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sorry it's me again!

Let $V$ be a $\mathbb K$ vector space with finite basis $B$ and $M \subset V$ is a finite linear independent subset.

Show that a subset $A \subset B$ exists, so that $(B \backslash A) \cup M$ is a basis from $V$.

Can I cheat and use the subset $\{\}$ from $B$? This would get me $B \cup M$, which should be a basis from V. Or did I get everything wrong? It's weird, because apparently my colleagues used linear span to show this.

Thanks a lot again!

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The definition of basis requires elements to be linearly independent. If you used $B \cup M$, and $M$ is nonempty, then that can't possibly be a basis (every element in $M$ is in the span of $B$). – Brandon Carter Dec 11 '11 at 19:14
$B$ is finite, right? – Ewan Delanoy Dec 11 '11 at 19:38
B is finite, @EwanDelanoy – Clash Dec 11 '11 at 19:48
up vote 3 down vote accepted

Brandon makes a good point about your "cheat" in the comments. We'll have to work harder than that.

Write $B = \{b_1, \ldots, b_n\}$ and $M = \{m_1, \ldots, m_k\}$. For each $0 \leq r \leq k$, consider the following statement.

After possibly relabeling the $b_i$, the set \[ \{m_1, \ldots, m_r, b_{r + 1}, \ldots, b_n\} \] is a basis for $V$.

Prove this by induction on $r$. When $r = 0$, this simply says that $B$ is a basis. By way of induction, assume that we've proven the statement for an $r < k$. Then \[ m_{r + 1} = a_1m_1 + \cdots + a_rm_r + a_{r + 1}b_{r + 1} + \cdots + a_nb_n. \] Note that in this expression some $b_i$, which might as well be $b_{r + 1}$, has a non-zero coefficient; otherwise, we would have a relation of linear dependence among elements of $M$. Now you can write $b_{r + 1}$ as an element of the span of $\{m_1, \ldots, m_{r + 1}, b_{r + 2}, \ldots, b_n\}$, and then show that this new set is a basis.

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hi dylan, thanks for the answer! do you mean, then $m_1$ can be written as a linear combination of elements of $B_1$? or did I get something wrong? Another question, can I just show that when A=M, then the statement I want to show is true? Or I have to show it for every M? – Clash Dec 11 '11 at 20:44
@Clash I think what I have here is right. You want to replace $b_1$ by $m_1$ and keep on doing that. This sort of proof appears to have a name. I don't see why you want to assume that $A = M$, could you explain? I can write more details later, if you like. – Dylan Moreland Dec 11 '11 at 20:47
ok, so I think I understood better what you mean. I think it's no mistake what you wrote. Let me see if I can show that $B_1$ is a basis for V – Clash Dec 11 '11 at 20:48
Dylan, about A=M, then I get (B\A) U M, but as A = M, then I get B again. – Clash Dec 11 '11 at 20:49
@Clash I don't see any reason why you can assume that. I'll try to be more explicit about the structure of the proof soon. – Dylan Moreland Dec 11 '11 at 20:50

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