How to find vectors $u,v$ such $Su=Tv$ where $S$ and $T$ are linearly independent?

Question

Let $X$ be a Banach space, and $S, T$ two surjective linearly independent bounded operators linearly.

How we can find two vectors $u,v\in X$ linearly independent such that $Su=Tv$?

$X$ has infinite dimension.

Answer 1

Hint: If $S$ and $T$ are surjective, then $Su$ is in the image of $T$.

Answer 2

I'm going to prove the converse, i.e. if $Su = Tv$ implies $u$ and $v$ are parallel, then $S$ and $T$ are linearly dependent. Sorry, but this turned out rather long.

Suppose that $Su = Tv$ implies $u$ and $v$ are parallel. Suppose $u_1, u_2 \in X$ such that $Su_1, Su_2 \neq 0$. Then, by surjectivity of $T$, there exist $v_1, v_2 \in X$ such that $Su_i = Tv_i$ for $i = 1, 2$. There also exist scalars $\lambda_1, \lambda_2$ such that $v_i = \lambda_i u_i$, by the hypothesis.

I wish to show that $\lambda_1 = \lambda_2$, i.e. there is a scalar $\lambda$, independent of $u$ and $v$, such that $Su = Tv \neq 0 \implies v = \lambda u$. First, consider the case where $u_1, u_2$ are linearly independent. Then, $$S(u_1 + u_2) = Su_1 + Su_2 = Tv_1 + Tv_2 = T(v_1 + v_2).$$ There must exist some $\lambda$ such that $$v_1 + v_2 = \lambda(u_1 + u_2).$$ But then, $$\lambda(u_1 + u_2) = \lambda_1 u_1 + \lambda_2 u_2 \implies (\lambda_1 - \lambda)u_1 + (\lambda_2 - \lambda)u_2 = 0.$$ Since $u_1, u_2$ are linearly independent, we have $\lambda = \lambda_1 = \lambda_2$.

On the other hand, suppose $u_1, u_2$ are linearly dependent. Since $Su_1, Su_2 \neq 0$, we have $u_1, u_2 \neq 0$, so there exists some $k \neq 0$ such that $u_2 = ku_1$. Note that $$T(kv_1) = kTv_1 = kSu_1 = Su_2,$$ and hence $kv_1$ and $u_2$ are parallel. Since $Tv_1 = Su_1 \neq 0$, we have $v_1 \neq 0$, hence $kv_1 \neq 0$. It follows by transitivity of non-zero parallel vectors that $kv_1$ (and hence $v_1$) is parallel to $v_2$. With a little tedious calculation, we see that $v_2 = kv_1$. We therefore have $$v_2 = \lambda_2 u_2 \implies kv_1 = \lambda_2 k u_1 \implies v_1 = \lambda_2 u_1 \implies \lambda_1 = \lambda_2.$$

Thus, in either case, the claim is true: there exists some scalar $\lambda$ such that, $$Su = Tv \neq 0 \implies v = \lambda u.$$ I claim that $S = \lambda T$, proving $S$ and $T$ are dependent. Suppose $x \in X$. If $Sx \neq 0$, then there exists some $v$ such that $Sx = Tv$, and by the given result, we have $v = \lambda x$. Therefore, $$(\lambda T - S)x = T(\lambda x) - Sx = Tv - Sx = 0.$$

Otherwise $Sx = 0$. If $Tx = 0$, then we are done and may assume $Tx \neq 0$. By the surjectivity of $S$, choose some $u$ such that $Su = Tx \neq 0$. But this means $u$ and $x$ are parallel, neither of which are equal to $0$, hence $Su$ is a multiple of $Sx = 0$, a contradiction.

Thus, $S$ and $T$ are linearly dependent.