# Alternating vector between bases makes it still a basis?

Let $V$ be a space over field $F$. $B=\{ v_1, v_2,...,v_n\}$, $C\{ u_1, u_2,...,u_n\}$ are bases of $V$. Show there is $i \in \{1,2,...,n \}$ so that the set $\{ v_2,...,v_n, u_i\}$ is a basis of $V$.

I tried proof by contradiction and my intuition tells me that's the correct approach, I just got stuck. Showing it's linear independent is sufficient to prove it's a basis if I'm not mistaken (due to number of vectors).

Thanks in advance for any assistance!

P.S. I would like if someone had a better idea of how to word my title.

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The question as stated is incorrect, since any two bases have the same cardinality. So if $\{v_1, \dotsc, v_n\}$ is a basis, then $\{v_1, \dotsc, v_n, u_i\}$ cannot be a basis, since it has one more element. You may mean $\{v_1, \dotsc, v_{n-1}, u_i\}$ is a basis? – Eric Auld Jul 29 '13 at 14:05
As written, your claim is false: If $\dim_F V=n$, no set of $n+1$ vectors can be a basis. – Andrea Mori Jul 29 '13 at 14:06
Oops, fixed! Thank you. – ohad Jul 29 '13 at 14:07
Don't forget that a set of linearly independent vectors always can be extended to a basis. – Sigur Jul 29 '13 at 14:09

I edited my answer because I accidentally changed the notation in the question.

1.: Every $u_i$ can be written as a linear combination of the $v_j$. Prove that there is an $i$ such that the coefficient $c_1$ of $v_1$ in $$u_i=\sum_{j=1}^n c_j v_j$$ is not zero. If there wasn't one then the $u_i$ would be contained in the span of $v_2,\ldots, v_n$ so they couldn't be a basis.

2.: Show that the vectors $v_1,\ldots,v_{n-1},u_i$ are linearly independent.

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As for 1., there must be such non-zero coefficient, otherwise $\{u_1,...,u_n\}$ is linearly dependent because there is $u_i=0$ for some $i$. Is that correct? And now you just say $u_i=c_1v_1+...+c_nv_n$. And so $$d_2v_2+...+d_nv_n+u_i=0 \rightarrow$$ $$d_2v_2+...+d_nv_n+c_1v_1+...+c_nv_n=0 \rightarrow$$ $$c_1v_1+(c_2+d_2)v_2+...+(c_n+d_n)v_n=0$$ And since $\{v_1,..v_n\}$ is linearly independent all the coefficients are zero. So the original expression is linearly independent and therefore a basis. Is this correct? – ohad Jul 29 '13 at 14:19
@ohad: Sorry I made a slight change in notation. I wanted to show that $v_1,\ldots,v_{n-1},u_i$ is linearly independent! – Michalis Jul 29 '13 at 14:23
Michalis isn't it equivalent in method? But is my way of solution generally correct? – ohad Jul 29 '13 at 14:25
@ohad: For the first step your argument is not quite correct. I was going for the following: If $c_n$ is zero for all $u_i$ then all $u_i$ would be in the span of $v_1,\ldots,v_{n-1}$ so they can't form a basis. You started well in the second step except that I wanted you to show that $v_1,\ldots,v_{n-1},u_i$ are linearly independent. Please adjust your solution, if you want I'll explain my concerns with your solution in the chat. – Michalis Jul 29 '13 at 14:25
If $\{v_1,...,v_n\}$ is a basis then $\{v_2,...,v_n\}$ are linearly independent. Now, what does it mean that $\{v_2,...,v_n,u_i\}$ is not a basis? What if this happens for all $i$?
We claim that there's a vector $u_i$ such that $$u_i=\sum_{k=1}^n\alpha_k v_k,\quad \alpha_1\neq 0$$ otherewise the family $(v_2,\ldots,v_n)$ span all the vectors $(u_1,\ldots,u_n)$ which's a contradiction. I think that the rest of reasoning is clear.