Linear Algebra: How can I demonstrate this lemma?

Question

I have a vector space $V$ with $\dim V = n$ and a base $\langle u_{1}, u_{2}, ... , u_{k}, ... , u_{n}\rangle$. I have to prove that $\langle u_{1}, ... , u_{k}\rangle \oplus \langle u_{k+1}, ... , u_{n}\rangle = V$. Any ideas? I've tried to show that taken $x \in \langle u_{1}, ... , u_{k}\rangle, x' \in \langle u_{k+1}, ... , u_{n}\rangle\Rightarrow x+x' = 0 \iff x = 0$ and $x'=0$ for the definition of direct sum, but I don't know how this could help.

Answer 1

Since$$u_1,\ldots u_k\in\langle u_1,\ldots u_k\rangle\text{ and }u_{k+1},\ldots u_n\in\langle u_{k+1},\ldots u_n\rangle,$$you know that$$u_1,\ldots,u_n\in\langle u_1,\ldots,u_k\rangle+\langle u_{k+1},\ldots,u_n\rangle$$ and, since $\{u_1,\ldots u_n\}$ is a basis, it generates $V$, and therefore$$\langle u_1,\ldots u_k\rangle+\langle u_{k+1},\ldots u_n\rangle=V.$$On the other hand, if $v\in\langle u_1,\ldots u_k\rangle\cap\langle u_{k+1},\ldots,u_n\rangle$, then $v$ can be written both as$$v=\sum_{j=1}^k\alpha_ju_j\text{ and }v=\sum_{j=k+1}^n\alpha_ju_j.$$This means that$$\alpha_1u_1+\cdots\alpha_ku_k-\alpha_{k+1}u_{k+1}-\cdots-\alpha_nu_n=0.$$Since $\{u_1,\ldots u_n\}$ is a basis, all the $\alpha_j$'s are $0$. This proves that$$\langle u_1,\ldots u_k\rangle\cap\langle u_{k+1},\ldots,u_n\rangle=\{0\}.$$