Vector Space external direct sum

Question

Question: Give an example to show that it is possible for $A \oplus B = A\oplus B'$ without having $B=B'$, where $A,B,B'$ are subspaces of $_FV$

I really can't imagine this, say let $A \oplus B = \left\{(a,b)|a \in A, b \in B \right\}$ and $A \oplus B' = \left\{(a,b')|a \in A, b' \in B' \right\}$\

Then $(a,b)=(a,b') \Rightarrow (a,b)-(a,b') = 0 \Rightarrow (a+(-a),(b+(-b')))=(0,0)$\

Which I can only conclude that $b=b'$. What example should I give?

Answer 1

Just look in $\Bbb R^2$. Consider the spaces

$$\begin{cases}A:\{(x,y)\in\Bbb R^2 : y=0\} \\ B:\{(x,y)\in\Bbb R^2 : x=0\} \\ B':\{(x,y)\in\Bbb R^2 : y=x\}\end{cases}.$$

That satisfies your conditions.

The sum is all of $\Bbb R^2$. If you want to get $(u,v)\in\Bbb R^2$ you just take $(a,b)\in A\oplus B$ where $a=(u,0), b=(0,v)$ and in $A\oplus B'$ you get $(c,d)\in A\oplus B'$ where $c= (u-v, 0), d=(v, v)$.

Then clearly $a+b=(u,v)$ and similarly $c+d= (u,v)$.