Question related to the justification of algorithm to find Jordan canonical form I am reading notes for my class for the algorithm to find Jordan canonical form at the moment. It doesn't give the justification and there is a spot I am confused with and I would greatly appreciate some clarification. 
Suppose $A$ is a matrix $n$ by $n$ complex entries, and for simplicity suppose it has only one distinct eigenvalue $\lambda$. 
We denote $E_{\lambda}^{m} = \ker (A - \lambda I)^m$.
The algorithm says to first find the smallest $k$ such that $E_{\lambda}^k$ has dimension $n$. Find a vector $v \in E_{\lambda}^{k} \backslash E_{\lambda}^{k-1}$ and form $B = \{ (A - \lambda I )^{k-1}v, .., (A - \lambda I)v, v \}$. If $k < n$, then find the largest $k' \leq k$ such that $E^{k'}_{\lambda} \not \subseteq
E^{k'-1}_{\lambda} +$ Span$(B)$ and take $u \in E^{k'}_{\lambda} \backslash \{ E^{k'-1}_{\lambda} +$ Span$(B)\}$.
And extend $B$ to be $B' = \{(A - \lambda I )^{k-1}v, .., (A - \lambda I)v, v, (A - \lambda I )^{k'-1}u, .., (A - \lambda I)u, u  \}$ and continue, etc.
My question is how do we know this set $B'$ is linearly independent? I know $B$ is linearly independent, but when extended to $B'$ I am not seeing how I can show it is linearly independent at the moment. I would appreciate some explanation. Thank you very much!
 A: The above comment is not correct for my carelessness. 
For example, we assume $k=3$ and $\lambda=0$ for simplicity, and then exist a generalized eigenvector $x$ of order 3 , it is evident that $ x, Ax , A^2 x$ are linearly independent. Suppose we have another $y$ such that $y$ is of order 3, and
Assumption: $y$ is not in
$$span\{x,Ax, A^2x\}+Ker(A^2).$$ Then we prove
$$\{y,Ay, A^2y,x, Ax, A^2x\}$$are linearly independent.
Since $\{x,Ax,A^2x\}$ is independent. So if $y$ is not in the span of ${x,Ax,A^2x}$, then $\{y,x,Ax,A^2x\}$ is linearly independent. Now prove:$$Ay\notin span\{y,x,Ax, A^2x\}$$. If otherwise, $$Ay=a_1 x+a_2 Ax+a_3 A^2x+b_1 y$$ where $a_1,a_2,a_3$ are scalars.
Obviously, $Ay$ belongs to $Ker(A^2)$ so, $b_1$ must be zero, otherwise, $$y\in span\{x,Ax,A^2x\}+Ker(A^2)$$. This is a contradiction. Then $a_1 $ should be zero, other wise , $x$ is in $Ker(A^2)$ which contradicts to that $A^2x\neq0$.Then we have $$A(y-a_2 x-a_3 Ax)=0$$, so
$$y-a_2 x-a_3 Ax\in Ker(A)\subseteq Ker(A^2)$$.This contradicts the above assumption again. So we showed that :
$$\{x,Ax, A^2x,y,Ay\}$$ is linearly independent.Then similarly, assume :
$$A^2y\in Span\{x,Ax,A^2x,y,Ay\}.$$
I.e., $$A^2y=a_1x+a_2Ax+a_3A^2x+b_1y+b_2Ay$$. We can see $b_1=0$ and $a_1=0$ similarly as the above. Then once again :
$$A(Ay-a_2x-a_3Ax-b_2 y)=0.$$ This implies $a_2=0, b_2=0$ since:
$$(Ay-a_2x-a_3Ax-b_2y)\in Ker(A)\subseteq Ker(A^2).$$
Then finally we have$ Ay-a_3Ax=A(y-a_3)\in Ker(A)$, so $$y-a_3x\in Ker(A^2)$$ which also contradicts to the assumption. So $$\{y,Ay,A^2y,x,Ax,A^2x\}$$ are linearly independent. I think the general case is similar.
