Eigenvectors and Eigenspace

  • Thread starter jjones1573
  • Start date
  • #1
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Homework Statement


Im looking at finding the eigenvectors of a matrix but also a basis for the eigenspace

A = [ 6 16 ]
[ -1 -4 ]

lambda = 4
lambda = -2


Homework Equations


(A - lambda I ) v = 0


The Attempt at a Solution



So with the above equation I get:

for lambda = 4

[ 6 - 4 16 ] [ v1 ] = [ 0 ]
[ -1 -4 - 4 ] [ v2 ] [ 0 ]

so

2 v1 + 16 v2 = 0
-v1 - 8v2 = 0

so v1 = 8v2

and the basis for the eigenspace is span [ 8 ]
[ 1 ]

First is that right? because when I put it into an eigenvector calculator on the web it gives me
-8 instead of 8 but I cant see how I could get to that.

Second if this is the basis for the eigenspace then how can I find the eigenvectors for the eigenvalue?

thanks,
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
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2 v1 + 16 v2 = 0
-v1 - 8v2 = 0

so v1 = 8v2

You made a sign error. v1=-8v2

ehild
 
  • #3
22
0
Oh yeah thats right thanks.

Is it as simple as the vector is:

[-8v2]
[v2]

and the eigenspace is: span

[-8]
[1]
 
  • #4
22
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Sorry I realised this should have been posted in the calculus section would it be possible to have it moved?

I think what I have put above for the eigenspace is correct? But what about the eigenvector I cant seem to understand what this is.
 
  • #5
ehild
Homework Helper
15,543
1,914
(-8,1) multiplied by any number is an eigenvector. You need to find the other one, which belongs to the other eigenvalue lambda=2.
The two eigenvectors are the basis of the "eigenspace". You can choose the normalised vectors as basis.

ehild
 
  • #6
22
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Oh thanks. Do I need to normalise the vectors or is it fine to just find the two vectors and give that?
 
  • #7
ehild
Homework Helper
15,543
1,914
You do not need to normalize in principle.

ehild
 
  • #8
22
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ok thanks.
 

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