linear transformation eigenvectors and eigenvalues 
Consider the operator defined by $T(x, y, z) = (-x+2y, 3y, 0)$. 
  
  
*
  
*Find the eigenvalues of T and all corresponding eigenvectors.
  
*Find each generalised eigenvector corresponding to each eigenvalue.

So for 1, I found the matrix with respect to the standard basis $(1,0,0), (0,1,0), (0,0,1)$. It was upper triangular so I read the eigenvalues from the diagonal entries, which gave me eigenvalues of $-1, 3$ and $0$. I'm just so lost with how to find the corresponding eigenvectors, I feel like this should be really simple.
I know once I have found these eigenvectors for 2, I can simply compute the generalized eigenvectors by solving the equation $(T- \lambda I)^j (v) = 0$ where $v$ is my particular eigenvector with corresponding eigenvalue $\lambda$.
Thanks in advance!
 A: Operator
$$
  \mathbf{T} =
\left[
\begin{array}{rrr}
 -1 & 2 & 0 \\
 0 & 3 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right]
$$
Eigenvalues 
To find the eigenvalues, compute $p(\lambda)$, the characteristic polynomial.
$$
 p (\lambda ) = \det \left( \mathbf{T} - \lambda \mathbf{I}_{3} \right) = \det
\left[
\begin{array}{ccr}
 -\lambda -1 & 2 & 0 \\
 0 & 3-\lambda  & 0 \\
 0 & 0 & -\lambda  \\
\end{array}
\right]
=
-\lambda \left( 3 - \lambda \right) \left( -1 - \lambda \right)
$$
The roots $p(\lambda) = 0$ are the eigenvalues: $\lambda = \left\{ 3, -1, 0 \right\}$.
Eigenvectors
Solve
$$
  \mathbf{T} v_{k} = \lambda_{k} v_{k} \qquad \Rightarrow \qquad \left( \mathbf{T} - \lambda_{k} \mathbf{I}_{3} \right) v_{k} = \mathbf{0}
$$
$\lambda = 3$
$$
\begin{align}
  \left( \mathbf{T} - 3 \mathbf{I}_{3} \right) v_{1} &= \mathbf{0} \\
\left[
\begin{array}{rrr}
 -4 & 2 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & -3 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 v_{1_{1}} \\
 v_{1_{2}} \\
 v_{1_{3}}
\end{array}
\right]
&=
\left[
\begin{array}{c}
 0 \\
 0 \\
 0
\end{array}
\right]
%
 \qquad \Rightarrow \qquad 
v_{1} =
\left[
\begin{array}{c}
 1 \\
 2 \\
 0
\end{array}
\right]
%
\end{align}
$$
$\lambda = -1$
$$
\begin{align}
  \left( \mathbf{T} + \mathbf{I}_{3} \right) v_{2} &= \mathbf{0} \\
\left[
\begin{array}{rrr}
 0 & 2 & 0 \\
 0 & 4 & 0 \\
 0 & 0 & 1 
\end{array}
\right]
%
\left[
\begin{array}{c}
 v_{2_{1}} \\
 v_{2_{2}} \\
 v_{2_{3}}
\end{array}
\right]
&=
\left[
\begin{array}{c}
 0 \\
 0 \\
 0
\end{array}
\right]
%
 \qquad \Rightarrow \qquad 
v_{2} =
\left[
\begin{array}{c}
 1 \\
 0 \\
 0
\end{array}
\right]
%
\end{align}
$$
$\lambda = 0$
$$
\begin{align}
  \left( \mathbf{T} + 0 \mathbf{I}_{3} \right) v_{3} &= \mathbf{0} \\
\left[
\begin{array}{rrr}
 -1 & 2 & 0 \\
 0 & 3 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 v_{3_{1}} \\
 v_{3_{2}} \\
 v_{3_{3}}
\end{array}
\right]
&=
\left[
\begin{array}{c}
 0 \\
 0 \\
 0
\end{array}
\right]
%
 \qquad \Rightarrow \qquad 
v_{3} =
\left[
\begin{array}{c}
 0 \\
 0 \\
 1
\end{array}
\right]
%
\end{align}
$$
