# Finding eigenvalues and eigenvectors 2x2 matrix

• andrey21

#### 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??

Yes, this is good!

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?

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...

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?

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

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) ?

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

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?

Correct! It seems you've got the idea!

thnx micromass:)

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...

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?

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!

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) ?

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.
The text passage contains a detailed discussion about finding eigenvalues and eigenvectors of a 2x2 matrix. It includes a step-by-step process, questions from the user, and answers from an expert. The user asks multiple mathematical questions, and the expert provides answers and confirms the user's solutions.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Finding eigenvalues and eigenvectors 2x2 matrix

• andrey21

#### 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??

Yes, this is good!

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?

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...

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?

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

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) ?

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

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?

Correct! It seems you've got the idea!

thnx micromass:)

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...

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?

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!

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) ?

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.
user
The text passage contains a detailed discussion about finding eigenvalues and eigenvectors of a 2x2 matrix. It includes a step-by-step process, questions from the user, and answers from an expert. The user asks multiple mathematical questions, and the expert provides answers and confirms the user's solutions.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS