# Basis for the intersection of subspaces

Prove or disprove the following; Let $V$ be a vector space and $U$ and $W$ two subspaces of $V$. If the set of vectors $\{b_1,\ldots b_n\}$ is a basis for $U\cap W$, then it is also a basis for both $U$ and $W$ themselves.

I am not even sure if this statement is true or not. I have tried to find a counter-example to disprove, which is what I feel needs to be done - but not sure.

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It is false. The simplest counterexample is to let $V$ be any vector space of dimension $\geq1$, and let $U=\{0\}$ (the trivial subspace) and $W=V$. We have $U=U\cap W=\{0\}$ is of dimension $0$ and $W=V$ is of dimension $\geq 1$, so a basis for one cannot be a basis for the other.
More generally, if $A$ is a vector space, $B\subseteq A$ is a subspace of $A$, and $\{x_1,\ldots,x_n\}$ is a basis for both $A$ and $B$, then in fact $A=B$. This is because $$\{x_1,\ldots,b_n\}\text{ is a basis for }A\implies A=\text{span}(x_1,\ldots,x_n)$$ $$\{x_1,\ldots,x_n\}\text{ is a basis for }B\implies B=\text{span}(x_1,\ldots,x_n)$$ hence $A=B$. Using this observation, we see that a basis for $U\cap W$ is a basis for $U$ if and only if $U=U\cap W$, and it is a basis for $W$ if and only if $W=U\cap W$, so it is a basis for both $U$ and $W$ if and only if $U=W$.
Recall that a basis is a maximal linearly independent set. If we remove the hypothesis of maximality then it is not true that every vector can be written as a linear combinations of the members of such a set. For if $\beta = \left\lbrace v_1, v_2, ..., v_k \right\rbrace$ is independent and $v_{k+1}$ is another vector such that $\beta \cup \left\lbrace v_{k+1} \right\rbrace$ is independent, then if for some scalars $\lambda_{i}$ we have $$v_{k+1} = \sum_{i=1}^k \lambda_i v_i$$ then would have $$0 = \sum_{i=1}^k \lambda_i v_i - v_{k+1}$$ which would be a contradiction since we assumed $\beta \cup \left\lbrace v_{k+1} \right\rbrace$ to be independent (and all the coefficients of such linear combination should be 0).
On the other hand, if a set is linearly independent in $U\cap V$ it most necessarily be independent in $V$ and $W$. The reason because the statement is FALSE is that even though it is independent it is not always maximal. And the only way it can be maximal, and thus a base, is in the way described by Zev Chonoles, that is, when both are equal.