How do I show a Jordan basis? I'm having a bit of trouble with this problem:
Let $β=\{e^{2x}, xe^{2x}, e^{x}\}$ and define $V=\mbox{span}(\beta)$. Let $T=D-2$ where $D=d/dx$. Show that $\beta$ is a Jordan basis for $T$.
How do I show that $\{e^{2x}, xe^{2x}, e^{x}\}$ is a Jordan basis for $D-2$?
 A: As  said in a comment what needs to be decide is if this  basis is formed by a chain of generalized eigen-vectors. 
To this end we check the effect of $T$ on the vectors. 


*

*$T(e^{2x}) = 0$, this is does an eigenvactor to the eigenvalue $0$. 

*$T(xe^{2x}) = e^{2x}$ and thus $T^2 (e^{2x})=0$, this is a generalized eigenvactor to the eigenvalue $0$, as $T^2 (e^{2x})=0$. 

*$T(e^x) = -e^x$. This is thus an actual eingenvector to eigenvalue $-1$.
Since $T(xe^{2x}) = e^{2x}$ we see that $(e^{2x}, x e^{2x})$ is a chain of eigenvectors to the eigenvalue $0$. And in addition we have one more actual eigenvector. Thus this is a Jordan basis.  
Alternatively, one could compute the matrix associate to the transformation with respect to this basis and note that it is in Jordan form. 
A: If the vector space is over a field K, then a basis is a jordan basis if and only if all eigenvalues of the matrix lie in K. 
Using this to solve your specific problem will simply involve finding the eigenvalues and showing that they are in the given field.
Have fun with the quiz. ;)
