) Verify that [11] and [2-1] are eigenvectors of A. Find the associated eigenvalues.

(c) Find closed formulas for n(t) and a(t).

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{array}{*{20}{c}}4\\6\end{array}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{array}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{array}{*{20}{c}}6\\{ - 2}\\3\end{array}} \right)\)

3. \(\frac{1}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{\mathop{\rm w}\nolimits} \)
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