Let V be the set of all $A_{3×3}$ matrices, where all of them have eigenvalue 0 and corresponding eigenvector $y=(1,2,3)^T$.
Assuming we have showed that V is a vector space, determine the dimension of V.
From the definition we know that: $x$ and $\lambda$ are eigenvector and eigenvalue iff $(A-\lambda I)x=0$ .
So for each matrix $A_i$ in our set$$ (A_i-0) \begin{pmatrix} 1 \\ 2 \\ 3 \\ \end{pmatrix} =0 $$ $$ A_i \begin{pmatrix} 1 \\ 2 \\ 3 \\ \end{pmatrix} =0 $$ $$ \begin{pmatrix} 1 \\ 2 \\ 3 \\ \end{pmatrix} \in N(A_i) $$ From this we find that $Nullity(A_i)\geq1 \Rightarrow $
$Rank(A_i) = n-Nullity(A_i)=3-1=2 \Rightarrow $
$Rank(A_i)=Dim(C(A_i)) \leq 2$
However, I am not sure how can I proceed from here?
1) How can determine the dimension of the matrix $A_i$ for sure?
2) How is that connected to the dimension of entire vector space formed by such matrices?