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Let $V$ be a $\mathbb K$ vector space, $M_1$ and $M_2$ subsets of $V$. ($[M] =$ linear span of $M$) Show that:

a) $[M_1 \cup M_2] = [[M_1 \cup M_2]]$ and specially $[[M]] = [M]$.

What I've tried: I guess once I've proven that $[[M]] = [M]$, it follows that $[M_1 \cup M_2] = [[M_1 \cup M_2]]$, as $[M_1 \cup M_2]$ is nothing more than a normal subset. Can I just say that by definition the linear span is already all linear combinations of vectors from $M$? Then getting again all linear combinations from all linear combinations isn't getting me anything new. Or is there a math way to write this?

b) $[M_1 \cap M_2] \subset [M_1] \cap [M_2]$

What I've tried: Playing with properties like the following:

$M \subset [M]$

$M_1 \subset M_2 \Rightarrow [M_1] \subset [M_2]$

But to no avail. So I have $M_1 \cap M_2 \subset [M_1 \cap M_2]$, $M_1 \subset [M_1]$ and $M_2 \subset [M_2]$, but couldn't prove anything with this.

c) no equality is valid in b).

What I've tried: I suppose they want me to show that $[M_1 \cap M_2] =[M_1] \cap [M_2]$ is false? I fail to understand how this is not true.

Many many thanks in advance!

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up vote 1 down vote accepted

Some hints:

For c) (I assume you mean no equality in b)), try to come up with two sets that both span $V$ (or more simply, span the same set), but have no elements in common.

In part a), for $[[M]]=[M]$:

What you said here is essentially correct. But you should write the details out. Show that if $a_1,a_2,\ldots,a_k$ are linear combinations of elements of $M$ (that is, in $[M]$), then a linear combination of these elements is also in $[M]$. This shouldn't be to hard to do (though it might be notationally messy). This gives the inclusion $[[M]]\subset [M]$. The other inclusion is easy...

For b), take a linear combination of elements in $M_1\cap M_2$. It should be clear that this is also in $[M_1]$ and in $[M_2]$ (since elements in $M_1\cap M_2$ are in both of the sets).

Note in all cases, the symbols $M_1$, $M_2$, $M_1\cap M_2$, etc. are just sets; while the symbols $[M_1]$, $[M_2]$ are the linear spans of sets.

For part b):

Suppose ${\bf v}\in [M_1\cap M_2]$. Then $\bf v$ has the form $${\bf v}= c_1 {\bf m_1}+c_2 {\bf m_2}+\cdots +c_k{\bf m_k}$$ for some ${\bf m}_1\in M_1\cap M_2$, ${\bf m}_2\in M_1\cap M_2$,$\ldots $, ${\bf m}_k\in M_1\cap M_2$ and for some scalars $c_1,c_2,\ldots,c_k$ .

Now each ${\bf m}_i\in M_1$, since it is in $M_1\cap M_2$. It follows then that $\bf v$ is a linear combination of vectors from $M_1$. Thus, ${\bf v}\in[M_1]$.

Similarly, ${\bf v}\in [M_2]$; and thus, ${\bf v}\in [M_1]\cap [M_2]$.

This shows $[M_1\cap M_2]\subset[M_1]\cap[M_2]$.

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sorry, corrected! yes I ment no equality in b) – Clash Dec 11 '11 at 16:30
hey david, could you help me by b)? It's not clear to me that this is also in [M1] and in [M2]. thanks! – Clash Dec 11 '11 at 17:23
@clash It's added. – David Mitra Dec 11 '11 at 17:33
thanks a lot david! – Clash Dec 11 '11 at 17:44

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