# Homework Help: Basis of the eigenspace

1. Sep 12, 2009

### ihumayun

1. The problem statement, all variables and given/known data

The matrix A=
2 0 4
-2 0 -4
-1 0 -2

has one real eigenvalue. Find this eigenvalue and a basis of the eigenspace.

Eigenvalue =

Basis ( , , )T , ( , , )T

2. Relevant equations

3. The attempt at a solution

I have found the eigenvalue to be 0, but I can't seem to figure out how to come up with 2 bases. I found the first one:

2x + 4z = 0
-2x - 4z = 0
-1x - 1z = 0

All three equations reduce to :
x + 2z = 0

So then,

x = t , y = 0, and z = -(1/2)t
which gives a basis of (1, 0, -1/2)T

I'm not completely sure that I have calculated this correctly, and I have no idea how to come up with the second basis. The eigenvalue I calculated was marked correct, though. If anyone could help me, I would greatly appreciate it. :)

2. Sep 12, 2009

### CFDFEAGURU

You only have one eigenvalue so you will have only one basis.

Your question states "Find this eigenvalue and a basis of the eigenspace".

Correct me if I am wrong but you should only need to find one basis.

Thanks
Matt

3. Sep 12, 2009

### Hurkyl

Staff Emeritus
Could you show your work here?

4. Sep 12, 2009

### ihumayun

I also thought that there should be only one basis, but the question asks for 2, and my prof confirmed this. I still don't understand how this would work though.

My work:
x+2z=0
2z = -x
z = - x/2

So if I set x with the parameter t,

x=t
y=0
and z = -t/2, or (-1/2) t

Sorry I didn't include this before .

5. Sep 12, 2009

### Hurkyl

Staff Emeritus
y doesn't appear in that equation -- how did you conclude something about it?

6. Sep 12, 2009

### ihumayun

Because the equation was

x +2z = 0 or
1x + 0y +2z = 0

y is nonexistent in this case. I assumed that would mean it would be equal to 0 in the basis, but like I said, I wasn't sure if I had calculated the basis correctly. How would I find a value for y?

7. Sep 12, 2009

### HallsofIvy

All equations reduce to x+ 2z= 0 so x= -2z. z can be anything at all and y can be anything at all. Your basis contains two vectors.

8. Sep 12, 2009

### ihumayun

So how would I find the bases in that case?

9. Sep 13, 2009

### ihumayun

Oh, I just got it.
x = -2t
y = s, not 0
and z = t

which means that the vector would be:

(-2t , s, t)T = (-2t, 0, t)T + ( 0, s, 0)T

which gives the bases

(-2, 0, 1)T, and (0, 1, 0)T

10. Sep 13, 2009

### ihumayun

Thank you for your help!