# [Linear Algebra] Pulling two vectors out of a one equation matrix.

1. Nov 16, 2013

### jeff.berhow

1. The problem statement, all variables and given/known data
Determine a basis for each eigenspace and whether or not the matrix is defective.

\begin{array}{ccc}
3 & -4 & -1 \\
0 & -1 & -1 \\
0 & -4 & 2 \end{array}

2. Relevant equations

3. The attempt at a solution
Ok, so I've found the eigenvalues first with no problem. λ_1 = 3 (with multiplicity 2) and λ_2 = -2. My misunderstanding comes with finding the eigenvectors with λ_1. Gauss-Jordan with the new augmented matrix gives me:

\begin{array}{ccc}
0 & 1 & 1/4 & | & 0\\
0 & 0 & 0 & | & 0\\
0 & 0 & 0 & | & 0\end{array}

Let x_3 = r and this tells me that x_1 = 0r, x_2 = -1/4r and so the eigenvector is (0, -1/4, 1). This then leads me to believe the matrix will be defective as I will only be able to get 1 more eigenvector out of λ_2 = -2 as this matrix has been exhausted.

Lo and behold! In the back of the book, they were able to extract another vector out of the matrix above: (1, 0, 0). This would then give us three total eigenvectors which is indeed a basis and makes the matrix nondefective. So, my question is, how did they get another vector out of that thing?

Jeff

2. Nov 17, 2013

### Dick

Uh, [1,0,0] is an eigenvector corresponding to the eigenvalue 3. Why do you think x1 must be 0?

3. Nov 17, 2013

### jeff.berhow

I guess, because it's not explicitly stated in the equation: 0x1 + x2 + (1/4)x3 = 0. If a variable has a coefficient of zero in an equation, is it assumed to be one? I feel I'm missing something very elementary here, haha.

4. Nov 17, 2013

### haruspex

(α, 0, 0) is a solution of 0x1 + x2 + (1/4)x3 = 0 for any α. If α nonzero then it is a nontrivial solution. You can pick what nonzero value you like for α, but if you want a vector magnitude 1 then clearly you'll pick ±1.

5. Nov 17, 2013

### jeff.berhow

Thanks! Of course it was something obvious. I seem to have a real problem with such obvious things.