## 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 . 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 then , too. But since this is a representation, we can use the fact that , because the polynomial algebra is commutative. Then we calculate

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

[...] 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 |

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

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[...] 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 |

[...] 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 |

[...] 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 |

[...] 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 |