# 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?

• 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\$. – lonza leggiera Apr 26 at 0:00
• Sorry, I meant $AJ_{3} = A_{2} + 2A_{3}$ since $J_{3}$ is : \begin{bmatrix} 0 \\ 1 \\ 2 \\ 0 \end{bmatrix} – JKM Apr 26 at 0:15
• 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. – JKM Apr 26 at 0:26
• This comes straight from the definition of generalized eigenvectors. – amd Apr 26 at 0:33
• 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: $x = P_{B}[x]_{B}$. – JKM Apr 26 at 3:07

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