Let $V \subset \mathbb{R}^n$ & $W \subset \mathbb{R}^m$ with set of basis $S_V=\{v_1,v_2,...,v_n\}$ and $S_W=\{w_1,w_2,...,w_m\}$. The vector space spanned by these basis vectors is the direct sum of $V$ and $W$ with a dimension of $n+m$.
How do I prove that $V \cap W=\{0\}$?
I understand how to draw it conceptually because the only intersection that both subsets share is the zero vector. But I don't quite know how to do the proof mathematically.
On
Since $V$ and $W$ are subspaces of different vector spaces, so $V\cap W=\emptyset$. But we can try to make the question meaningful by introducing the maps \begin{align} F\colon V\subset\mathbf{R}^m&\to\mathbf{R}^{m+n}\\ v&\mapsto(v,\underbrace{0,\ldots,0}_{n\text{ copies}}) \end{align} and \begin{align} G\colon W\subset\mathbf{R}^n&\to\mathbf{R}^{m+n}\\ w&\mapsto(\underbrace{0,\ldots,0}_{m\text{ copies}},w) \end{align} $F$ and $G$ are injective and if we define $V':=F(V)\subset\mathbf{R}^{n+m}$ and $W':=G(W)\subset\mathbf{R}^{n+m}$, the maps $V\ni v\mapsto F(v)\in V'$ and $W\ni w\mapsto G(w)\in W'$ are vector space isomorphisms. You can easily prove that $V'\cap W'=\{0\}$ and if $V$ and $W$ span $\mathbf{R}^m$ and $\mathbf{R}^n$, respectively, then $\mathbf{R}^{n+m}=V'\oplus W'$.
Thanks for the responses, everyone. This is what I came up with: $\forall u \in \mathbb{R}^{m+n}, u=v_i \in V \lor u=w_i \in W$ Therefore, because of the definition of the direct sum and subspaces, we know that $u_i \neq v_i \land u_i \neq w_i$ unless $u_i=0$ therefore $V \cap W = \{0\}$.