PlanetMath: finding eigenvalues

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (205)
Orphanage
Unclass'd
Unproven (490)
Corrections (8)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

finding eigenvalues

(Example)

This example investigates eigenvalues and the similarity transformation used to diagonalize matrices. We seek the eigenvalues of the matrix below. Afterward, we can transform this matrix into a diagonal matrix which has many useful applications.

$\displaystyle A=\left( \begin{array}{cc} 2 & 1 \ 1 & 2 \end{array} \right )$

Here, we need to solve the corresponding matrix equation;

$\displaystyle \left( \begin{array}{cc} 2 & 1 \ 1 & 2 \end{array} \right ) \le... ...rray} \right )=\lambda \left( \begin{array}{c} x_1 \ x_2 \end{array} \right )$

$\displaystyle AX=\lambda X$

rearranging gives

$\displaystyle AX-\lambda X=0$

$\displaystyle (A-\lambda I) X=0$

We seek the values for $\lambda$ and

. First, we need to solve the characteristic equation of

. We do this by finding $det(A-\lambda I)$ . First, calculating $A-\lambda I$ gives;

$\displaystyle A -\lambda I= \left( \begin{array}{cc} 2-\lambda & 1 \ 1 & 2-\lambda \end{array} \right )$

Next, calculating $det(A-\lambda I)$ yields

$\displaystyle det(A-\lambda I)=(2-\lambda)^2-1=\lambda^2-4\lambda+3=(\lambda-1)(\lambda-3)=0$

Substituting $\lambda=1$ into $(A-\lambda I)X$ gives...

$\displaystyle \left\{ \begin{array}{c} x_1+x_2=0 \ x_1+x_2=0 \end{array} \right.$

so that

and the corresponding eigenvector is

$\displaystyle \left( \begin{array}{c} t \ -t \end{array} \right )=t \left( \begin{array}{c} 1 \ -1 \end{array} \right )$

where $t\ne 0.$
Substituting $\lambda=3$ gives...

$\displaystyle \left\{ \begin{array}{c} -x_1+x_2=0 \ x_1-x_2=0 \end{array} \right.$

so that

and the corresponding eigenvector is

$\displaystyle \left( \begin{array}{c} t \ t \end{array} \right )=t \left( \begin{array}{c} 1 \ 1 \end{array} \right )$

where $t\ne 0.$
Finally, to diagonalize

we let the eigenvectors be the columns of a new matrix

$\displaystyle P=\left( \begin{array}{cc} 1 & -1 \ 1 & 1 \end{array} \right )$

and then since our eigenvectors are linearly independent we can also find;

$\displaystyle P^{-1}=\frac{1}{2} \left( \begin{array}{cc} -1 & 1 \ 1 & 1 \end{array} \right )$

then we create a diagonal matrix as follows...

$\displaystyle D=P^{-1}AP=\left( \begin{array}{cc} 1 & 0 \ 0 & 3 \end{array} \right )$

Computing powers of

is a very useful application of

. Solving for

lets us compute powers of

$\displaystyle A=PDP^{-1}$

so that

$\displaystyle A^n=PD^nP^{-1}$

$\displaystyle A^n=P\left( \begin{array}{cc} 1^n & 0 \ 0 & 3^n \end{array} \right )P^{-1}$

"finding eigenvalues" is owned by PrimeFan. [ full author list (2) | owner history (2) ]

(view preamble)

Keywords:

eigenvalue, eigenvector

This object's parent.

Log in to rate this entry.
(view current ratings)

Cross-references: powers, linearly independent, columns, eigenvector, characteristic equation, equation, diagonal matrix, Transform, matrices, diagonalize, similarity transformation, eigenvalues
There is 1 reference to this entry.

This is version 2 of finding eigenvalues, born on 2006-04-27, modified 2008-02-15.
Object id is 7873, canonical name is FindingEigenvalues.
Accessed 7484 times total.

Classification:

AMS MSC:

15A18 (Linear and multilinear algebra; matrix theory :: Eigenvalues, singular values, and eigenvectors)

Pending Errata and Addenda

None.[ View all 1 ]

Discussion

forum policy

No messages.

Interact

post | correct | update request | add example | add (any)