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 […]
Pingback by Jordan Normal Form « The Unapologetic Mathematician | March 4, 2009 |
[…] 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 |