# Finding eigenvectors- am I arranging it wrong?

• Apr 8th 2012, 07:05 AM
rpowell
Finding eigenvectors- am I arranging it wrong?
Hello, I am trying to get to grips with eigenvectors and seem to do ok until I get to one question, so I am wondering if perhaps the lecturer has given the wrong answer. The matrix is:

5 0 6
0 11 6
6 6 -2

And the eigenvalues are 14, 7 and -7. The lecturer gets the following eigenvectors:

14
2
6
3

7
6
-3
2

-7
3
2
-6

I can't seem to get these, mainly the y values. For example, I will do for lambda=14

-9 0 6 1)-9x +6z=0
0 -3 6 2) -3y+6z=0
6 6 -16 3)6x+6y-16z=0

1) 6z=9x (2z=3x)
2) 6z=3y (2z=y)

So I think that should be

3
1
2

I'm sure my problem lies with how I arrange the numbers in the eigenvector answer, but I just don't know, it always seems to be the part I get wrong. Would someone mind explaining how to do it properly? Thanks in advance, I have been banging my head against this for days now and will be very thankful if someone can enlighten me!
• Apr 8th 2012, 07:46 AM
Deveno
Re: Finding eigenvectors- am I arranging it wrong?
Quote:

Originally Posted by rpowell
Hello, I am trying to get to grips with eigenvectors and seem to do ok until I get to one question, so I am wondering if perhaps the lecturer has given the wrong answer. The matrix is:

5 0 6
0 11 6
6 6 -2

And the eigenvalues are 14, 7 and -7. The lecturer gets the following eigenvectors:

14
2
6
3

7
6
-3
2

-7
3
2
-6

I can't seem to get these, mainly the y values. For example, I will do for lambda=14

-9 0 6 1)-9x +6z=0
0 -3 6 2) -3y+6z=0
6 6 -16 3)6x+6y-16z=0

1) 6z=9x (2z=3x)
2) 6z=3y (2z=y)

you're fine up to here.

Quote:

So I think that should be

3
1
2

I'm sure my problem lies with how I arrange the numbers in the eigenvector answer, but I just don't know, it always seems to be the part I get wrong. Would someone mind explaining how to do it properly? Thanks in advance, I have been banging my head against this for days now and will be very thankful if someone can enlighten me!
suppose we pick x = 1. then 2z = 3x, means 2z = 3, and z = 3/2.

then 2z = y means 2(3/2) = y, so that y = 3. this gives us the eigenvector (1,3,3/2). since any scalar multiple of this is also an eigenvector, we can multiply by 2 to clear out the fractions, giving us the eigenvector (for the eigenvalue λ = 14), 2(1,3,3/2) = (2,6,3), which is the same eigenvector your instructor got.
• Apr 8th 2012, 08:00 AM
rpowell
Re: Finding eigenvectors- am I arranging it wrong?
Literally just punched the air. Thank you very much, much appreciated!