When we considered the representation theory of the algebra of polynomials, we saw that all it takes to specify such a representation is choosing a single endomorphism 
. That is, once we pick a transformation 
we get a whole algebra of transformations 
, corresponding to polynomials 
in one variable over the base field 
. Today, I want to outline one useful fact about these: that their kernels are invariant subspaces under the action of .
First, let’s remember what it means for a subspace 
to be invariant. This means that if we take a vector 
then its image 
is again in 
. This generalizes the nice situation about eigenspaces: that we have some control (if not as complete) over the image of a vector.
So, we need to show that if =0](http://s0.wp.com/latex.php?latex=%5Cleft%5Bp%28T%29%5Cright%5D%28u%29%3D0&bg=e6e6e6&fg=333333&s=0 "\left[p(T)\right](u)=0")
then ![\left[p(T)\right]\left(T(u)\right)=0](http://s0.wp.com/latex.php?latex=%5Cleft%5Bp%28T%29%5Cright%5D%5Cleft%28T%28u%29%5Cright%29%3D0&bg=e6e6e6&fg=333333&s=0 "\left[p(T)\right]\left(T(u)\right)=0")
, too. But since this is a representation, we can use the fact that 
, because the polynomial algebra is commutative. Then we calculate
![\displaystyle\begin{aligned}\left[p(T)\right]\left(T(u)\right)&=T\left(\left[p(T)\right](u)\right)\\=T(0)&=0\end{aligned}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Cleft%5Bp%28T%29%5Cright%5D%5Cleft%28T%28u%29%5Cright%29%26%3DT%5Cleft%28%5Cleft%5Bp%28T%29%5Cright%5D%28u%29%5Cright%29%5C%5C%3DT%280%29%26%3D0%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\displaystyle\begin{aligned}\left[p(T)\right]\left(T(u)\right)&=T\left(\left[p(T)\right](u)\right)\\=T(0)&=0\end{aligned}")
Thus if is a linear transformation which is built by evaluating a polynomial at , then its kernel is an invariant subspace for .
February 24, 2009
Posted by John Armstrong |
Algebra, Linear Algebra |
6 Comments
