Exploring Vector Spaces U, V, and W: Solving U⊕V = U⊕W for U, V, and W

[srn]

U⊕V = U⊕W; find U, V and W

I need to give an example of different vectorspaces U, V, W so that $U \oplus V = U \oplus W$.

Can anyone give a hint please? It's basically asking for V and W such that $u_i + v_i = u_i + w_i$ yet V and W have to be different. How?

 

[micromass]

You are working with tensor products right?? You didn't mean to type $\oplus$ for direct sum??

 

[I like Serena]

I need to give an example of different vectorspaces U, V, W so that U⊗V = U⊗W.

Can anyone give a hint please? It's basically asking for V and W such that $u_i + v_i = u_i + w_i$ yet V and W have to be different. How?

Hi srn!

So... U, V and W have to be different, such as U=<(1,0)>, V = <(1,1)> and W=<(0,1)>?

 

[srn]

Thanks for the replies. And sorry, clearly posted this too late because I messed up the symbol in the question. :( Meant to say direct sum indeed...

Hi srn!

So... U, V and W have to be different, such as U=<(1,0)>, V = <(1,1)> and W=<(0,1)>?

Hey. :) Yes they do. But here the direct sums are not equal though, right? I guess you meant the tensor product? Sorry :(

 

[micromass]

Hey. :) Yes they do. But here the direct sums are not equal though, right? I guess you meant the tensor product? Sorry :(

Are you sure they are not equal (when taking the direct sum)??
What are the direct sums??

 

[srn]

If $(R,S, +)$ is a vectorspace with $U, W$ as subspaces, then $U \oplus W = \{u + w | u \in U, w \in W\}$ and every $s \in S$ can only be written in one possible way (as the sum of vectors of U and W). I.e. it's every possible combination of elements in $(R, U, +)$ and $(R, W, +)$.

Suppose U=<(1,0)>, V = <(1,1)> and W=<(0,1)> are subspaces, then

$U \oplus W = R^2$. But how is $U \oplus V = R^2$? I'm imagining $R^2$. V is every possible vector through $\stackrel{\rightarrow}{o}$ with $arg(v) = 1$. Then $U \oplus V$ would be the area under y = x for $x > 0, y > 0$. How can you form (0,1) for example?

edit: come to think of it, would $(R, V, +)$ also contain (0,1) and (1,2) etc? I sort of assumed from "$\forall v \in V$ and $\forall r \in R: rv \in V$" that $(R, V, +)$ would only contain (1,1), (2,2) etc, is that incorrect?

I'm sort of confused because my book says that if $U \cap V \neq (0,0)$ then $U \oplus V$ cannot exist. From the example, $U \cap V$ would be $\{((x,0) | x \in R\}$, but then (1,0) would be both in U and V?

 

[kru_]

If V = <(1,1)> then how can (1,0) be in V? There is no scalar a such that a*(1,1) = (1,0). Similarly, there is no scalar b such that b*(1,0) = (1,1). So the intersection of U and V is indeed (0,0).

 

[srn]

Uh, right. So the intersection is (0,0) but $U + V \neq R^2$. There's no scalars so that $a\cdot (1,0) + b\cdot (1,1) = (0,1)$, for example. So U and V cannot form $R^2$ and the direct sums are hence not equal? edit: ooops, a = -1 and b = 1 :) so they do actually form R^2

Sidenote:

If U=<(1,0) then V=<(0,1)> and W=<(0,-1)> would form the same direct sum space, and that also answers the question I think? edit: eh, no, (R, V, +) is equal to (R, W, +) now.

 

[HallsofIvy]

Let U be the subspace of R² spanned by <1, 0>. That is U is the set of all vectors of the form <x, 0> for any real number x. Let V be the vector space spanned by <0, 1> and let W be the subspace spanned by <1, 1>.

You can then show that U⊕V= U⊕W.