What is the Eigenspace for a 2x2 matrix with eigenvalues -5 and 3?

[g.lemaitre]

Homework Statement

Find the Eigenspace of the following matrix:
$$\begin{bmatrix}<br /> 1 & 3 \\<br /> 4 & -3<br /> \end {bmatrix}$$
I'm skipping a few steps but the Eigenvalues are -5 and 3. Let's starts with -5. Skip a few more steps, I know I'm right, just trust me.
We now have the following matrix:
$$\begin{bmatrix}<br /> -6 & -3 \\<br /> -4 & -2<br /> \end {bmatrix}$$
Then you find the null space, which starts with putting it in reduced row echelon form:
$$\begin{bmatrix}<br /> -6 & -3 \\<br /> 0 & 0<br /> \end {bmatrix}$$
you can reduce that further to
$$\begin{bmatrix}<br /> -2 & -1 \\<br /> 0 & 0<br /> \end {bmatrix}$$
This is where I'm confused. This nullspace calculator http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi
says that the basis of the null space is
$$\begin{bmatrix}<br /> -1 \\<br /> 2<br /> \end {bmatrix}$$
My textbook confirms that. How do I get from here
$$\begin{bmatrix}<br /> -2 & -1 \\<br /> 0 & 0<br /> \end {bmatrix}$$
to there
$$\begin{bmatrix}<br /> -1 \\<br /> 2<br /> \end {bmatrix}$$
I would think you would just eliminate the 2nd row and transpose the first row but that would give.
$$\begin{bmatrix}<br /> -2 \\<br /> -1<br /> \end {bmatrix}$$

 

[LCKurtz]

you can reduce that further to
$$\begin{bmatrix}<br /> -2 & -1 \\<br /> 0 & 0<br /> \end {bmatrix}$$
This is where I'm confused. This nullspace calculator http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi
says that the basis of the null space is
$$\begin{bmatrix}<br /> -1 \\<br /> 2<br /> \end {bmatrix}$$
My textbook confirms that. How do I get from here
$$\begin{bmatrix}<br /> -2 & -1 \\<br /> 0 & 0<br /> \end {bmatrix}$$
to there
$$\begin{bmatrix}<br /> -1 \\<br /> 2<br /> \end {bmatrix}$$
I would think you would just eliminate the 2nd row and transpose the first row but that would give.
$$\begin{bmatrix}<br /> -2 \\<br /> -1<br /> \end {bmatrix}$$

The matrix corresponds to -2x - y = 0 or y = -2x So$$
\begin{bmatrix}x\\ y \end{bmatrix}=\begin{bmatrix}x\\ -2x \end{bmatrix}
=x\begin{bmatrix}1\\ -2 \end{bmatrix} $$
Any nonzero constant works, so take $x=-1$ to get their answer.