Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 0 down vote accepted

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.

share|cite|improve this answer
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
@ohad: I adjusted my answer. The second part of your solution is fine. – Michalis Jul 29 '13 at 14:33


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 $i$?

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.