in Linear Algebra (Third Edition), Springer by Serge Lang there is a proof of Jordan normal form that I don t quite understand. He gives the following reasoning:
${\bf 1)}$ Given a vector space $V$ over $\mathbb{C}$, a polynomial $P\in K[t], \ P=(t-\alpha_1)^{m_1}\ldots(t-\alpha_n)^{m_r}$ and a linear operator $A:V\rightarrow V$ such that $P(A)=0$, then $V= W_1\oplus\ldots\oplus W_r$, where $W_i=Ker (A-\alpha_iI)^{m_i}, \ i\in \{1,2,\ldots ,r \}$
${\bf 2)}$ For a vector space $V$ over $\mathbb{C}$ and for a linear operator $A:V\rightarrow V$, we say $v\neq 0$ is cyclic under $(A-\alpha I)$ of perior $r$ iff $(A-\alpha I)^k\neq 0,\forall k\in\{0,\ldots,r-1\}$ and $(A-\alpha I)^k= 0.$ Then we can show that $\mathbb{B}=\{v,(A-\alpha I)v,\ldots,(A-\alpha I)^{r-1}v\}$ are linearly independent. Call $V$ cyclic if $\exists \alpha$ such that $\mathbb{B}$ is a basis of $V.$
${\bf Proof}$
All we need to do is express $V= V_1\oplus\ldots\oplus V_m$, where $V_i$ is cyclic, as if we define a basis ${\mathbb{B'}}=\bigcup{\mathbb{B_i}}$, where $\
{\mathbb{B_i}}$ are defined as cyclic bases (as above),we obtain that $[A]_{\mathbb{B'}}$ is in Jordan normal form.
By ${\bf 1)}$, $\exists\alpha,r.$ such that $B:=(A-\alpha I)^r=0$ ( take $r$ minimal). It' easy to see that the subspace $BV$ is strictly included in $V$, and by an inductive argument we can write $BV$ as a direct sum of $A-invariant \ subspaces$ that are cyclic, say
$BV= W_1\oplus\ldots\oplus W_m$
( the author says here '' or $B$ invariant''. should this ring a bell to me?)
Then define bases $\{w_i,(A-\alpha I)w_i,\ldots,(A-\alpha I)^{r-1}w_i\}$ of $W_i$.
LET $v_i\in V$ SUCH THAT $Bv_i=w_i$. THEN EACH $v_i$ IS A CYCLE VECTOR. LET $V_i$ THE SUBSPACE OF $V$ GENERATED by the elements of ${B^kv_i}$. How do we know $V_i$ is ${\bf uniquely \; defined}$?
I appreciate any help. Thank you.