Proof with intersection of subspaces

In summary, the equation L \cap (M+N) = (L \cap M)+(L \cap N) is not necessarily true, but it does hold when L \subset (M+N). For (b), we need to 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)) by considering the definitions of subspaces and vector addition.
  • #1
Dafe
145
0

Homework Statement


Suppose L, M, and N are subspaces of a vector space.

(a)
Show that the equation
[tex] L \cap (M+N) = (L \cap M)+(L \cap N) [/tex]
is not necessarily true.

(b)
Prove that
[tex] L \cap (M+(L \cap N))=(L \cap M) + (L \cap N) [/tex]

Homework Equations


N/A

The Attempt at a Solution



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

Then,

[tex]
\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}
[/tex]

and the equation is not true.

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

(b)

[tex] L \cap M + L \cap (L \cap N) = (L \cap M) + (L \cap N) [/tex]

That's all I can come up with on my own.
Any suggestions are appreciated, thanks!
 
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  • #2
For (b), show that [itex]L \cap (M+(L \cap N))\subset (L \cap M) + (L \cap N)[/itex] and [itex](L \cap M) + (L \cap N) \subset L \cap (M+(L \cap N))[/itex].

For example, assume [itex]x \in L \cap (M+(L \cap N))[/itex]. Then [itex]x \in M+(L \cap N)[/itex], so you can write [itex]x = m + n[/itex] where [itex]m \in M[/itex] and [itex]n \in L \cap N[/itex]. If you can show that [itex]m \in L[/itex], then you'll have that [itex]m \in (L \cap M)[/itex] and consequently that [itex]x \in (L \cap M)+(L \cap N)[/itex], so [itex]L \cap (M+(L \cap N))\subset (L \cap M) + (L \cap N)[/itex].
 
  • #3
vela said:
If you can show that [itex]m \in L[/itex], then you'll have that [itex]m \in (L \cap M)[/itex] and consequently that [itex]x \in (L \cap M)+(L \cap N)[/itex], so [itex]L \cap (M+(L \cap N))\subset (L \cap M) + (L \cap N)[/itex].

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

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

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

Thank you!
 
  • #4
Dafe said:
From our assumption that [tex]x\in L \cap (M+(L\cap N)) [/tex], we have that [tex]x\in L[/tex].
Since [tex]x=m+n[/tex] we have that [tex]m\in L[/tex] and [tex]n \in L [/tex], so
[tex] L \cap (M+(L\cap N)) \subset (L\cap M)+(L\cap N) [/tex].
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 [tex] (L\cap M)+(L\cap N)\subset L\cap (M+(L\cap N)) [/tex]?
You have to show both directions to prove equality.

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

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

Hint: Remember that L, M, and N are subspaces.
 
  • #5
vela said:
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 [tex]m+n \in L,\;n\in L[/tex] and [tex]L[/tex] is a subspace (closed under vector addition), we know that [tex]m \in L[/tex]?

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

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

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

Is this any good?

Thanks!
 
  • #6
Dafe said:
Since [tex]m+n \in L,\;n\in L[/tex] and [tex]L[/tex] is a subspace (closed under vector addition), we know that [tex]m \in L[/tex]?
Yes.
From [tex](L \cap N)[/tex] I know that [tex]n \in L[/tex], and from [tex](L \cap M)[/tex] I know that [tex]m \in L[/tex].
[tex]m+n[/tex] must also be in [tex]L[/tex] since it is a subspace.
Now, [tex]m+n \in (M+(L \cap N))[/tex] and [tex]m+n \in L[/tex] and so [tex]x[/tex] is an element of [tex]L \cap (M+(L\cap N))[/tex].

Is this any good?
Perfect.
 
  • #7
Thank you very much :)
 
  • #8
One other thing that I didn't notice coming up in this thread.
Dafe said:
[tex]\begin{aligned}M=&\;span\{(0,0),(1,0)\}\\N=&\;span\{(0,0),(0,1)\}\\L=&\;span\{(0,0),(1,1)\}\end{aligned}[/tex]
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
That makes sense and looks prettier. Thanks.
 

FAQ: Proof with intersection of subspaces

1. What is the definition of an intersection of subspaces?

An intersection of subspaces is the set of all elements that are common to two or more subspaces. It is denoted by the symbol ∩ and is read as "intersection".

2. How is the intersection of subspaces calculated?

The intersection of subspaces is calculated by finding all the elements that are common to all the subspaces. This can be done by solving the system of equations that represent each subspace and finding the values that satisfy all of them.

3. Can the intersection of subspaces be empty?

Yes, the intersection of subspaces can be empty if there are no common elements between the subspaces. This can happen when the subspaces are disjoint or when one subspace is contained within the other.

4. What is the significance of the intersection of subspaces in linear algebra?

The intersection of subspaces is a fundamental concept in linear algebra as it helps in understanding the relationship between different subspaces. It also plays a crucial role in solving systems of linear equations and finding solutions that satisfy all the equations simultaneously.

5. Can the intersection of subspaces be used to prove linear independence?

Yes, the intersection of subspaces can be used to prove linear independence. If the intersection of two subspaces is only the zero vector, then it implies that the two subspaces are linearly independent. This can be extended to prove linear independence of a set of vectors by considering all the possible intersections of subspaces formed by those vectors.

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