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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 $S+T$.

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Hint:

$S+T = \{ s+t : s\in S\text{ and } t\in T\}$

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?

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  • $\begingroup$ I can write in s+t, s and t as linear combinations of basis vectors? $\endgroup$ – Matematică Jan 15 at 20:17
  • $\begingroup$ 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 $x_1,\dots,x_k,y_{k+1},\dots, y_n,z_{k+1},\dots, z_m$. $\endgroup$ – Santana Afton Jan 15 at 21:46
  • $\begingroup$ They are linear independent , but after adding them, the result is still linear independent? $\endgroup$ – Matematică Jan 16 at 8:31
  • $\begingroup$ @Matematică Linear (in)dependence isn’t a property of vectors, but rather of a set of a vectors. What do you mean? $\endgroup$ – Santana Afton Jan 16 at 17:35

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