- #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??

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- Thread starter andrey21
- Start date

- #1

- 466

- 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

- 22,089

- 3,296

Yes, this is good!

- #3

- 466

- 0

lambda = 2 1-2 1

1 1-2

giving me: -1 1

1 -1

Where do I go next?

- #4

- 22,089

- 3,296

- #5

- 466

- 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

- 22,089

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- #7

- 466

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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

- 22,089

- 3,296

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

- #9

- 466

- 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

- 22,089

- 3,296

Correct! It seems you've got the idea!!

- #11

- 466

- 0

thnx micromass:)

- #12

Mark44

Mentor

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- 6,598

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

- 466

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- #14

- 22,089

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- #15

- 466

- 0

3,1

0,0

eigenvector is (-1/3,1) ???

- #16

Mark44

Mentor

- 34,866

- 6,598

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

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