Let $T$ and $U$ be non-zero linear transformations from $V$ to $W$. If $R(T)\cap R(U) = \{0\}$, prove that $\{T,U\}$ is LI

Let $$V$$ and $$W$$ be vector spaces, and let $$T$$ and $$U$$ be non-zero linear transformations from $$V$$ to $$W$$. If $$R(T)\cap R(U) = \{0\}$$, prove that $$\{T,U\}$$ is a linearly independent subset of $$\mathcal{L}(V,W)$$.

My solution

Let us consider the following linear combination: \begin{align*} \alpha T + \beta U = 0 \end{align*}

If they were linear dependent, we could assume without loss of generality that $$\alpha \neq 0$$. Thus we would have \begin{align*} T = -\frac{\beta}{\alpha}U \end{align*}

Consequently, if we consider a basis $$\mathcal{B}_{V} = \{v_{1},v_{2},\ldots,v_{n}\}$$, we get the following relation \begin{align*} T(v_{j}) = -\frac{\beta}{\alpha}U(v_{j}) \Rightarrow T(v_{j}) \in R(T)\cap R(U) \Rightarrow T(v_{j}) = 0 \Rightarrow T = 0 \end{align*} which contradicts the given assumption. Hence the proposed result holds.

Could someone please double-check my solution? Is there another way to approach just for the sake of curiosity?

Your proof looks fine. One refinement might be to avoid taking a basis (after all, no guarantee that $$V$$ has finite dimension). Rather you could just note that $$\beta \neq 0$$ since neither mapping is zero, and then see that any vector in the range of $$U$$ is also in the range of $$V$$. Also, you might emphasise that you move the $$-\beta/\alpha$$ inside the operator $$U$$‘S argument noting that linearity allows you do that.