So I have this matrix
$$ \begin{pmatrix} -2 & 0 & 0 \\ 3 & 1 & -6 \\ 0 & 0 & -2 \\ \end{pmatrix} $$
Finding Nul(A+2Id) gives me
$$ \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & -6 \\ 0 & 0 & 0 \\ \end{pmatrix} $$
On the answers sheet it says to solve the homogeneous system associated with A + 2Id which gives us
Nul(A+2Id)=Span $$ \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 1 & 0 \\ \end{pmatrix} $$
I am completely lost on how to find the Span Nul(A+2Id)=Span $$ \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 1 & 0 \\ \end{pmatrix} $$
Can somebody please help me with the steps in order to get to this solution ?
Moreover, on the answers sheet is states
Since the dimension of the eigenspace E(λ1) is equal to 2 we observe that the ma- trix A is diagonalizable.
How do i know from this solution that the eigenspace E(λ1) is equal to 2?
Thanks!
You know that $\dim E_{-2}=2$ because you computed $\operatorname{Nul}(A+2\operatorname{Id})$ and it turned out that it has dimension $2$: it is spanned by two linearly independent vectors.