# Thread: union of vector space

1. ## union of vector space

I could use some help with this problem,
Suppose that V is a vector space over any field F and that U and W are subspaces of V.
Show that the only time the union of U and W is a subspace of V is when either U is a subset of W or W is a subset of U.

2. Suppose that neither is a subset of the other.
This means $a \in U\backslash V\quad \& \quad b \in V\backslash U.$
If $U \cup V$ is a subspace then $a + b = c \in \left( {U \cup V} \right).$
But if $c \in U$ then $c - a \in U$ or $b \in U.$
That is a contradiction. Similarly if $c \in V$ we get a contradiction.
Thus one of U or V is a subset of the other.

Clearly, if one of U or V is a subset of the other then their union is a subspace.

3. Hello math helpers...
I cant understand exactly why that is a contradiction...maybe if you explained me with less formal language I could get it.
Thank you, and sorry for annoying you all..

4. Suppose that neither is a subset of the other.
This means

What is meat here is that there is an element 'a' which is in U but not in V
Similarly there is an element 'b' which is in V but not in U
This statement follows directly from the fact that U is not a subset of V and V is not a subset of U

If is a subspace then

both 'a' and 'b' (defined above) will belong to which we assume to be a sub-space. As subspace is closed under addition so a+b =c will also belong to the union of U and V.

But if then or
Thus one of U or V is a subset of the other.

As c belong to the union it has to belong to (at least) one of U,V. Let the chosen one be U. Now c and a both belong to U. Thus c-a=b will again below to U (under the closure and inverse axiom). But we always maintained b doesn't belong to U. Hence contradiction !!

Same can be argued for V.

Hence the result. Thanks

5. Suppose that neither is a subset of the other.
This means

What is meant here is that there is an element 'a' which is in U but not in V
Similarly there is an element 'b' which is in V but not in U
This statement follows directly from the fact that U is not a subset of V and V is not a subset of U

If is a subspace then

both 'a' and 'b' (defined above) will belong to which we assume to be a sub-space. As subspace is closed under addition so a+b =c will also belong to the union of U and V.

But if then or
Thus one of U or V is a subset of the other.

As c belong to the union it has to belong to (at least) one of U,V. Let the chosen one be U. Now c and a both belong to U. Thus c-a=b will again below to U (under the closure and inverse axiom). But we always maintained b doesn't belong to U. Hence contradiction !!

Same can be argued for V.

Hence the result. Thanks

6. Suppose that neither is a subset of the other.
This means

What is meant here is that there is an element 'a' which is in U but not in V
Similarly there is an element 'b' which is in V but not in U
This statement follows directly from the fact that U is not a subset of V and V is not a subset of U

If is a subspace then

both 'a' and 'b' (defined above) will belong to which we assume to be a sub-space. As subspace is closed under addition so a+b =c will also belong to the union of U and V.

But if then or
Thus one of U or V is a subset of the other.

As c belong to the union it has to belong to (at least) one of U,V. Let the chosen one be U. Now c and a both belong to U. Thus c-a=b will again below to U (under the closure and inverse axiom). But we always maintained b doesn't belong to U. Hence contradiction !!

Same can be argued for V.

Hence the result. Thanks

7. sorry about the multiple posts above. browser got stuck and made multiple submits, i guess.
@admin: if there is a way i can delete them plz do let me know. i cldn't find an option under edit. thanks

Originally Posted by aman_cc
Suppose that neither is a subset of the other.
This means

What is meant here is that there is an element 'a' which is in U but not in V
Similarly there is an element 'b' which is in V but not in U
This statement follows directly from the fact that U is not a subset of V and V is not a subset of U

If is a subspace then

both 'a' and 'b' (defined above) will belong to which we assume to be a sub-space. As subspace is closed under addition so a+b =c will also belong to the union of U and V.

But if then or
Thus one of U or V is a subset of the other.

As c belong to the union it has to belong to (at least) one of U,V. Let the chosen one be U. Now c and a both belong to U. Thus c-a=b will again below to U (under the closure and inverse axiom). But we always maintained b doesn't belong to U. Hence contradiction !!

Same can be argued for V.

Hence the result. Thanks

8. I realize it has been a while, but I was thinking about the same problem trying to prove that it either has to be U is a subset of V or V is a subset of U. However, I am confused. Are a,b,c, vectors?
Also, could you explain a little more about how you got to a+b=c and how that actually gives us the desired result.

Thanks!

9. $a, b$ and $c$ are vectors. If you assume that neither is a subspace of the other one, then there exist $a$ and $b$ as described above.

Since all live in a vector space, you can add $a$ and $b$, call the result c, so define $c=a+b$.
Now we assume that the union of $U$ and $V$ is a subspace. Then since $a,b \in (U \cup V)$, their sum (that is c) is also inside $U \cup V$.

Clear?