Prove or give a counterexample: if $U_1, U_2, W$ are subspaces of $V$ such that $V= U_1\oplus W$ and $V= U_2\oplus W$ then $U_1=U_2$.

direct-sumlinear algebrasolution-verification

I've read some proofs about this problem in the Mathstack, but I could not understand them all. I probably need some more time to digest Direct Sum completely.

So, I want to share my approach to show my level of understanding.

My Approach:

Suppose $U_1=\{(x,0) \in F^2: x \in F\}$, $U_2=\{(0,y) \in F^2: y \in F\}$, $W=\{(x,y) \in F^2: x,y \in F\}$, $V=F^2$.

Note that, $U_1 + W$, $U_2 + W$ are direct sum. And, $U_1 \neq U_2$.

If the statement is true, this contrapositive statement should give us $V \neq U_1 \oplus W$ or $ V \neq U_2 \oplus W$.

However, $U_1 \oplus W = F^2 = V = U_2 \oplus W$.
This is a contradiction.

Thus, the contrapositive statement is false. Hence, the statement is false.

End of Proof.

But, I am not confident with this proof since I think I am missing some important things about Direct Sum.

I hope my approach could show which part I need to focus on more, or which part I should not forget about Direct Sum.

Best Answer

Consider the following particular case: \begin{align*} \begin{cases} V = \mathbb{R}^{2}\\ W = \{(x,y)\in\mathbb{R}^{2}\mid y = 0\}\\ U_{1} = \{(x,y)\in\mathbb{R}^{2}\mid x = 0\}\\ U_{2} = \{(x,y)\in\mathbb{R}^{2}\mid y = x\} \end{cases} \end{align*} Then we have that $V = U_{1}\oplus W = U_{2}\oplus W$, but $U_{1}\neq U_{2}$.

Hopefully this helps!

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