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Question: $V$ is a vector space over field $F$ , $U,W$ are subspaces of $V$. Is the next statement true or false? "$K$ is a basis for $W$, and $L$ is a basis for $U$. Therefore $K\cup L$ is a basis for $U + W$".

What I did: (I "proved" it's true. The book says it's false)

$K \cup L$ is an union of two linear independent groups, and is therefore linear independent. We mark: $$K = \{ v_1, ... , v_n, w_1, ... , w_k \} \ ; \ L=\{ v_1, ... , v_n, u_1, ... , u_l \} \ ; \\ K\cup L = \{ v_1, ... , v_n, w_1, ... , w_k,u_1,...,u_l \}$$ Therefore, for every $u\in U, w\in W$: $$w = a_1v_1 + ... + a_nv_n + a'_1w_1 +...+ a'_kw_k\\ u=a_1v_1 + ... + a_nv_n + a'_1u_1 +...+ a'_lu_l\\ w + u= a_1v_1 + ... + a_nv_n + a'_1w_1 +...+ a'_kw_k + b_1v_1 + ... + b_nv_n + b'_1u_1 +...+ b'_lu_l = (a_1+b_1)v_1 + ... + (a_n+b_n)v_n + a'_1w_1 +...+ a'_kw_k+ b'_1u_1 +...+ b'_lu_l$$ Note that: $$\text{span}\{K \cup L\}= c_1v_1 +...+c_nv_n +c'_1w_1 +...+c'_kw_k +c''_1u_1+...+c''_1u_l$$ Therefore for $c_h=(a_h+b_h)\ , \ c'_i=a_i \ , \ c''_j=b_j$: $$\text{span}\{K \cup L\} = w + u$$ Therefore $K\cup L$ is a basis for $U + W$

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What if $U\cap W\not=\emptyset$? –  metacompactness Jul 6 '13 at 14:43
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Or, even more interesting, what if $U \cap W \ne \{0\}$ –  GEdgar Jul 6 '13 at 15:08
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2 Answers 2

up vote 3 down vote accepted

For another counter example assume that $U,V\leq \mathbb R^4$ such that $$U=\langle (1,1,0,-1),(1,2,3,0),(2,3,3,-1)\rangle,~V=\langle (1,2,2,-2),(2,3,2,-3),(1,3,4,-3)\rangle $$ You can easily show that $\dim(U+V)=3\neq6$.

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$\quad _+^+ +_+^+$ –  amWhy Jul 7 '13 at 2:36
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You claim that the union of two linearly independent sets is linearly independent. This is false. For example, let $V=F=\Bbb{R}$. Then $\{1\}$ and $\{2\}$ are both linearly independent, but $\{1,2\}$ is not.

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