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!

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.

Re: Finding eigenvectors- am I arranging it wrong?

Literally just punched the air. Thank you very much, much appreciated!