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Finding eigenvalues and eigenvectors 2x2 matrix

  • Thread starter andrey21
  • Start date
  • #1
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Find the eigenvalues and corresponding eigenvectors of the following matrix.

1,1
1,1


Here is my attempt to find eigenvalues:


1-lambda 1
1 1-lambda

Giving me:

(Lambda)^2 -2(lambda) = 0

lambda = 0 lambda = 2

Is this correct??
 

Answers and Replies

  • #2
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Yes, this is good!
 
  • #3
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Great from there I get a little confused finding the eigenvectors. I sub the lambda values back into matrix but dont know where to go there.

lambda = 2 1-2 1
1 1-2

giving me: -1 1
1 -1

Where do I go next?
 
  • #4
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So you sub lambda into your matrix. This gives you a matrix A. Now you need to find a vector x such that Ax=0. This involves solving a system of equations. The solutions of that system should give you the eigenvectors...
 
  • #5
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Rite so I have the two following matrices from lambda values:

lambda = 2
-1, 1
1,-1

so that means -x1 + X2 = 0 AND x1 -x2 = 0 therefore x1 = x2

Am i on the rite track???
 
  • #6
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3,280
Yes, so all vectors of the form [tex](x_1,x_1)[/tex] are eigenvectors. Specifically, (1,1) is an eigenvector with eigenvalue 2...
 
  • #7
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Great so when eigenvalue = 2 eigenvector = (1,1)

When lambda = 0 I get the following matrix:

1 1
1 1

Now is this just saying x1 + x2 = 0 so x1 = -x2

eigenvector be (1,-1) ?????
 
  • #8
22,097
3,280
Yes, that is correct!! (1,-1) is indeed an eigenvector of the matrix!
 
  • #9
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Fantastic thank you micromass.

Can I just ask you if this is correct?

matrix A = 2, 1
0, -1

Eigenvalues I get are:

2 and -1:

when lambda = 2:

0, 1
0, 0 So x2 = 0 therefore eigenvector is (1,0)

when Lambda = -1

3,1
0,0 n so 3x1 + x2 = 0 3x1 = -x2 so eigenvector is (-1/3, 1)

Correct?
 
  • #10
22,097
3,280
Correct! It seems you've got the idea!!
 
  • #11
466
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thnx micromass:)
 
  • #12
33,503
5,190
So you sub lambda into your matrix. This gives you a matrix A.
Actually that gives you A - [itex]\lambda[/itex]I, which is different from A (unless [itex]\lambda[/itex] happens to be zero).
Now you need to find a vector x such that Ax=0.
I know what you mean, but what you really want is to find a solution of (A - [itex]\lambda[/itex]I)x = 0.
This involves solving a system of equations. The solutions of that system should give you the eigenvectors...
 
  • #13
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Just quick question micromass for the post 7 Ive sed eigenvector is (1,-1) should it be (-1,1) or are they both acceptable?
 
  • #14
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3,280
Both are good. If x is an eigenvector and if [tex]\lambda\neq 0[/tex], then [tex]\lambda x[/tex] is an eigenvector as well. In this case: (1,-1) is an eigenvector. So take [tex]\lambda=-1[/tex], then (-1,1) is an eigenvector as well!
 
  • #15
466
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Ah rite I see so when lambda = 0 eigenvector is (1,-1)? Its just confusing as when matrix wa:

3,1
0,0

eigenvector is (-1/3,1) ???
 
  • #16
33,503
5,190
Ah rite I see so when lambda = 0 eigenvector is (1,-1)? Its just confusing as when matrix wa:

3,1
0,0

eigenvector is (-1/3,1) ???
Assuming this is A - [itex]\lambda[/itex]I, yes, <-1/3, 1> is an eigenvector, and so are <-1, 3>, <1/3, -1>, <1, -3>, and many more. You can easily check whether a vector x is an eigenvector with associated eigenvalue [itex]\lambda[/itex] by verifying that Ax = [itex]\lambda[/itex]x, or equivalently, that (A - [itex]\lambda[/itex]I)x = 0.
 

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