I am staying in downtown Toronto (near the Church/Adelaide St crossing). I will have to catch a national flight from Pearson Airport (YYZ). The flight is leaving at a proper time (around noon) on a weekday. How can I effectively (quick, cheap) get from downtown Toronto to YYZ by public transportation?


Question: Is there an analogue of the jordan normal form of an nilpotent linear transform of a polynomial ring?
Answer: Let $f:V\to V$ be an idempotent endomorphism of any vector space over any field. Let $V_0=\ker f$ and $V_1=\{v\in V:f(v)=v\}$. Then it is immediate that $V_0\cap V_1=\{0\}$, and if $v\in V$, then $v=v_0+v_1$ with $v_0=v-f(v)$ and $v_1=f(v)$, so that $f(v_0)=f(v)-f(f(v))=0$ and $f(v_1)=f(f(v))=f(v)=v_1$, that is, $v_0\in V_0$ and $v_1\in V_1$. This shows that $V=V_0\oplus V_1$.

It follows that every idempotent endomorphism of a vector space is diagonalizable, and that its only possible eigenvalues are zero and one. Finite dimensionality plays no role in this context.

This information in no way helps you find the kernel of $f$, by the way... Finding the «friendly basis» you talk about (if by that we understand «basis which diagonalizes the map») is equivalent (if not actually harder!) to finding the kernel.