If U is a T-invariant subspace of dim n+1, and if there is no subspace W ⊆≠U such that T(w) ∈ W for all w ∈ W, then the set of vectors {u,T(u),T^2(u),...,T^n(u)}, u ∈ U, u ≠ 0 forms a basis for U.
To show the set of vectors forms a basis. I need to show
- The set is a linearly independent set
- The span of this set is U.
I am stuck at this point. How can I show the vectors in this set is linearly independent? Also, I am not sure how to show the span(set) = U. I know the dimension of this set is also n, which matches to the dimension of the subspace. Can I claim since all the T-invariant subspaces are in U, the set of vectors span U? Thank you so much!
Edit: added u is non-zero
Since you're given that $\;\dim U=n+1\;$ and we have $\;n+1\;$ vectors in $\;B:=\{\,u=T^0u, Tu,...,T^nu\,\}\;$ , all we have to do is to prove $\;B\;$ is linearly independent . So assume it isn't, and this means there's a vector in $\;B\;$ which is a linear combination of the preceeding ones:
$$T^mu=\sum_{k=0}^{m-1}a_kT^ku:=r$$
But this then means the subspace $\;W:=\text{Span}\,\{u,Tu,...,T^{m-1}u\}\;$ is $\;T\,-$ invariant, since
$$x\in W\implies x=\sum_{i=0}^{m-1}b_iT^iu\;\;\implies\;\;Tx=b_0Tu+\ldots+b_{m-1}T^mu=$$
$$=b_0Tu+\ldots+b_{m-2}T^{m-1}u+b_{m-1}r\in W\ldots!$$
This means either $\;W=\{0\}\;$ or $\;W=U\;$, and we're done (fill in details...)