I've been given the matrix: $$B = \begin{bmatrix}4& 2&-2\\3&-1&2\\2&2&0\end{bmatrix}$$
And found its Eigenvalues to be $\lambda=-3,2,4.$ I now need to calculate its eigenspace, but am unsure of wether I'm doing it correctly. I set up the matrix to be:
$$B-(-3) I=\begin{bmatrix}4-(-3)& 2&-2\\3&-1-(-3)&2\\2&2&0-(-3)\end{bmatrix}=\begin{bmatrix}7& 2&-2\\3&2&2\\2&2&3\end{bmatrix}$$
I know that I need to set these equal to $B\vec{x}=\vec{0}$ to find the eigenspace. I start by reducing $B$, but keep getting the identity matrix implying $x_1,x_2,x_3=0.$ I don't think this is correct, but am also lost on what comes next. If it is correct, does that mean that my basis could be:
$$\begin{bmatrix}1\\1\\1\end{bmatrix}\text{ or }\begin{bmatrix}2\\2\\2\end{bmatrix}\text{ etc…}$$
Sorry if this is a lot to ask or even a simple thing. I am struggling to wrap my head around this and could use any help!
Best Answer
$$ \begin{pmatrix} 7 & 2 & -2 \\ 3 & 2 & 2 \\ 2 & 2 & 3 \end{pmatrix}\to \begin{pmatrix} 1 & 2/7 & -2/7 \\ 0 & 8/7 & 20/7 \\ 0 & 10/7 & 25/7 \end{pmatrix} \to \begin{pmatrix} 1 & 2/7 & -2/7 \\ 0 & 1 & 5/2 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 5/2 \\ 0 & 0 & 0 \end{pmatrix}$$ Thus the kernel is given by $(x,y,z)$ with $z=x$ and $y=-5/2 z=-5/2x$, which gives you the basis $\{ (1,-5/2,1\}$ or $\{(2,-5,2)\}$, if you prefer this.
For the other eigenvalues proceed the same way.