# 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 $T$.

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February 24, 2009 - Posted by | Algebra, Linear Algebra

## 6 Comments »

1. […] it as the sum . Since , the same is true of the scalar multiple . And since is a polynomial in , its kernel is invariant. Thus we have as well. And thus is invariant under the transformation , and we can assume […]

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2. […] generalized eigenspaces do not overlap, and each one is invariant under . The dimension of the generalized eigenspace associated to is the multiplicity of , which […]

Pingback by Jordan Normal Form « The Unapologetic Mathematician | March 4, 2009 | Reply

3. […] can regard this as a polynomial in applied to , which has real coefficients. We can factor it to […]

Pingback by Real Invariant Subspaces « The Unapologetic Mathematician | March 31, 2009 | Reply

4. […] of the characteristic polynomial of . That is, the characteristic polynomial has a factor . We can evaluate this polynomial at to get the linear transformation . Vectors in the kernel of this space are the eigenvalues […]

Pingback by Eigenvectors of an Eigenpair « The Unapologetic Mathematician | April 3, 2009 | Reply

5. […] off, the generalized eigenspace of an eigenpair is the kernel of a polynomial in . Just like before, this kernel is automatically invariant under , just like the generalized eigenspace […]

Pingback by Generalized Eigenspaces are Still Invariant and Disjoint « The Unapologetic Mathematician | April 7, 2009 | Reply

6. […] tells us that all of our eigenvalues are , and the characteristic polynomial is , where . We can evaluate this on the transformation to find […]

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