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?
Best Answer
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.