What is the easiest proof you know for the Jordan Canonical Form

I read numerous demonstration of the existence of the Jordan Canonical Form, but all of them involve more than 2 pages of demonstration with numerous lemmas in between.

I'm writing some notes for some students, but the subject is only tangentially related to Jordan Normal Form and so I was wondering if anybody knew a simple 1-page demonstration of the existence of this form! If the demonstration is original I will find a way to cite you :)

• I'd say the answer depends greatly on the background of the students. For instance, are they well-versed in the theory of finitely generated modules over a PID? If so, then half a page would suffice. Are they only familiar with the basic operations on real and complex matrices? Something in between? – Noah Olander Jan 17 '16 at 21:28
• The Wikipedia article's presentation seems pretty straightforward to me, at least if you are already comfortable with the rank theorem. The gist: the range of $A-\lambda I$ is invariant under $A$; if $\lambda$ is nondefective, then you're done (by induction); if $\lambda$ is defective, then introduce $r-s$ generalized eigenvectors; argue that the collection of eigenvectors and generalized eigenvectors are linearly independent, and that the action of $A$ on the span of the eigenvectors of generalized eigenvectors is given by a Jordan block. – Ian Jan 17 '16 at 21:31
• The difficult thing, that I really still don't have a good feel for myself, is: why introduce the generalized eigenvectors? Clearly you have to introduce some additional basis elements, but why those in particular? – Ian Jan 17 '16 at 21:37
• The generalized eigenvectors, and the associated Jordan chains, can be understood, I think, just by looking at some examples and understanding them geometrically as linear transformations, starting with the simplest, namely the $2 \times 2$ matrix of the form $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. – Lee Mosher Jan 17 '16 at 21:44
• Are notes on basic Operator Theory and the environment it would be the one of the spectral theorem so maybe your demonstration Ian could be interesting. Which article is of Wikipedia? – Dac0 Jan 17 '16 at 22:42

$$\newcommand{\span}{\operatorname{span}}$$Are you assuming an algebraically closed closed ground field? If so then here is a proof. Let $$W_{\lambda}=\{v :(T-\lambda)^kv=0, \text{for some k} \}$$. Its not hard to show that $$V=\bigoplus W_{\lambda}$$, ill skip this but it does take some time. So look at the operator $$N=T-\lambda$$ in the space $$W=W_{\lambda}$$. $$N$$ is nilpotent. Chose $$v_1, \ldots , v_p$$ basis for $$\ker N$$, which has to be non zero. Now consider those $$w$$ such that $$Nw\in \span\{ v_1\}$$ these $$w$$ are one dimensional, modulo the $$\ker N$$ : since if there were $$w$$ and $$w^{\prime}$$ ,linearly independent, with a choice of constant $$N(w-bw^{\prime})=0$$ and so $$w-bw^{\prime}\in \ker N$$. So now chose some vectors such that $$Nw_i \in \span\{ v_i\}$$ where possible say $$w_1\ldots w_q$$ for $$q\leq p$$. Next chose $$x_i$$ such that $$Nx_i\in \span\{ w_i\}$$ and so on. Now your basis for $$W$$ is $$v_1,w_1,x_1,y_1,\ldots v_2,w_2,x_2, \ldots v_3,w_3,x_3,\ldots$$ its easy to see that under this basis $$N$$ and thus $$T$$ has the desired form on $$W$$. Its not 2 pages but it is also quite terse.
• I don't understand this answer. According to me, the main difficulty in the proof is to "unravel" the vectors as we build the basis. I don't see how it is done here. To be more specific, what if there is no $w$ such that $Nw \in \operatorname{span}\{v_1\}$? It could happen for instance that $\dim(W) = 3$, $N^2 = 0$ with a basis $v_1, v_2$ for $\ker(N)$, but $\operatorname{im}(N) ⊆ \ker(N)$ is not included in the span of neither $v_1$ nor $v_2$. We have to do some "unraveling" if we want to build the basis "bottom-up". – Idéophage Dec 17 '18 at 21:04