Finding eigenvalues and eigenvectors 2x2 matrix

[andrey21]

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??

 

[micromass]

Yes, this is good!

 

[andrey21]

Great from there I get a little confused finding the eigenvectors. I sub the lambda values back into matrix but don't know where to go there.

lambda = 2 1-2 1
1 1-2

giving me: -1 1
1 -1

Where do I go next?

 

[micromass]

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...

 

[andrey21]

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?

 

[micromass]

Yes, so all vectors of the form $$(x_1,x_1)$$ are eigenvectors. Specifically, (1,1) is an eigenvector with eigenvalue 2...

 

[andrey21]

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) ?

 

[micromass]

Yes, that is correct! (1,-1) is indeed an eigenvector of the matrix!

 

[andrey21]

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?

 

[micromass]

Correct! It seems you've got the idea!

 

[andrey21]

thnx micromass:)

 

[Mark44]

So you sub lambda into your matrix. This gives you a matrix A.

Actually that gives you A - $\lambda$I, which is different from A (unless $\lambda$ 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 - $\lambda$I)x = 0.

This involves solving a system of equations. The solutions of that system should give you the eigenvectors...

 

[andrey21]

Just quick question micromass for the post 7 I've sed eigenvector is (1,-1) should it be (-1,1) or are they both acceptable?

 

[micromass]

Both are good. If x is an eigenvector and if $$\lambda\neq 0$$, then $$\lambda x$$ is an eigenvector as well. In this case: (1,-1) is an eigenvector. So take $$\lambda=-1$$, then (-1,1) is an eigenvector as well!

 

[andrey21]

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) ?

 

[Mark44]

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 - $\lambda$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 $\lambda$ by verifying that Ax = $\lambda$x, or equivalently, that (A - $\lambda$I)x = 0.