# Prove that basis of sum of subspaces contains bases of subspaces and the intersection

We have a vector space $$V$$ and two subspaces $$S$$, and $$T$$. We know that $$\{x_1, x_2, ..., x_k\}$$ is a basis of $$S\cap T$$, such that $$\{x_1, x_2, ..., x_k, y_{k+1}, ..., y_n\}$$ is a basis of $$S$$, and $$\{x_1, x_2, ..., x_k, z_{k+1}, ..., z_m\}$$ is a basis of $$T$$, we have to prove that $$\{x_1, x_2, ..., x_k, y_{k+1}, ..., y_n, z_{k+1}, ..., z_m\}$$ is a basis of $$S+T$$.

I tried to write the condition of linear independence of this base and extracting bases of $$S$$, $$T$$, and $$S\cap T$$ which we know that are linear independent but I have to use base vectors of $$S\cap T$$ in three places after separating the linear combination of big basis, and I'm not sure if this works, and I don't know how to prove linear independence and that this basis is generating $$S+T$$.

$$S+T = \{ s+t : s\in S\text{ and } t\in T\}$$
Given a basis for $$S$$ and a basis for $$T$$, you can write $$s$$ and $$t$$ as a linear combination of these two. What do you get when you write out $$s+t$$ using these linear combinations?
• If you have an expression of $s$ in terms of $x_1,\dots,x_k,y_{k+1},/dots, y_n$, and an expression of $t$ in terms of $x_1,\dots,x_k,z_{k+1}, z_m$, then you can add them together to get an expression of $s+t$ in terms of $x_1,\dots,x_k,y_{k+1},\dots, y_n,z_{k+1},\dots, z_m$. Jan 15, 2019 at 21:46