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

Question

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.

Answer 1

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, $w=0$.