Understanding the proof of Jordan normal form, by matrix trick I read a proof for JNF which I am unclear.
Proof:
For any complex matrix $A$. Assume $Av_1=\lambda v_1$ for some $v_1 \in \mathbb{C}^n$, then $A(v_1, \cdots, v_n)=(v_1, \cdots, v_n)\begin{pmatrix}\lambda & *\\ & A_0\end{pmatrix}$.
So $A$ is similar to $A(v_1, \cdots, v_n)=(v_1, \cdots, v_n)\begin{pmatrix}\lambda & *\\ & A_0\end{pmatrix}$. By induction, we can assume that $A_0$ is in form of Jordan norm form. Assume more that $A_0=\begin{pmatrix}A_1 & \\ & A_2\end{pmatrix}$. Where $A_1$ arranges Jordan block belonging to $\lambda$, and $A_2$ arranges the others.
Existence:
Using the following trick:
\begin{equation}\label{eq:1}
\begin{pmatrix}1 & & x & \\ &\ddots &&\\
 & & 1&\\ &&&\ddots\end{pmatrix}\begin{pmatrix}\lambda &0 & a & * \\ &\ddots &&\\ & & \mu&\\&&&\ddots\end{pmatrix}\begin{pmatrix}1 & & -x & \\ &\ddots &&\\
 & & 1&\\ &&&\ddots\end{pmatrix}=\begin{pmatrix}\lambda &0 & a+\mu x-\lambda x & * \\ &\ddots &&\\ & & \mu& * \\&&&\ddots\end{pmatrix}
\end{equation} to eliminate all the entries over $A_2$ in the first row.
Using the following trick:
\begin{equation}\label{eq:2}
\begin{pmatrix}1 & & x & \\ &\ddots &&\\
 & & 1&\\ &&&\ddots\end{pmatrix}\begin{pmatrix}\lambda &0 & a & b \\ &\ddots &&\\ & & \lambda&\\&&&\ddots\end{pmatrix}\begin{pmatrix}1 & & -x & \\ &\ddots &&\\
 & & 1&\\ &&&\ddots\end{pmatrix}=\begin{pmatrix}\lambda &0 & a & b+x \\ &\ddots &&\\ & & \mu& 1 \\&&&\ddots\end{pmatrix}
\end{equation} to eliminate all the entries over $A_1$ except the first column of each Jordan block.
Using the following trick:
\begin{equation}\label{eq:3}
\begin{pmatrix}1 & &  & \\ & I & & xI \\
 & & \ddots &\\ &&& I \end{pmatrix}\begin{pmatrix}\lambda & ae_1 & 0 & be_1 \\ &J & \Delta &\\ & & J'&\\&&& J\end{pmatrix}\begin{pmatrix}1 & &  & \\ &I&&-xI\\
 & & \ddots&\\ &&&I\end{pmatrix}=\begin{pmatrix}\lambda & ae_1 & 0 & (b-xa)e_1 \\ &J & \Delta &\\ & & J'&\\&&& J\end{pmatrix}
\end{equation} to eliminate all the entries at the first column of each Jordan block and the first row except at most one entry. Where $e_1=(1,0,\cdots,0)$, $\begin{pmatrix}J & \Delta \\ & J'\end{pmatrix}$ is a Jordan block bigger than J.
Last step is
\begin{equation}
\begin{pmatrix}1/a&\\&1\end{pmatrix} \begin{pmatrix}\lambda/a&\\&\lambda\end{pmatrix}\begin{pmatrix}a&\\&1\end{pmatrix}=\begin{pmatrix}\lambda&1\\&\lambda\end{pmatrix}
\end{equation} the proof is complete.

So far, I have a question: 
why can we assume $A_0$ is in Jordan normal form?
 A: (original) all matrices in the proof of existence are upper triangular :
That's because the product of two upper triangular matrices is also upper triangular.
$A = (a_{i,j}) \in \mathcal{M}_n(K)$ is upper triangular if $j>i \Rightarrow a_{i,j} = 0$.
Suppose $A, B$ are upper triangular, let $C = AB$, then $C = (c_{i,j})_{1\leqslant i,j \leqslant n}$ where
$$ c_{i,j} = \sum_{1\leqslant k \leqslant n} a_{i,k}b_{k,j}$$
Suppose $j>i$. Since $a_{i,k} = 0$ for $k>i$, one gets
$$c_{i,j} = \sum_{1\leqslant k \leqslant i} a_{i,k}b_{k,j}$$
But when $k\leqslant i$, one has $b_{k,j} = 0$.
Hence, $c_{i,j} = 0$ for all $j>i$ : $C$ is upper triangular.
(edit) wlog $A_0$ is in Jordan normal form :
Assume $A(v_1, \cdots, v_n)= \begin{pmatrix}\lambda & *\\ & B\end{pmatrix}$ for some $B \in \mathcal{M}_{n - 1}(K)$.
(edit_2) Here we use the following induction hypothesis :
For every $k < n$, every endomorphism of $\mathbb{C}^k$ can be put is Jordan normal form. 
One should think of $B$ as the matrix representing an endomorphism of $\operatorname{Span}(v_2, \dots, v_n) \cong \mathbb{C}^{n-1}$. By induction hypothesis, we can find a new basis $(v'_2,\dots,v'_n)$ of $\operatorname{Span}(v_2, \dots, v_n)$ in which $B$ is in Jordan normal form.
Now we have :
  1. $(v_1, v'_2, \dots, v'_n)$ is a basis of $\mathbb{C}^n$.
  2. $A_0 := B(v'_2,\dots,v'_n)$ is in Jordan normal form.
  3. $A(v_1, v'_2, \cdots, v'_n)=\begin{pmatrix}\lambda & *\\ & A_0\end{pmatrix}$
NB: $*$ is not the same as before, but who cares?
And everything is fine! :-)
