Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Finding Eigenvectors from Eigenvalues

Date: 08/24/2005 at 01:15:17
From: Matthew
Subject: finding eigenvector from eigenvalue

Given the matrix

      | 0  0 -2|
  A = | 0 -2  0|
      |-2  0  3|

find the eigenvalues, and then the corresponding eigenvectors.

I can find the eigenvalues, but my eigenvectors turn out to be just
zero, and I thought the whole point of an eigenvector was that it
didn't equal 0?

I have solved the determinant of A-lI to get eigenvalues of -1, 4, and
-2, and I checked that they are correct by checking that the product
of those eigenvalues equals the determinant of A (-1*4*-2 = 8 = detA).

However, plugging the values in to find the eigenvectors I get
something like (for l = -1):

  |-1  0 -2| |x|   |0|
  | 0 -3  0| |y| = |0|
  |-2  0  2| |z|   |0|

which follows through to give me x = 0, y = 0, z = 0.

I am confused, I thought there were supposed to be an infinite number
of eigenvectors for each eigenvalue, or at least they are not supposed
to be zero.  Pointing out what I am doing wrong would be most appreciated.



Date: 08/24/2005 at 08:19:37
From: Doctor George
Subject: Re: finding eigenvector from eigenvalue

Hi Matthew,

Thanks for writing to Doctor Math.

Let's work out an eigenvector for the eigenvalue of -2.  We want to
find the column vector (x,y,z) such that

  _         _  _   _        _   _
  | 0   0  -2| | x |        | x |
  | 0  -2   0| | y |  =  -2 | y |
  |-2   0   3| | z |        | z |
  -         -  -   -        -   -

If we multiply the first row by the vector we get

  -2z = -2x
    z = x           (1)

If we multiply the second row by the vector we get

  -2y = -2y         (2)

If we multiply the third row by the vector we get

  -2x + 3z = -2z
       -2x = -5z    (3)       

Equations (1) and (3) require that x = 0 and z = 0.  Equation (2) is
true for all y, so we can pick y = 1 for convenience.  Now we have an
eigenvector for the eigenvalue of -2.
  _   _
  | 0 |
  | 1 |
  | 0 |
  -   -

Since the eigenvector is only unique to within a scale factor, we
will always find them by making convenient choices for one of the
values, and then solving for the others.

Does that make sense?  See if you can work out the other eigenvectors now.

Write again if you need more help.

- Doctor George, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 08/24/2005 at 09:04:16
From: Matthew
Subject: Thank you (finding eigenvector from eigenvalue)

Wow, thank you so much.  That problem was driving me crazy.  I have
worked out all my eigenvectors now! :-)
Associated Topics:
College Linear Algebra

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/