# Finding an eigenvector of 3x3 matrix

1. Feb 1, 2013

### enc08

Hi,

I'm trying to find an eigenvector of a matrix. I know that λ = 1, so my matrix (A - λI) is
$$[-0.5253, 0.8593, -0.1906; -0.8612, -0.5018, 0.1010; 0.1817, 0.1161, -0.0236]$$

And from rows 2 and 3 I get these simultaneous equations

$$-0.8612t_{1}-0.5018t_{2}+0.1010t_{3}=0$$
$$0.1817t_{1}+0.1161t_{2}+0.0236t_{3}=0$$

I eliminate to find $t_{2} = -4.02t_{3}$ and $t_{1}=-2.23t_{3}$

Thus the eigenvector is

t=$k [-2.23, -4.02, 1]$

But using an online solver gives the eigenvector as (-0.016, 0.206, 0.978).

Thanks for any pointers.

Last edited: Feb 1, 2013
2. Feb 1, 2013

### mathman

Before you do anything else, check to see if it is an eigenvector (At = λt ?).
Also how do know λ = 1?

3. Feb 1, 2013

### enc08

Hi,

The reason I'm assuming λ = 1 is because A is a rotation matrix (I didn't mention this in the original post), and I'm looking for the equivalent axes eigenvector.

4. Feb 1, 2013

### enc08

As a sanity check, I have checked the eigenvalues and λ = 1 is one of them.

5. Feb 1, 2013

### enc08

I realised I typo in my original matrix (matrix term 3,3 in the first post should be negative). I've correct it.

I get a closer answer, but my first term appears to be incorrect:

6. Feb 1, 2013

### Staff: Mentor

Since your matrix is a rotation matrix, the entries are sin and cos values. I would try using more precision than the 4 decimal places you show, and see if that makes a difference in your resulting eigenvector.

7. Feb 2, 2013

### enc08

I've just been given the numbers as is, and told that it's a rotation matrix.

Can you tell if my method for computing the eigenvectors is correct?

My approach is

λ = 1, so (A - λI) is
$$[-0.5253, 0.8593, -0.1906; -0.8612, -0.5018, 0.1010; 0.1817, 0.1161, -0.0236]$$

From rows 2 and 3:

$$-0.8612t_{1}-0.5018t_{2}+0.1010t_{3}=0$$

$$0.1817t_{1}+0.1161t_{2}-0.0236t_{3}=0$$

Eliminate to find $$t_{2}=0.225t_{3}$$ and $$t_{1}=-0.0137t_{3}$$

Thus the eigenvector is

t=$$k (-0.0137, 0.225, 1)$$

But the actual answer is given as (-0.0088, 0.216, 1).

8. Feb 2, 2013

### Staff: Mentor

Here's your matrix in a nicer form. To see what I did, right-click the matrix to view the underlying script.
$$A - I =\begin{bmatrix} -0.5253 & 0.8593 & -0.1906 \\ -0.8612 & -0.5018 & 0.1010 \\ 0.1817 & 0.1161 & -0.0236\end{bmatrix}$$

To check your work, I multiplied A - I above times your eigenvector x. That multiplication should produce a zero vector, but what I got was off by a little. My result was approximately < .0099, .0088, .00003>.

When you were row-reducing A - I, if you weren't careful with your arithmetic, you could have introduced a certain amount of imprecision in your results.

This is easy to check. Just multiply (A - I) times <-0.0088, 0.216, 1>.

9. Feb 2, 2013

### enc08

Thanks Mark44. I have used more precise values, and using your technique for testing the eigenvector, I get an answer much closer to a zero vector.

That you've confirmed my method is correct will do, especially since the discrepancies seem only due to rounding errors. I would have much preferred the question to use round numbers, but never mind :)