Proof with intersection of subspaces

[Dafe]

Homework Statement

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)

Homework Equations

N/A

The Attempt at a Solution

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

Then,

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

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!

 

[vela]

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).

 

[Dafe]

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).

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!

 

[vela]

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).

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.

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

You have to show both directions to prove equality.

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!

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.

 

[Dafe]

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.

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

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.

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!

 

[vela]

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

Yes.

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?

Perfect.

 

[Dafe]

Thank you very much :)

 

[Mark44]

One other thing that I didn't notice coming up in this thread.

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

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.

 

[Dafe]

That makes sense and looks prettier. Thanks.