# Finding a basis of eigenvectors

1. Aug 2, 2012

### Catchfire

1. The problem statement, all variables and given/known data
$A = \left( \begin{array}{ccc} 2 & 0 & -1 \\ 4 & 1 & -4 \\ 2 & 0 & -1 \end{array} \right)$

Find the eigenvalues and corresponding eigenvectors that form a basis over R3

2. Relevant equations

3. The attempt at a solution

OK so I've found the characteristic polynomial: -λ(λ-1)2
so I know my eigenvalues are 0,1,1

Then to find the eigenvectors I sub the eigenvalues in to the matrix A - λI

$A - 0I = \left( \begin{array}{ccc} 2 & 0 & -1 \\ 4 & 1 & -4 \\ 2 & 0 & -1 \end{array} \right)$

then I solve:
2x -z = 0
4x + y -4z = 0

z = 2x
y = 4x

so my eigenvector is (1,4,2)

$A - 1I = \left( \begin{array}{ccc} 1 & 0 & -1 \\ 4 & 0 & -4 \\ 2 & 0 & -2 \end{array} \right)$

x = z

so the eigenvector is (1,0,1)

Now I'm out of new eigenvalues to substitute. How do I find the last eigenvector?

2. Aug 2, 2012

### voko

The rank of the second matrix is 1, so there are two independent eigenvectors to get from it.

3. Aug 2, 2012

### Catchfire

So are you saying dim(A) = rank(A) + nullity(A)
and since dim(A-I) = 3 and rank(A-I) = 1, then nullity(A-I) = 2, implying I need two vectors from A-I.

Ahh I see where I messed up, since there is no y values I can write y = a for any a in R.
So really my eigenvectors are (1,a,1) and (0,b,0).

Is all of this correct?

4. Aug 2, 2012

### Ray Vickson

What is preventing you from substituting these vectors into the equation to check if they work?

RGV

Last edited: Aug 2, 2012
5. Aug 2, 2012

### Catchfire

Nothing, but I already know the answers to the problem. I just wanted to make sure my reasoning was correct and that I understood what theorem voko was citing.

Can you lend a hand with any of that? It would much appreciated if you could.

6. Aug 3, 2012

### voko

I do not remember whether that theorem has any particular name, but it states that the dimension of the solution space of a homogeneous linear system is the dimension of the system minus its rank. And you got that correctly.

7. Aug 3, 2012

### Ray Vickson

In the present case the 3x3 matrix is so simple that one can spot immediately that it has rank 1. In more complex cases (say a 5x5 or a 10x10 matrix), finding the rank involves essentially the same algorithm that one uses to solve the system Ax=0 (Gaussian elimination/row-reduction), so you determine the dimension of the null space (and find a basis for it) at the same time that you find the rank.

RGV

8. Aug 3, 2012

### voko

Well, in the case of eigenvectors one knows the rank even before that.

9. Aug 3, 2012

### Ray Vickson

No. The algebraic and geometric miltiplicities may differ, in which case the dimension of the null space can be less than the multiplicity. That is why the Jordan form is sometimes non-diagonal.

RGV

10. Aug 3, 2012

### Catchfire

Thanks for the responses, I appreciate the help.