I have the following
How can I show that $\bf{A}$ is diagonalizable and then find the matrix $\bf{P}$ (which I know is made up of the eigenvectors) that diagonalizes $\bf{A}$. So far, I have already deduced that $\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} $ and $\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} $ are eigenvectors since $\bf{A}\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}=\bf{0}\ $ and $\bf{A}\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix}=\bf{0}$.
How can this be done in an easy manner that follows the order of the question?
Note: I am able to solve the rest of the problem and just got stuck on this minor thing as it differs from what we have done earlier in class.
It's easier to do (iii) first: $$c_A(x)=x((x-10)(x-2)-4\cdot5)=x^2(x-12)$$ Now (i) tells us that there are two linearly independent eigenvectors for $0$. The one for $12$ can be determined as $(-5,2,0)^T$, which is linearly independent of the other two. Hence $A$ is diagonalisable.