Question: Here's some extra information:

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> Let $T:\mathbb R^4→\mathbb R^4$ denote a linear transformation. Set $p(x)=(x−2)^2 + 4$ and $q(x)=(x+1)^2 + 5$. Suppose the characteristic polynomial of $T$ is the product of $p$ and $q$.
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> a) Show that $T$ has no eigenvectors.
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> b) Show that $\dim(\ker(p(T))=2$ and $\dim(\ker(q(T))=2$.
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I was able to answer a) with no issues. Since p and q are not able to be split into real roots, there can't possibly be eigenvalues for our transformation. Thus, T has no eigenvalues so it does not have any eigenvectors.

The trouble I'm having is proving the proposition in the title. My professor gave a very unsatisfactory proof of this. :( It is true that we have two solutions that will produce 0 for both functions $p,q$, but these solutions only exist in the complex plane, not over the real line! So my question for you guys is: how is this actually true and how do I go about proving this?

Answer: 
We have by the hypothesis
$$\chi\_T(x)=p(x)q(x)$$
and by the Cayely-Hamilton theorem and since the two polynomials are co-prime so using the [lemma of kernels$^{(1)}$](http://fr.wikipedia.org/wiki/Lemme_des_noyaux) we have

$$\Bbb R^4=\ker(p(T))\oplus\ker (q(T))$$
Moreover we know that if $\lambda$ is an eigenvalue of $T$ then $p(\lambda)$ is an eigenvalue of $p(T)$ so we conclude that $0$ is an eigenvalue of multiplicity $2$ of $p(T)$ and then $\dim \ker(p(T))=2$.

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$(1)$ This page is in French language.
A simple consideration of dimensions can be used to deduce point b) from point a), using only the basic fact that the kernel or image of any polynomial in $T$ will be a $T$-stable subspace (both statements have a one-line proof which uses only that the polynomial in $T$ *commutes* with $T$), and the Cayley-Hamilton theorem.

Point a) is equivalent to the fact $T$ has no $1$-dimensional $T$-stable subspaces: the space spanned by a nonzero vector $v$ is $T$-stable is and only if $v$ is an eigenvector. This means $P[T]$ and $Q[T]$ cannot have rank $1$ (since their image is $T$-stable) or rank $3$ (since their kernel is $T$-stable). Neither can they have rank $0$, since for instance $P[T]=0$ would contradict that $q$ (an irreducible quadratic polynomial distinct from$~p$) divides the characteristic polynomial, and vice versa. Now
$$
 P[T]\circ Q[T]=0=Q[T]\circ P[T]
$$
(obtained from the Cayley-Hamilton theorem) means that $\def\Im{\operatorname{Im}}\ker(P[T])\supseteq\Im(Q[T])$ and $\ker(Q[T])\supseteq\Im(P[T])$, and given what we saw about their ranks this is only possible if both $P[T]$ and $Q[T]$ have rank$~2$, with $\ker(P[T])=\Im(Q[T])$ and $\ker(Q[T])=\Im(P[T])$.