# Eigenvector kernel problem

1. Jun 20, 2005

### amcavoy

If you have the following kernel (I think that's what it's called):

$$A-\lambda I=\begin{pmatrix}4 & 1 \\ 4 & 1\end{pmatrix}$$

You could write the eigenvector as:

$$\operatorname{span}\begin{pmatrix}1 \\ -4\end{pmatrix}$$

My question is: does it matter how you write the "span" part of it?

For instance, would $$\operatorname{span}\begin{pmatrix}-1 \\ 4\end{pmatrix}$$ be preferred (or different) than what I have above?

2. Jun 20, 2005

### HallsofIvy

Staff Emeritus
$$A-\lambda I=\begin{pmatrix}4 & 1 \\ 4 & 1\end{pmatrix}$$

doesn't make sense- there's no &lambda; on the right hand side! Did you mean
$$A=\begin{pmatrix}4 & 1 \\ 4 & 1\end{pmatrix}$$ and that is the linear operator you want to find eigenvectors for?

In that case, there are two eigenvalues: 0 and 2.

Taking 0 as the eigenvalue leads to 4x-y=0 or y= 4x. The eigenvectors are spaned by [1, 4] (not [-1, 4]).
Taking 2 as the eigenvalue leads to 4x-y= 2x or y= -2x. The eigenvectors are spanned by [1, -2].

Because the "span" of a single vector is all real multiples of the vector, and -1 is a real number, it doesn't matter whether you use [-1, 4] or [1, -4] or [-1,2] instead of [1,-2].

3. Jun 20, 2005

### amcavoy

Sorry, I should have elaborated. The original matrix (A) was:

$$\begin{pmatrix}3 & 1 \\ 4 & 0\end{pmatrix}$$

And the eigenvalue I was using was -1. I didn't think that mattered in relation to my question (because my ? dealt with notation), but I don't really know all of the terms, so I appologize.