# Prove that two subspaces of a vector space intersect only at 0

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.

• What definition do you use for direct sums? Commented Feb 6, 2019 at 5:16
• Doesn't it follow directly from definition of direct sum? Commented Feb 6, 2019 at 5:54
• How can we sum subspaces of different vector spaces ? Commented Feb 6, 2019 at 6:15
• Think about the dimension of the direct sum. If the intersection had anything else in it.... Commented Feb 6, 2019 at 6:50
• I gave it a shot below, any feedback would be greatly appreciated! Commented Feb 7, 2019 at 7:09

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 $$V \cap W = \{0\}$$.
• $u$ is not equal to $v_i$ or $w_i$ but instead a linear combination of the $v_i, w_i$. Put differently y, you could write $u = v + w$, where $v \in V, w \ in W$ Assume that some vector $v$ is in both $V$ and $W$; try and use the linear independence of $\{v_1, …, v_n, w_1, …, w_m\}$ to conclude that $v = 0$. Commented Aug 6, 2023 at 13:24
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 $$\mathbf{R}^{n+m}=V'\oplus W'$$.