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Q3SE

Expert-verifiedFound in: Page 267

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.

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.

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.

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