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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?

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    $\begingroup$ I don't understand why you write "$\ A\mathbf{v_{3}} = A_{2} +2A_{3}\ $". Are not $\ A_{2}, A_{3}\ $, and $\ \mathbf{v_{3}}\ $ the second and third columns of $\ A\ $, and the third column of $\ P\ $, respectively? When I calculate $\ A\mathbf{v_{3}}\ $ I get $$ A\mathbf{v_{3}} = A_3= \mathbf{v_{2}}+2\mathbf{v_{3}}\ .$$ Note that $\ \mathbf{v_{3}}\ $ is not an eigenvector of $\ A\ $, although it does belong to the invariant subspace of $\ A\ $ belonging to the eigenvalue $\ 2\ $. $\endgroup$ – lonza leggiera Apr 26 at 0:00
  • $\begingroup$ Sorry, I meant $AJ_{3} = A_{2} + 2A_{3}$ since $J_{3}$ is : \begin{bmatrix} 0 \\ 1 \\ 2 \\ 0 \end{bmatrix} $\endgroup$ – JKM Apr 26 at 0:15
  • $\begingroup$ Sorry about that typo, so the question then is precisely why $A_{3} = \mathbf{v_{2}} +2\mathbf{v_{3}}.$ I mean I see it by looking at $A_{3}$, but formally where does this linear combination come from. $\endgroup$ – JKM Apr 26 at 0:26
  • $\begingroup$ This comes straight from the definition of generalized eigenvectors. $\endgroup$ – amd Apr 26 at 0:33
  • $\begingroup$ It took me some time to realize simply that these are the coordinates of $A\mathbf{v_{3}}$ with respect to the basis $B$ . $P^{-1}_{B} [A \mathbf{v_{3}}] = [A \mathbf{v_{3}}]_{B} = \begin{bmatrix} 0 \\ 1 \\ 2 \\ 0\\ \end{bmatrix}$ ... coming from the change of basis formula: $x = P_{B}[x]_{B}$. $\endgroup$ – JKM Apr 26 at 3:07
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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.

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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: $x = P_{B}[x]_{B}$.

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