Restriction of a Linear Transformation

[dward1996]

Given a linear tansformation T of a vector space V (over a field K) with eigenbasis {v_{1},...,v_{n}}, and a (non-trivial) subspace W of V such that T(W) is a subset of W, a lecturer keeps using the result that W will contain an eignvector for T. I can see why this would be the case if the field K were algebraically closed, but how do we know that W will have any eigenvectors for T if K is an arbitrary field?

 

[henry_m]

It follows from the fact that the operator has an eigenbasis. Here's a proof:

W is a nontrivial subspace so it contains some nonzero vector $v=a_{i_1}v_{i_1}+\cdots +a_{i_k}v_{i_k}$, where the $v_{i_k}$s are some subset of the eigenbasis, all of the $a_{i_k}$s are nonzero, and $1\leq k\leq n$. Since $T(W)\subseteq T$, we have $T(v)\in W$. If all the eigenvalues of the $v_{i_k}$s are equal, we've found an eigenvector so we're done. If not, we can use the fact that W is a subspace to construct a linear combination of v and T(v), in W, that eliminates one of the $v_{i_k}$s (concretely, $\lambda_{i_1}v-T(v)$, where $\lambda_{i}$ is the eigenvalue of $v_{i}$. It's nonzero if the eigenvalues aren't all equal.). So we construct another nonzero vector in W, but with k reduced by 1.

We can now repeat the process. It must eventually terminate, at k=1 if not before. So we've constructed an eigenvector for T in W.

 

[dward1996]

Thanks for that. It makes perfect sense now!