I was assigned the following problem:

Let $W$ and $Z$ be two subspaces of $V$ such that $W \cap Z = \emptyset$ and let $K = W + Z$. Prove that every element of $K$ can be written uniquely as the sum of an element of $W$ and an element of $Z$.

I did actually manage to prove this. That is not my question. What seems odd to me about this problem though, is the nature of $W$ and $Z$. I don't understand how $W \cap Z$ could ever equal $\emptyset$ given that the zero vector of vector space $V$ will certainly always be in $W \cap Z$. ($W$ and $Z$ are subspaces; closed under scalar mult.; just multiply any element by $0$ in each). Am I missing something? Or is this situation just impossible?

Thanks, and sorry for the poor symbol formatting.

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    $\begingroup$ Yes, that is a sloppy typo. The hypothesis should have been that $W$ and $Z$ have "trivial intersection," i.e., precisely $\{0\}$. P.S. Welcome to MSE. Work on writing more informative titles for your questions :) $\endgroup$ – Ted Shifrin Feb 18 at 1:17
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    $\begingroup$ You are correct that any subspace must contain the zero vector $\endgroup$ – J. W. Tanner Feb 18 at 1:19
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    $\begingroup$ Which textbook are you using? It’s certainly sloppy notation, but I wonder if the author makes some statement at the beginning of the book clarifying that the notation “means” trivial intersection. $\endgroup$ – Clayton Feb 18 at 1:21
  • $\begingroup$ Thanks all, looks like this was just a typo. The question was professor-generated (not sure if from a book) and I do indeed remember my prof explicitly clarifying that {zero vector} is not the same as {∅}. And I will work on my titling in the future :) $\endgroup$ – Albert Elling Feb 18 at 1:48

Notice that the definition of direct sum determines that given a vector space $V$, and two sub-vector spaces of $V$, let them be $U,W$, then:

$$V = U\oplus W \iff \forall v\in V \exists u\in U, w\in W: v=u+w$$

Also, for every $v\in V$ there is exactly \one $u\in U, w\in W$ such that $v=u+w$ is unique.

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