Vector Subspaces - Direct Sum

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Homework Statement



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

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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Answers and Replies

  • #2
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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.
 
  • #3
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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.

Well it is true but how do I explicitly show it?
 
  • #4
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I don't think it is true...
 
  • #5
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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] ?
 
  • #6
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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??
 
  • #7
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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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