# Invariant subspaces of vector space.

I'm confused with the next exercise:

Take $$\left\{ 1,e^t,e^{-t}\right\}$$ a basis for a vector space $$V$$ (note that $$V$$ is a space of continuous functions). Take a linear transformation $$T:V\to V$$ defined by $$T(f)=f'$$ (derivative). The question is: find all the invariant subspaces of $$V$$ under $$T$$.

After a lot of time thinking, I think that the answers is: every subspace generated by a non-empty subset of the basis, i.e., take $$W$$ an invariant subspace, then, there exist $$\emptyset\neq B\subseteq \left\{ 1,e^t,e^{-t}\right\}$$ such that $$\text{span}(B)=W$$. But, how to prove? I can prove that if $$\emptyset\neq B\subseteq \left\{ 1,e^t,e^{-t}\right\}$$ then $$\text{span}(B)$$ is invariant under T but the other direction seems too dificult for me and I can't find how to prove. Any hint? I really appreciate any help you can provide.

• Hint: consider how the endomorphisms $\frac{1}{2}T(T-I)$, $-(T-I)(T+I)$, $\frac{1}{2}T(T+I)$ act on the basis. – Mindlack Nov 4 '19 at 22:49

There are several ways to do that. But in this exercise in particular, the basis $$B$$ is a basis of eigenvectors of $$T$$ then. In particular, the matrix associated to it in that basis is diagonal
$$[T]_B=\left(\begin{array}{ccc}0&0&0\\ 0&1&0\\ 0&0&-1\end{array}\right)$$
And the eigenspaces of $$T$$ are simply $$V_0=\mathbb R, V_1=\mathbb Re^t,V_{-1}=\mathbb Re^{-t}$$. It is a well known fact (and I encourage you to prove it) that an invariant subspace of a diagonalizable operator is an invariant subspace if and only if it is the direct sum of some of its eigen-spaces.
Suppose that $$W$$ is an invariant subspace and let $$w=\alpha+\beta e^t+\gamma e^{-t}\in W$$. Then $$T(w)\in W$$ too. But $$T(w)=\beta e^t-\gamma e^{-t}$$. And $$T\bigl(T(w)\bigr)\in W$$. But $$T\bigl(T(w)\bigr)=\beta e^t+\gamma e^{-t}$$. So $$\alpha\bigl(=w-T\bigl(T(w)\bigr)\bigr)\in W$$. Therefore, either $$\alpha=0$$ or the constant functions belong to $$W$$. In either case, $$\beta e^t+\gamma e^{-t}\in W$$. So$$\beta e^t=\frac12\bigl((\beta e^t+\gamma e^{-t})+(\beta e^t-\gamma e^{-t})\bigr)\in W.$$So, either $$\beta=0$$ or every multiple of $$e^t$$ belongs to $$W$$. Finally, from$$\gamma e^{-t}=\frac12\bigl((\beta e^t+\gamma e^{-t})-(\beta e^t-\gamma e^{-t})\bigr)\in W$$you get that either $$\gamma=0$$ or every multiple of $$e^{-t}$$ belongs to $$W$$.