I have a matrix that looks like this: $$ \begin{pmatrix} 1 & 2 & 4 \\ 2 & 4 & 2 \\ 4 & 1 & 1 \end{pmatrix} $$
Now the calculated eigenvalues are: $-3$, $2$ and $7$. For the eigenvalue $-3$, I want to calculate the eigenspace and the basis. So after a little bit of rearranging I get to this matrix: $$ \left(\begin{array}{ccc@{}|@{}c} 4 & 2 & 4 & 0 \\ 0 & 12 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) $$
So now, the rank is two and the dimension of the nullsapce is $1$. Second line: $12 v_2 + 0 v_3 = 0$, everything goes to zero and my basis is then $(0, 0, 0)$. But can this be within the eigenspace or did I just have make a fault?
You are reduced to solve the system \begin{cases} 4x_1+2x_2+4x_3=0\\[4px] 12x_2=0 \end{cases} This implies $x_2=0$ and so $x_1+x_3=0$. So a vector in the eigenspace must be of the form \begin{bmatrix} a\\0\\-a \end{bmatrix} so the vector \begin{bmatrix} 1\\0\\-1 \end{bmatrix} forms a basis of the eigenspace.