- #1

- 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??

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter andrey21
- Start date

- #1

- 476

- 0

1,1

1,1

1-lambda 1

1 1-lambda

Giving me:

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

lambda = 0 lambda = 2

Is this correct??

- #2

- 22,178

- 3,304

Yes, this is good!

- #3

- 476

- 0

lambda = 2 1-2 1

1 1-2

giving me: -1 1

1 -1

Where do I go next?

- #4

- 22,178

- 3,304

- #5

- 476

- 0

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

- 22,178

- 3,304

- #7

- 476

- 0

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

- 22,178

- 3,304

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

- #9

- 476

- 0

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

- 22,178

- 3,304

Correct! It seems you've got the idea!

- #11

- 476

- 0

thnx micromass:)

- #12

Mentor

- 37,236

- 9,405

Actually that gives you A - [itex]\lambda[/itex]I, which is different from A (unless [itex]\lambda[/itex] happens to be zero).So you sub lambda into your matrix. This gives you a matrix A.

I know what you mean, but what you really want is to find a solution of (A - [itex]\lambda[/itex]I)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...

- #13

- 476

- 0

- #14

- 22,178

- 3,304

- #15

- 476

- 0

3,1

0,0

eigenvector is (-1/3,1) ?

- #16

Mentor

- 37,236

- 9,405

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

Share:

- Replies
- 10

- Views
- 687

- Replies
- 5

- Views
- 395

- Replies
- 4

- Views
- 449

- Replies
- 8

- Views
- 543

- Replies
- 5

- Views
- 424

- Replies
- 8

- Views
- 435

- Replies
- 8

- Views
- 959

- Replies
- 19

- Views
- 3K

- Replies
- 5

- Views
- 389

- Replies
- 3

- Views
- 1K