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

\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 [...]

    Pingback by Generalized Eigenspaces are Disjoint « The Unapologetic Mathematician | February 25, 2009 | Reply

  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 [...]

    Pingback by A Lemma on Reflections « The Unapologetic Mathematician | January 19, 2010 | Reply


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