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Let $U,V$ be subspaces of $W$.

Prove that $U,V$ are linearly independent iff $U \cap V = \{0\}$

I understand that for a subspace to be linearly independent it means that no linear combination of one element from each of $U,V$ is equal to $0$ except for the trivial one. I am having difficulty applying this to the proof though.

Is it true that the two subspaces are dependent if they have a non-empty intersection because this means that the span of each subspace overlaps with the other? This implies there are only the scalars $c_1 = c_2 = ... = c_n = 0$ that allow the elements of the subspace to equal $0$?

My understanding of the notion of linearly independent subspaces is clearly shaky. Any insight would be much appreciated.

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  • $\begingroup$ Small caution: As stated, your assertion is not correct. Probably you mean that if $U$ and $V$ are linearly independent subsets of $W$, then $U \cup V$ is linearly independent if and only if $\operatorname{span}(U) \cap \operatorname{span}(V) = \{0\}$. The existing answers have implicitly reworded your question in this way. The set of vectors in a subspace is never a linearly independent set. $\endgroup$ – Andrew D. Hwang Jan 31 '14 at 11:50
  • $\begingroup$ I edited my question to state $U \cap V = \{0\}$ instead of $U \cap V = 0$. $\endgroup$ – Alan Jan 31 '14 at 18:13
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Hint

If a non zero vector $v\in U\cap V$ then $$\{\lambda v\;|\; \lambda \in \Bbb R\}\subset U\cap V$$

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$Hint:$ if $au+bv = 0$, then $u = -\frac{b}{a}v$. Conversely, if $w \ne 0$ is in $U \cap V$, then $w-w = 0$ gives the desired relation.

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Assume there is some $0\neq v\in U\cap V$, fix bases $a_i$ and $b_j$ for $U$ and $V$. Then you can write $v=\sum_i\alpha_ia_i=\sum_j\beta_jb_j$ for some $\alpha_i,\beta_j\in\mathbb{R}$ (or whatever field you're on). Substracting the two summations, you get an expression of linear dependence between $U$ and $V$.

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