[Math] Proof verification: If $U$ is a subspace of $V$ that is invariant under every operator on $V$, then $U=\{0\}$ or $U=V$

invariant-subspacelinear algebraproof-verificationvector-spaces

I've to prove or produce a counter-example of the following statement:

If $U$ is a subspace of $V$ that is invariant under every operator on $V$, then $U=\{0\}$ or $U=V$.

My approach is to prove the contrapositive): Fix a subspace $U$ of $V$, that is neither $\{0\}$ nor $V$. We'll show that we can always construct an operator $S$ such that $U$ is not invariant under $S$. Since $U \neq \{0\}$, we can form a basis of $U$ as $\{u_1,\ldots,u_p\}$ for some $p \in \mathbb{N}$. One can extend this basis to form a basis for the whole space $V$. Since $U \neq V$, we can have a vector $w \in V$ which is not in $U$. Recall that we can define a linear map $S$ by defining the value that $S$ takes at every element of the basis of $V$ (for any $v \in V$, $Sv$ can be calculated by expressing $v$ in terms of the basis vectors and using linearity of $S$). So we define $S$ by setting $Su_i=w$ for $i=1(1)p$ and setting $S$ arbitrarily (in $V$) for other elements in the basis of $V$ (obtained by extending the basis of $U$). Clearly $S$ is an operator on $V$ but its restriction on $U$ not an operator on $U$. Hence the proof.

Can anyone please verify this solution? My doubt is that is it okay to assume $U$ to be finite-dimensional here?

Best Answer

This idea will work, but you can clean it up (and get rid of the basis). Here's a sketch. Assume that $U\not=\{0\}$ and $U\not=V$. We will prove that $U$ is not invariant.

Since $U\not=V$, there is some $v\in V\setminus U$. Since $U\not=\{0\}$, there is a nonzero $u\in U$. Let $T$ be a linear operator on $\operatorname{Span}(u)$ such that $T(u)=v$. Now, extend $T$ to a linear operator on all of $V$ (depending on how you do this, you might need the axiom of choice / Zorn's lemma).

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