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Given $V$ a K-vector space, and $E_1, E_2$ subspaces of V. If $B_1=\{v_1,\dots,v_m\}$ and $B_2=\{w_1,\dots,w_s\}$ are two basis of $E_1$ and $E_2$ and the vectors of the basis are linearly independent, that is, the set ${v_1,...,v_m,w_1,...,w_s}$ is linearly independent, then $E_1 \cap E_2=\{0\}$. Suppose that $E_1 \cap E_2 \neq \{0\}$. Then there is a non null vector $u \in E_1 \cap E_2$. Then $u \in E_1$ and $u \in E_2$. Then there's an expression of u in terms of the basis $B_1, B_2$. Then $u=a_1v_1+\dots+a_mv_m$ and $u=b_1w_1+\dots+b_sw_s$, with $a_1,\dots, a_m,b_1, \dots,b_s \in K$. Then $a_1v_1+\dots+a_mv_m=b_1w_1+\dots+b_sw_s$. Then $a_1v_1+\dots+a_mv_m-(b_1w_1+\dots+b_sw_s)=0$. The vectors are linearly independent, so $a_1=\dots=a_m=b_1=\dots=b_s=0$. Hence $v=0$, rising contradiction.

Is that proof correct?

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  • $\begingroup$ Your solution looks right to me. Note that you do not really need to argue by contradiction: let $u\in E_1\cap E_2$ then your argument shows that $u=0$, which proves that $E_1\cap E_2=\{0\}$. $\endgroup$
    – DKal
    Commented Jun 20, 2014 at 8:04

1 Answer 1

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You can conclude the result earlier than you are. The equation $$a_1\vec{v_1} + \cdots + a_m\vec{v_m} - \left(b_1\vec{w_1} + \cdots + b_s\vec{w_s}\right) = \vec{0}$$ does justify $$a_1 = a_2 = \cdots = a_m = b_1 = b_2 = \cdots = b_s = 0$$ with the assumption of linear independence. Instead, however, observe that $$\vec{0} = \begin{cases} a_1\vec{v_1} + \cdots + a_m\vec{v_m} \\ b_1\vec{w_1} + \cdots + b_s\vec{w_s} \end{cases}$$ has only the trivial solutions by the linear independence assumption. Also, you know $$\vec{u} = \begin{cases} a_1\vec{v_1} + \cdots + a_m\vec{v_m} \\ b_1\vec{w_1} + \cdots + b_s\vec{w_s} \end{cases} \,\,.$$ It follows that $\vec{u} = \vec{0}$, so that $E_1 \cap E_2 = \{0\}$, as desired.

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