Is the Direct Sum of Vector Subspaces U, W, and V Valid?

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Homework Help Overview

The discussion revolves around the validity of the direct sum of vector subspaces U, W, and V in a vector space context. Participants are examining the conditions under which the sum of these subspaces equals the larger space V, particularly focusing on the dimensions and intersection properties of the subspaces.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore whether every vector in V can be expressed in the form derived from the sum of vectors from U and W. There are attempts to establish the relationship between the dimensions of the subspaces and the overall space.

Discussion Status

Some participants express uncertainty about the validity of the direct sum, questioning whether the conditions for the direct sum are met. There is a recognition of the intersection being trivial, but doubts remain regarding the dimensionality of the combined subspaces compared to V.

Contextual Notes

Participants note the dimensions of U and W are not explicitly stated, which is crucial for determining the validity of the direct sum. The discussion also references the need to verify specific properties related to the dimensions and intersections of the subspaces.

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Homework Statement



[PLAIN]http://img571.imageshack.us/img571/1821/subspaces.png

Homework Equations





The Attempt at a Solution



Is my solution correct?:

For [itex]a,b\in \mathbb{C}[/itex]

let [itex]A=\begin{bmatrix} a \\ a \\ 0 \end{bmatrix}\in U[/itex] and [itex]B=\begin{bmatrix} 0 \\ b \\ b \end{bmatrix}\in W[/itex]

Then [itex]A+B=\begin{bmatrix} a \\ a \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ b \\ b \end{bmatrix} = \begin{bmatrix} a \\ a+b \\ b \end{bmatrix}\in \mathbb{C}^3[/itex]

How do I get from this that [itex]U+W=V[/itex] ?

Clearly the only vector in the intersection of U and W is the zero vector when [itex]a=b=0[/itex] so [itex]U\cap W = \{\bf 0} \}[/itex]

[itex]v = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \in A \cap B \Rightarrow \begin{cases} a=b, c=0 \quad v \in A \\ a = 0, b=c \quad v \in B \end{cases} \Rightarrow a=b=c=0[/itex]

[itex]\therefore V = U\oplus W[/itex]
 
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It is indeed correct that the vectors in A+B have the form (a,a+b,b). Is it now true that every vector in V can be written in this form. I mean, if (x,y,z) is arbitrary, does there always exists a and b such that (a,a+b,b)=(x,y,z)?

If this is true, then A+B=V. If this is not true, then A+B is not V.
 
micromass said:
It is indeed correct that the vectors in A+B have the form (a,a+b,b). Is it now true that every vector in V can be written in this form. I mean, if (x,y,z) is arbitrary, does there always exists a and b such that (a,a+b,b)=(x,y,z)?

If this is true, then A+B=V. If this is not true, then A+B is not V.

Well it is true but how do I explicitly show it?
 
I don't think it is true...
 
micromass said:
I don't think it is true...

I can use the following can't I?

[itex]V = U\oplus W \iff U\cap W = \{0\}\;\text{and}\;\text{dim}(U) + \text{dim}(W) = \text{dim}(V)[/itex]

Now [itex]\text{dim}(V) = 3[/itex] but what's the dimensions of [itex]U[/itex] and [itex]W[/itex] ?
 
Take an arbitrary (x,y,z), does there always exist a and b such that (x,y,z)=(a,a+b,b)? If yes, what do a and b have to be??
 
micromass said:
Take an arbitrary (x,y,z), does there always exist a and b such that (x,y,z)=(a,a+b,b)? If yes, what do a and b have to be??

I've proved [itex]V\neq U\oplus W[/itex] by using the fact that

[itex]V = U\oplus W \iff U\cap W = \{{\bf 0}\}\;\text{and}\;\text{dim}(U+W) = \text{dim}(V)[/itex]

Although [itex]U\cap W = \{0\}[/itex] we have [itex]\text{dim}(V)=3[/itex]

and an obvious basis for [itex]U+V[/itex] is [itex]\begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix}[/itex]

so [itex]\text{dim}(U+V)=2 \neq \text{dim}(V)[/itex] .

How do I check the two things in my other thread?
- if [A]=, then the assigned number is also the same
- if [A] and are different classes, then the assigned number is different.
 

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