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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

# Question: Is $$\lambda = - 2$$ an eigenvalue of $$\left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)$$? Why or why not?

Yes, $$\lambda = - 2$$ is the eigenvalue of the given matrix $$\left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)$$, because there exists a nontrivial solution of $$A{\bf{x}} = - 2{\bf{x}}$$ and columns of the matrix $$\left( {A + 2I} \right)$$ are linearly dependent.

See the step by step solution

## Step 1: Definition of 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 an eigenvalue of the matrix $$A$$ if there exists a nontrivial solution $${\bf{x}}$$ of $$A{\bf{x}} = \lambda {\bf{x}}$$.

## Step 2: Determine whether $$\lambda = 2$$ is the eigenvalue of given matrix

Denote the given matrix by $$A$$.

$$A = \left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)$$

According to the definition of eigenvalue, $$\lambda = - 2$$ is the eigenvalue of the matrix $$A$$, if satisfies the equation $$A{\bf{x}} = \lambda {\bf{x}}$$.

Substitute $$\lambda = - 2$$ into $$A{\bf{x}} = \lambda {\bf{x}}$$.

$$A{\bf{x}} = - 2{\bf{x}}$$

The obtained equation is equivalent to $$\left( {A + 2I} \right){\bf{x}} = 0$$, which is a homogeneous equation.

So, first, solve the matrix $$\left( {A + 2I} \right)$$ by using $$A = \left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)$$.

$$\begin{array}{c}\left( {A + 2I} \right) = \left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}2&0\\0&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}9&3\\3&1\end{array}} \right)\end{array}$$

It can be observed that the columns of the obtained matrix are linearly dependent, where elements of the second column is three multiple of the elements of the first column, which can be written as:

$$\left( {A + 2I} \right) = \left( {\begin{array}{*{20}{c}}{3\left( 3 \right)}&3\\{3\left( 1 \right)}&1\end{array}} \right)$$

Hence, $$\left( {A + 2I} \right){\bf{x}} = 0$$ has a nontrivial solution, so $$\lambda = - 2$$ is an eigenvalue of the given matrix $$\left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)$$.