# show there exist non zero vector which is linear combination of other

sLet $a_1, \ldots , a_n$ be a basis of linear space $V$ let $W \le V$ be a $k$ dimensional subspace $k \ge 1$ Show for each subset $\displaystyle a_{i_i}, \ldots a_{i_m}$ for $m>n-k$ exist non zero vertor $\beta \in W$ which is linear combination of $\displaystyle a_{i_i}, \ldots, a_{i_m}$

My try: from cond. $m>n-k$ we have at least one vector from $W$ belongs to $\displaystyle a_{i_i}, \ldots, a_{i_m}$ so we can pick this vector and we have $\beta \in \operatorname{span}( a_{i_i}, \ldots, \beta , \ldots, a_{i_m})$. Is my reasoning OK ?

No. A $\beta \in W$ such that $\beta \in \{a_1,\ldots,a_m\}$ does not necessarily exist. Nonetheless the assertion is true because of Grassman's formula as follows.
Define $U= \operatorname{span}(a_1, \ldots, a_m)$. Since $a_1,\ldots,a_m$ are linearly independent, we have $\dim(U)=m$ and, since $W+U<V$, also $\dim (W+U) \leq \dim V = n$. From Grassman's formula we now have: $$\dim (W \cap U) = \dim W + \dim U - \dim (W+U) \geq k+m-n > 0$$ since $m>n-k$ by hypothesis, and this proves the assertion.
• one more question why $dim(W+U)\le dim V$ ?
• Because $W+U<V$ as pointed out in the answer. Nov 23, 2014 at 17:45
• The notation $W+U<V$ means that $W+U$ is a subspace of V. Nov 23, 2014 at 17:47