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Let $A,B$ be linear subspaces of $\mathbb R^n$. Show that if $$ \dim(A+B) = \dim(A \cap B) + 1, \tag{1} $$ then one of the space is a subset of the other, $$ A \subset B \text{ or } B \subset A. \tag{2} $$

By the dimension theorem, then (1) $$ \dim(A+B)=\dim(A)+\dim(B)-\dim(A \cap B) = \dim(A \cap B) + 1, \tag{3} $$ so that $$ \dim(A)+\dim(B)= 2\dim(A \cap B) + 1. \tag{4} $$

I can't see where to go next.

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up vote 5 down vote accepted

Since $A \cap B \subseteq A \subseteq A+B$, it follows that $$ \dim A \cap B \leq \dim A \leq \dim(A+B) = \dim(A \cap B) + 1. $$ Hence, either $\dim A = \dim(A \cap B)$ or $\dim A = \dim(A+B)$. What follows from each of these two cases?

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Hint: To prove the contraposative, suppose neither $A \subset B$ or $B\subset A$. Then $A\cap B$ is strictly contained in both $A$ and $B$, so $\dim(A\cap B) < \dim(A), \dim(B)$.

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@CameronBuie Sorry, I don't see how that's relevant, could you please explain what you mean? – Tom Oldfield Apr 10 '13 at 18:18
Oops! As I was writing out my explanation, I saw that it wasn't necessary to note that, at all. Your hint is really all that is needed. (+1) – Cameron Buie Apr 10 '13 at 18:33
@CameronBuie No worries,thanks anyway! When I wrote the hint I couldn't find a way to indicate the direction I was thinking in without giving the answer away entirely. Since I like to leave more of a solution hidden rather than less, I left it as it was, although I appreciate that it's not immediately clear where the argument's going! – Tom Oldfield Apr 10 '13 at 18:38

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