# Suppose that $(u_1,\ldots, u_m, v_1, \ldots, v_n)$ is linearly independent. Prove that $W_1 ∩ W_2 = \{0\}$.

Let $$V$$ be a real vector space and $$W_1$$ and $$W_2$$ two finite dimensional subspaces. Let $$(u_1,\ldots, u_m)$$ be a basis for $$W_1$$ and $$(v_1,\ldots, v_n)$$ be a basis for $$W2$$.

Suppose that $$(u_1,\ldots, u_m, v_1,\ldots, v_n)$$ is linearly independent. Prove that $$W_1 ∩ W_2 = \{0\}$$.

I know that if $$w ∈ W_1 ∩ W_2$$, then $$w$$ is a linear combination of the $$u_i$$ and also a linear combination of the $$v_j$$). I am however not sure how to proceed next.

Since $$w$$ is a linear combination of the $$u_i$$'s and also of the $$v_j$$'s, then you have$$w=\alpha_1u_1+\cdots+\alpha_mu_m=\beta_1v_1+\cdots+\beta_nv_n.$$But then$$\alpha_1u_1+\cdots+\alpha_mu_m-\beta_1v_1-\cdots-\beta_nv_n=0.$$Therefore, the linear independence implies that all coefficients are $$0$$. In particular, $$w=0$$.