Finding eigenvalues and eigenvectors 2x2 matrix

In summary: So (-1/3, 1) is just one possible eigenvector for this particular eigenvalue 0, but there are infinitely many others. Does that make sense?In summary, to find the eigenvalues and corresponding eigenvectors of a matrix, you first solve for lambda by setting the determinant of the matrix minus lambda times the identity matrix equal to 0. Then, for each lambda value, you solve a system of equations by substituting lambda back into the matrix and finding a vector x such that (A - lambdaI)x = 0. This vector x is the corresponding eigenvector for that lambda value.
  • #1
andrey21
476
0
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??
 
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  • #2
Yes, this is good!
 
  • #3
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?
 
  • #4
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
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
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
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
Yes, that is correct! (1,-1) is indeed an eigenvector of the matrix!
 
  • #9
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
Correct! It seems you've got the idea!
 
  • #11
thnx micromass:)
 
  • #12
micromass said:
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).
micromass said:
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.
micromass said:
This involves solving a system of equations. The solutions of that system should give you the eigenvectors...
 
  • #13
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?
 
  • #14
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
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
andrey21 said:
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.
 

1. What is an eigenvalue and eigenvector?

An eigenvalue is a scalar value that represents a specific property of a matrix. An eigenvector is a non-zero vector that is associated with a particular eigenvalue and is only scaled by that eigenvalue when the matrix is multiplied by it.

2. How do I find the eigenvalues of a 2x2 matrix?

To find the eigenvalues of a 2x2 matrix, you can use the characteristic equation. Set the determinant of the matrix minus the identity matrix to equal 0 and solve for the eigenvalues.

3. Can a 2x2 matrix have more than two eigenvalues?

No, a 2x2 matrix can only have two eigenvalues. This is because the characteristic equation of a 2x2 matrix is a quadratic equation, which can only have two solutions.

4. How do I find the eigenvectors of a 2x2 matrix?

To find the eigenvectors of a 2x2 matrix, you can use the eigenvalue-eigenvector equation. Substitute each eigenvalue into the equation and solve for the corresponding eigenvector.

5. What is the significance of eigenvalues and eigenvectors?

Eigenvalues and eigenvectors have many applications in various fields such as physics, engineering, and computer science. They help in understanding the behavior and properties of a matrix, and can be used to solve systems of linear equations, analyze data, and perform transformations on vectors and matrices.

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