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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 \(\lambda = 4\).

See the step by step solution

Step by Step Solution

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 \(\lambda = 4\).

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