# Vector Subspaces - Direct Sum

## Homework Statement

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

## The Attempt at a Solution

Is my solution correct?:

For $a,b\in \mathbb{C}$

let $A=\begin{bmatrix} a \\ a \\ 0 \end{bmatrix}\in U$ and $B=\begin{bmatrix} 0 \\ b \\ b \end{bmatrix}\in W$

Then $A+B=\begin{bmatrix} a \\ a \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ b \\ b \end{bmatrix} = \begin{bmatrix} a \\ a+b \\ b \end{bmatrix}\in \mathbb{C}^3$

How do I get from this that $U+W=V$ ?

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

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

$\therefore V = U\oplus W$

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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.

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...

I don't think it is true...

I can use the following can't I?

$V = U\oplus W \iff U\cap W = \{0\}\;\text{and}\;\text{dim}(U) + \text{dim}(W) = \text{dim}(V)$

Now $\text{dim}(V) = 3$ but what's the dimensions of $U$ and $W$ ?

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??

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 $V\neq U\oplus W$ by using the fact that

$V = U\oplus W \iff U\cap W = \{{\bf 0}\}\;\text{and}\;\text{dim}(U+W) = \text{dim}(V)$

Although $U\cap W = \{0\}$ we have $\text{dim}(V)=3$

and an obvious basis for $U+V$ is $\begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix}$

so $\text{dim}(U+V)=2 \neq \text{dim}(V)$ .

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.