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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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    B is finite, @EwanDelanoy2011-12-11

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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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    Dylan, this was very use$f$ul, tha$n$$k$s a lot! They gave the solution to this exercise today and a student actually $f$ou$n$d a $f$law on the solution given, so I guess it isn't easy at all to prove this.2011-12-12