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  • Let $S=\operatorname{Span}\{v_1,\ldots,v_k\}$ and $M=\operatorname{Span}\{w_1,\ldots,w_l\}$ be a set in $\mathbb{R^3}$ such that $S\cap M=\{0\}$. Show that $k+l\leq3$ and $\{v_1,\ldots,v_k,w_1,\ldots,w_l\}$ is a basis of $S+M=\{u+v: u\in S,v\in T\}$.

My question is why $S\cap M=\left\{ 0\right\}$?

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    $\begingroup$ Why? Because it's given so. Or are you asking how this information is useful for the desired proof? More importantly, the claim is stated here is false. $\endgroup$ – zipirovich Jan 4 '18 at 2:24
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$S\cap M=\left\{ 0\right\}$ is a necessary assumption. It means that the only way to obtain $\alpha_1v_1+\ldots \alpha_kv_k=\beta_1w_1+\ldots \beta_lw_l$ is by setting $\alpha_1=\ldots=\alpha_k=\beta_1=\ldots=\beta_l$=0. If the intersection was not $\{ 0\}$, then we had linear dependent vectors, and linear dependent vectors cannot form a basis, because a basis must always contain linear independent elements.

Note that we cannot have $S\cap M=\varnothing$, as we can always choose the scalars equal to $0$, so both spans always contain the zero vector.

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If they intersected nontrivially, the basis given wouldn't be linearly independent, so it wouldn't be a basis.

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