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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 # Question: Is $$\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)$$ an eigenvalue of $$\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)$$? If so, find the eigenvalue.

No, $$\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)$$ is not the eigenvector of the given matrix $$\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)$$.

See the step by step solution

## Step 1: Definition of eigenvector

If there exists a non-zero vector $${\bf{x}}$$ which satisfies $$A{\bf{x}} = \lambda {\bf{x}}$$ for some scaler $$\lambda$$, then $${\bf{x}}$$ be the eigenvector of an $$n \times n$$ matrix $$A$$.

## Step 2: Determine whether $$\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)$$ is the eigenvalue of the given matrix

Denote the given matrix by $$A$$ and the given vector by $${\bf{x}}$$.

$$A = \left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)$$, $${\bf{x}} = \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)$$

According to the definition of eigenvalue, $${\bf{x}} = \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)$$ is the eigenvector of the matrix $$A$$, if $$A{\bf{x}} = \lambda {\bf{x}}$$.

Find the product of $$A$$, and $${\bf{x}}$$.

$$\begin{array}{c}A{\bf{x}} = \left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1\\{29}\end{array}} \right)\end{array}$$

It can be observed that $$\left( {\begin{array}{*{20}{c}}1\\{29}\end{array}} \right) \ne \lambda \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)$$

So, $$\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)$$ is not the eigenvector of the given matrix $$\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)$$. ### Want to see more solutions like these? 