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Prove or give a counterexample:

If $U_1, U_2, W$ are subspaces of $V$ such that $U_1 \bigoplus W = U_2 \bigoplus W$, then $U_1 = U_2$.

Proof

Let $u_1+w \in U_1 \bigoplus$ W. Since $U_1\bigoplus W= U_2\bigoplus W, u_1+w \in U_2\bigoplus W$. Since $w \in U_2 \bigoplus W, -w \in U_2 \bigoplus W$. Therefore $ u_1 \in U_2 \bigoplus W $.Since $U_1$ and $W$ have no elements in common,$u_1 \not\in W$ . Therefore $u_1 \in U_2$. Therefore$ U_1 \subseteq U_2.$ Similarly other side inclusion works. Therefore $U_1=U_2$.

What is wrong with this proof?

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Bluey
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    There are lots of elements in $U_2\oplus W$ that are in neither $U_2$ nor $W$. Specifically, if $u\in U_2$ is nonzero, and $w\in W$ is nonzero, then $u+w$ is in neither $U_2$ nor in $W$ (check that). So your claim that “Therefore $u_1\in U_2$” does not follow. – Arturo Magidin Mar 06 '19 at 04:52
  • By the way, you show use the simple $+$ if you're referring to Minkowski sum, ie., $A+B={a+b\mid a\in A~,~b\in B}$ (the $\bigoplus$ is generally used for direct sum which doesn't seem to be the case here) – learner Mar 06 '19 at 04:54
  • @learner: Have you heard of internal direct sums? – Arturo Magidin Mar 06 '19 at 04:57
  • @Arturo Magidin Thanks. I can see my mistake now. – Bluey Mar 06 '19 at 05:02
  • @ArturoMagidin: I had, but I wasn't under the impression that it would be isomorphic to the (external) direct sum in this case. I stand corrected. – learner Mar 06 '19 at 05:02
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    Possible duplicate of Cancellation law for direct sums – learner Mar 06 '19 at 05:28
  • @learner i have seen that question, but i think my question is related to this particular erroneous proof of mine. – Bluey Mar 06 '19 at 05:31

1 Answers1

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This is not true. The error is in this line:

Since $U_1$ and $W$ have no elements in common, $u_1 \not\in W$. Therefore $u_1 \in U_2$.

It is not true in general that if $x \in A \oplus B$ then $x \in A$ or $x \in B$. (Perhaps you are confusing the properties of the direct sum $U_2 \oplus W$, with the properties of the union $U_2 \cup W$, which in general is not a subspace.) An instructive example is taking $A$ to be the $x$-axis in $\Bbb R^2$ and $B$ the $y$-axis. Then, $\Bbb R = A \oplus B$ but, e.g., $(1, 1)$ is in neither $A$ nor $B$.

Travis Willse
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  • What do you mean “the direct sum, which in general is not a subspace”? Was that clause meant to apply to the union? – Arturo Magidin Mar 06 '19 at 05:36
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    Yes, of course---I originally had the two concepts in the other order. I'm grateful for the catch, and I've fixed the error. – Travis Willse Mar 06 '19 at 05:48