Projection, Jordan form and a new basis to $R^{n}$

Question

I'm reading the book (Introduction to Applied Nonlinear Dynamical System and Chaos, Stephen wiggins, Theorem 19.5.3 ) in this Theorem the author give the idea of how create a new basis to $\mathbb{R}^{n}$, The idea is the following.

Assume that $J$ is in the canonical Jordam form. Then it has $r$ Jordam blocks, whith each blocks $J_{i}$ corresponding to invariant subspace of $\mathbb{R}^{n}$, denoted $E_{j}$ and to eigenvalue $\lambda_{j}$, $j=1,2,\cdots,r$. Let $v_{j}$ denoted of dimension of $E_{j}$ and $P_{j}$ denote de projection onto $E_{j}$. Then $$\sum P_{j}= Id$$ To each Jordan block we associated $v_{j}$ linear operators. $$P_{j}, (J_{i}-\lambda_{j}Id)P_{j},\cdots,(J_{i}-\lambda_{j}Id)^{v_{j}-1}P_{j}, \hspace{3mm}j=1,2,\cdots,r $$ This is set a set of $n$ linearly independent operators, which each commuting with $J$, which we denoted by $\zeta_{j}$. Now we choose any $x \in \mathbb{R}^{n}$ having the property that it has a non zero component in each $E_{j}$. then $\zeta_{j}x$, $j=1,2,\cdots,n$ forms a basis to $\mathbb{R}^{n}$.

Now, My question is the following, I try to do an example but I have problems. The exammple is the following:

Consider $$J=\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0& 0 \\ 0 & 0&-1& 1 \\ 0 & 0&0& -1 \\ \end{bmatrix}$$

So, We have

$$J_{1}=\begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix}, \hspace{2mm} J_{2}=\begin{bmatrix} -1& 1 \\ 0 & -1 \\ \end{bmatrix}$$

The Eigenspaces $$E_{1} = Span \{e_{1},e_{2}\}, \hspace{2mm} E_{2}= Span\{e_{3}, e_{4}\}$$,

here $e_{i}, i=1,2,3,4$ are the elements of canonical basics to $\mathbb{R}^{n}$, $v_{j}=2$, $j=1,2$, and the projection (Here I'm not very sure)

$$P_{1}(x,y,z,w)= (x,y,0,0), \hspace{3mm} P_{2}(x,y,z,w)= (0,0,z,w)$$

Thus, We have the 4 linearly independents operators

$$P_{1}, (J_{1}-\lambda_{1}Id)P_{1},P_{2},(J_{2}-\lambda_{2}Id)P_{2}. $$ Here is my problem, since $J_{i}$, $i=1,2$ are matrices $2\times2$, while $P_{j}$ are $4-vectors$. And if for instance I put $$P_{1}(x,y,z,w)= (x,y), \hspace{3mm} P_{2}(x,y,z,w)= (z,w)$$ The dimension is fine, but when I compute $\zeta_{i}x$, $i=1,2,3,4$ the resultant vectors are $2-vectors$ which are not basis to $\mathbb{R}^{4}$.

What is wrong? Thanks in advance.

Answer 1

we identify the blocks $J_{1}$ and $J_{2}$ with : $$J_{1}=\begin{pmatrix} 1 & 1& 0 & 0 \\ 0 & 1& 0 & 0\\ 0 & 0 & 0 &0 \\ 0 & 0 & 0 &0 \end{pmatrix}$$ $$J_{2}=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0& 0 & 0\\ 0 & 0 & -1 &1 \\ 0 & 0 & 0 & -1 \end{pmatrix}$$

Answer 2

@Skinner this is too long for a comment. To simplify and in order to avoid projections, let's consider the case of a single Jordan block $J\in M_{v}(\mathbb{R})$ of size $v$ with the eigenvalue $0$,i.e a nilpotent matrix (remark that the general case is similar since $J_{i}-\lambda_{i}$ is nilpotent with nilpotency index $v_{j}$).So we have $J^{v}=0 \land J^{v-1}\neq 0$.U can prove the following claim:a vector $x$ is cyclic iff $x \not \in Ker(J^{v-1})$,where a vector $x$ is said to be cyclic if $\{ J^{0}x=x,J^{1}x,J^{2}x,\cdots J^{v-1}x \}$ spans the space (in fact it is a basis),So I think that the hypothesis $x$ has non zero coordinate along $E_{j}$ (i.e $P_{i}(x)\neq 0$) must be replaced by $P_{j}(x) \not \in Ker(J_{i}-\lambda_{i})^{v_{j}-1}$