# Endomorphisms and Invariant Subspaces

I have a question or two regarding the following exercise:

Let $\alpha$ be the endomorphism of $\Bbb{Q}^4$ defined by: $$\alpha : \left[\begin{matrix}a \\ b \\ c \\ d \end{matrix}\right] \mapsto \left[\begin{matrix}2a-2b-2c-2d \\ 5b-c-d \\ -b+5c-d \\ -b-c+5d \end{matrix}\right].$$ Find nontrivial proper subpsaces of $\Bbb{Q}^4$ which are invariant under $\alpha$.

I know what is meant by $\alpha$-invariant, and I know what constitutes trivial subspaces that are invariant under this transformation. I've also read discussions about using the eigenvectors of the linear transformation to determine these invariant subspaces, and that made sense, more or less. But this particular textbook is broaching endomorphisms and invariant subspaces much earlier than any mention of eigenvalues/eigenvectors and even before matrix algebra or any number of matrix manipulations. How does the author intend for us to find invariant subpsaces by other methods besides those relying upon eigenspace techniques? The chapter of this book provides only one concrete example, and it is for a much simpler scenario. Interestingly, the introduction to the book states that the book "does not assume any previous knowledge of formal linear algebra... and thus the volume is self-contained." I assumed that this could mean the reader does not need to know what an eigenvector is by this stage (however, I am quite familiar).

Furthermore, while I understand the basic definition of endomorphism, how does one determine if a certain linear transformation qualifies as an endomorphism in particular? Does it have something to do with the fact that addition and multiplication (given by composition) are defined on endomorphisms of a vector space? At a glance, some of these endomorphisms appear indistinguishable from other linear transformations.

I appreciate the input.

• I would say that in this context endomorphisms (of a vector space) are the same as linear maps. Jun 12, 2014 at 16:59

If it goes from a linear space to the same linear space, it's an endomorphism. That is, $T:V\to V$ is an endomorphism, but $T:V\to W$, where $W$ is another vector space, is not.
No idea. To find all invariant subspaces is essentially same thing as to find the Jordan form of $\alpha$.