# The Unapologetic Mathematician

## Kernels of Polynomials of Transformations

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 $T:V\rightarrow V$. That is, once we pick a transformation $T$ we get a whole algebra of transformations $p(T)$, corresponding to polynomials $p$ in one variable over the base field $\mathbb{F}$. Today, I want to outline one useful fact about these: that their kernels are invariant subspaces under the action of $T$.

First, let’s remember what it means for a subspace $U\subseteq V$ to be invariant. This means that if we take a vector $u\in U$ then its image $T(u)$ is again in $U$. 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 $\left[p(T)\right](u)=0$ then $\left[p(T)\right]\left(T(u)\right)=0$, too. But since this is a representation, we can use the fact that $p(X)X=Xp(X)$, 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}

Thus if $p(T)$ is a linear transformation which is built by evaluating a polynomial at $T$, then its kernel is an invariant subspace for $T$.

February 24, 2009 Posted by | Algebra, Linear Algebra | 6 Comments