The Unapologetic Mathematician

Mathematics for the interested outsider

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


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


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  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 […]

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  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 […]

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