• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

### Select your language

Suggested languages for you:

Americas

Europe

Q10E

Expert-verified
Found in: Page 267

### 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

### Answers without the blur.

Just sign up for free and you're in.

# 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 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 $$\lambda = 4$$.

### Want to see more solutions like these?

Sign up for free to discover our expert answers

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.