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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

# Suppose $${\bf{x}}$$ is an eigenvector of $$A$$ corresponding to an eigenvalue $$\lambda$$.a. Show that $$x$$ is an eigenvector of $$5I - A$$. What is the corresponding eigenvalue?b. Show that $$x$$ is an eigenvector of $$5I - 3A + {A^2}$$. What is the corresponding eigenvalue?

(a) It is proved that $$x$$ is an eigenvector of $$5I - A$$. Therefore,$$\;5 - \lambda$$ is the eigenvalue.

(b) It is proved that $$x$$ is an eigenvector of $$5I - 3A + {A^2}$$. Therefore, $$5 - 3\lambda + {\lambda ^2}$$ is the eigenvalue.

See the step by step solution

## Step 1: Definition of matrix and eigenvalue

A matrix (plural matrices) is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns, which is used to represent a mathematical object or a property of such an object.

The eigenvalue $$\lambda$$ is the real or complex number of a matrix $$A$$ which is a square matrix that satisfies the following equation

$$\det \left( {A - \lambda I} \right) = 0$$

.

This equation is called the characteristic equation.

## Step 2: Find the eigenvector of matrix product

Let $$\lambda$$ be an eigenvalue of the matrix $$A$$. Then we have

$$Ax = \lambda x$$

(a)

Simplify $$(5I - A)x$$.

$$\begin{array}{r}(5I - A)x = 5x - Ax\\ = 5x - \lambda x\\ = (5 - \lambda I)x\\ = (5 - \lambda )x\end{array}$$

Therefore$$\;5 - \lambda$$ is the eigenvalue.

b)

Consider,

$$\begin{array}{c}\left( {5I - 3A + {A^2}} \right)x = 5x - 3Ax + {A^2}x\\ = 5x - 3\lambda x + A\lambda x\\ = 5x - 3\lambda x + \lambda Ax\\ = 5x - 3\lambda x + {\lambda ^2}x\\ = \left( {5 - 3\lambda + {\lambda ^2}} \right)x\end{array}$$

Therefore $$5 - 3\lambda + {\lambda ^2}$$ is the eigenvalue.