Jordan normal as transformation with respect to the basis of eigenvectors I have the following matrix
$$A = \begin{pmatrix}
      2 & 0 & 1 & -3 \\
      0 & 2 & 10 & 4 \\
      0 & 0 & 2 &  0 \\
      0 & 0 & 0 &  3 \\
      \end{pmatrix}$$
and its Jordan normal form is 
$$J = \begin{pmatrix}
      2 & 0 & 0 & 0 \\
      0 & 2 & 1 & 0 \\
      0 & 0 & 2 &  0 \\
      0 & 0 & 0 &  3 \\
\end{pmatrix}$$
with the linearly independent set of eigenvectors:
$$P = \begin{pmatrix}
       0 & 1 & 0 & -3 \\
       1 & 10 & 0 & 4 \\
       0 & 0 & 1 &  0 \\
       0 & 0 & 0 &  1 \\
      \end{pmatrix}$$
where $J = P^{-1} A P$. I am told that the following relations hold:
$$A\mathbf{v_{1}} = 2\mathbf{v_{1}}, \qquad A\mathbf{v_{2}} = 2\mathbf{v_{2}}, \qquad A\mathbf{v_{3}} = \mathbf{v_{2}} + 2\mathbf{v_{3}}, \qquad A\mathbf{v_{4}} = 3\mathbf{v_{4}}$$
In the case of $\mathbf{v_{1}}, \mathbf{v_{2}}, \mathbf{v_{4}}$, it makes total sense since this is due to the fundamental relation $A\mathbf{v} = \lambda \mathbf{v}$. Also, I see how $A\mathbf{v_{3}} = A_{2} +2A_{3}$, where $A_{2}$ and  $A_{3}$ are the second and third column vectors of thre matrix $A$. This is simply multiplying $\mathbf{A}$ by $\mathbf{v_{3}}$. However, my intuition is failing me in seeing how 
$$A\mathbf{v_{3}} = \mathbf{v_{2}} + 2\mathbf{v_{3}}$$
I am told that this is simply because $J$ represents the transformation corresponding to $A$ with respect to the basis $\big\{\mathbf{v_{1}}, \mathbf{v_{2}}, \mathbf{v_{3}}, \mathbf{v_{4}}\big\}$. Any pointers that can help me understand this?
 A: One way of seeing where equations like  $\ A\mathbf{v_{3}} = \mathbf{v_{2}} + 2\mathbf{v_{3}}\ $ come from, which I found helpful when first introduced to Jordan forms, is the definition of the invariant subspace corresponding to an eigenvalue $\ \lambda\ $ as that spanned by the non-zero vectors $\ \mathbf{v} $ for which $\ \left(A-\lambda I\right)^{\,k}  \mathbf{v}=0\ $ for some positive integer $\ k\ $.  If $\ k=1\ $, then $\ \mathbf{v}\ $ is an eigenvector, but if $\ k>1\ $ and $\ \left(A-\lambda I\right)^{\,k-1}  \mathbf{v}\ne0\ $ then it isn't.  However, if we put $\ \mathbf{v}_i=\left(A-\lambda I\right)^{\,i}  \mathbf{v}\ $ for $\ i=0,1,\dots,k-1\ $, then we have $\ A\mathbf{v}_i=\lambda \mathbf{v}_i + \mathbf{v}_{i+1}\ \ $, with $\ \mathbf{v}_{k-1}\ $ being an eigenvector.
A: B= $\big\{ v_{1},v_{2}, v_{3}, v_{4} \big\} $
$P^{-1}_{B} [A \mathbf{v_{3}}] = [A \mathbf{v_{3}}]_{B}  =  \begin{bmatrix}
       0 \\
       1 \\
       2 \\
       0\\
 \end{bmatrix}$ 
... from the change of basis formula: $x = P_{B}[x]_{B}$. 
