4
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

(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! :-)

$\endgroup$
  • $\begingroup$ But how can we conclude $A_0$ is upper triangular from the start? $\endgroup$ – Tengerye Sep 5 at 13:56
  • 1
    $\begingroup$ Well my answer is obsolete since you edited your question. You should leave the original question, then add the other question after a markup like Edit $\endgroup$ – Olivier Roche Sep 5 at 14:00
  • $\begingroup$ As to this second question, you should reformulate it as "why can we assume $A_0$ is in Jordan normal form?". $\endgroup$ – Olivier Roche Sep 5 at 14:02
  • $\begingroup$ Thank you for your kind suggestion. $\endgroup$ – Tengerye Sep 5 at 14:22
  • $\begingroup$ Oh, and don't forget to mark the answer if it answers your question(s). ;) $\endgroup$ – Olivier Roche Sep 5 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.