Show that if we exchange elements in 2 different basis will still give us a basis. I came across a proof of the following theorem:
Let V be a free module over a division ring D. Suppose X and Y are two finite basis of V, then |X| = |Y|.
that uses the fact that if we exchange elements in 2 different basis, we will still get a basis. Say $X = \{x_1, x_2, ..., x_n\}$ and $Y = \{y_1, ..., y_k\}$, then $\{x_1, x_2, ..., x_{n-1}, y_k\}$ is still a basis. 
However, I am having trouble proving that, I can't show that if $0 = a_1 x_1 + ... a_n y_k$, then all the coefficients are 0. Can someone teach me how to solve this? Thanks

Let $X=\{x_1, x_2\}$, $Y=\{y_1, y_2\}$ with $y_1=x_2$, $y_2=x_1$. Now $\{x_1, y_2\} = \{x_1\}$ is not a basis in general. Hence, your statement is wrong.
