# Proof with intersection of subspaces

1. Feb 6, 2010

### Dafe

1. The problem statement, all variables and given/known data
Suppose L, M, and N are subspaces of a vector space.

(a)
Show that the equation
$$L \cap (M+N) = (L \cap M)+(L \cap N)$$
is not necessarily true.

(b)
Prove that
$$L \cap (M+(L \cap N))=(L \cap M) + (L \cap N)$$

2. Relevant equations
N/A

3. The attempt at a solution

(a)
I let
\begin{aligned} M=&\;span\{(0,0),(1,0)\}\\ N=&\;span\{(0,0),(0,1)\}\\ L=&\;span\{(0,0),(1,1)\} \end{aligned}

Then,

\begin{aligned} M+N=&\;span\{(0,0),(1,0),(0,1),(1,1)\}\\ L \cap (M+N)=&\;span\{(0,0),(1,1)\}\\ L \cap M=&\;span\{(0,0)\}\\ L \cap N=&\;span\{(0,0)\} \end{aligned}

and the equation is not true.

This in fact leads me to believe that the equation does not hold when $$L \subset (M+N)$$, because then $$L \cap (M+N) = L$$ and $$L \cap M$$ and $$L \cap N$$ are something else.
I would guess they turn out to be something like $$L-L \cap N$$ and $$L-L \cap M$$, respectively..

(b)

$$L \cap M + L \cap (L \cap N) = (L \cap M) + (L \cap N)$$

That's all I can come up with on my own.
Any suggestions are appreciated, thanks!

2. Feb 6, 2010

### vela

Staff Emeritus
For (b), show that $L \cap (M+(L \cap N))\subset (L \cap M) + (L \cap N)$ and $(L \cap M) + (L \cap N) \subset L \cap (M+(L \cap N))$.

For example, assume $x \in L \cap (M+(L \cap N))$. Then $x \in M+(L \cap N)$, so you can write $x = m + n$ where $m \in M$ and $n \in L \cap N$. If you can show that $m \in L$, then you'll have that $m \in (L \cap M)$ and consequently that $x \in (L \cap M)+(L \cap N)$, so $L \cap (M+(L \cap N))\subset (L \cap M) + (L \cap N)$.

3. Feb 6, 2010

### Dafe

From our assumption that $$x\in L \cap (M+(L\cap N))$$, we have that $$x\in L$$.
Since $$x=m+n$$ we have that $$m\in L$$ and $$n \in L$$, so
$$L \cap (M+(L\cap N)) \subset (L\cap M)+(L\cap N)$$.

Is that it or do I have to show that $$(L\cap M)+(L\cap N)\subset L\cap (M+(L\cap N))$$?

Suppose $$x \in (L\cap M)+(L\cap N)$$ and $$x=m+n$$ where $$m\in (L\cap M)$$ and $$n\in (L\cap N)$$.
Then $$m\in L$$ and $$n\in L$$.
I do not know how to proceed..

Thank you!

4. Feb 6, 2010

### vela

Staff Emeritus
Wait, how do you know that m is in L from knowing x is in L? M is not necessarily contained in L, so isn't it possible that m is in M but not in L. You need to explain why m must also be in L.

You have to show both directions to prove equality.

You know that $m \in M$ and $n \in L\cap N$, so it follows that $x=m+n \in M+(L \cap N)$. If you can show that x is in L, you can conclude it's an element of $L\cap (M+(L\cap N))$.

Hint: Remember that L, M, and N are subspaces.

5. Feb 7, 2010

### Dafe

Since $$m+n \in L,\;n\in L$$ and $$L$$ is a subspace (closed under vector addition), we know that $$m \in L$$?

From $$(L \cap N)$$ I know that $$n \in L$$, and from $$(L \cap M)$$ I know that $$m \in L$$.
$$m+n$$ must also be in $$L$$ since it is a subspace.
Now, $$m+n \in (M+(L \cap N))$$ and $$m+n \in L$$ and so $$x$$ is an element of $$L \cap (M+(L\cap N))$$.

Is this any good?

Thanks!

6. Feb 7, 2010

### vela

Staff Emeritus
Yes.
Perfect.

7. Feb 7, 2010

### Dafe

Thank you very much :)

8. Feb 8, 2010

### Staff: Mentor

One other thing that I didn't notice coming up in this thread.
There is no need to include (0,0) as a vector in your span{} sets. It's not incorrect to include it, but it needlessly clutters things up. You could have more simply said that
M = span{(1, 0)}
N = span{(0, 1)}
L = span{(1, 1)}
All of these are one-dimensional subspaces of R2, and for each the entire subspace is generated by scalar multiples of the given vector. Choose the scalar to be 0 for each, and you get the zero vector.

9. Feb 8, 2010

### Dafe

That makes sense and looks prettier. Thanks.