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If $U_{1}\oplus...\oplus U_{m}=U_{1}+...+U_{m}$ implies that $U_{1}\cap...\cap U_{m}=\{0\}$, how is it that we can say that dim $(U_{1}+...+U_{m})=$ dim $U_{1}+...+$ dim $U_{m}$?

Wouldn't this go against the the true statement dim $(U_{1}+...+U_{m})=$ dim $U_{1}+...+$ dim $U_{m} -$ dim $U_{1}\cap...\cap U_{m}$?

Unless dim $\{0\}$ is zero?

Is the dimension of a set containing only the zero vector equal to zero? I don't see that in my book.

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    Yes, the dimension of the trivial vector space $\{0\}$ is zero. – 2017-02-12
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    Yes, $\dim \{0\}=0$, because you can't pick up a single linearly independent vector. So the basis of $\{0\}$ would have no elements. Recall that the dimension of a space is the number of elements of their basis. – 2017-02-12
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    @flytothesurface Should I delete this question then? Your comment reminding me to remember the definition of dimension was all I needed. Perhaps the question is too silly? – 2017-02-12
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    No, there's no need to. The question is never too silly, it's quite frequent to ask questions about the trivial space because, as it's trivial, everyone takes for granted that is simple. But sometimes, things like this can happen. – 2017-02-12
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    By the way, your statement about dimensions of (not necessarily direct) sums of subspaces only holds when $m=2$. There isn't even a good generalization for $m=3$. For direct sums it always holds. – 2017-02-13

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