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I am just starting to learn about direct sums and every definition I have read about direct sums say that the intersection of subspaces must be zero. So, how can $$\mathbb{R} \oplus \mathbb{R}$$ be a direct sum when they are identical? Thanks in advance!

David South
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    There is a difference between an internal and external direct sum. Internal direct sums are when you take two subspaces of a larger space and combine them (this is what you are referring to). External direct sums are when you take two vector spaces and combine them. It took me a while to realize this because professors can sometimes be sloppy and a little handwavy with their terminology (they know what they're referring to, but it happens) and at least a couple texts I know are not super clear on the distinction. Fundamentally the two ideas are pretty much the same if you think about it. – Cameron Williams Mar 21 '16 at 17:55
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    This is handled by what is called an external direct sum vs. internal direct sum. We are writing $\mathbb{R} \oplus \mathbb{R}$ with the understanding that the points are ordered pairs $(x,y)$ of real numbers. – hardmath Mar 21 '16 at 17:55
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    ${(x,0) : x\in \mathbb{R}} \oplus {(0,y) : y \in \mathbb{R}}$, each is a subspace of $\mathbb{R}^2$ – Xiao Mar 21 '16 at 17:56

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The direct sum depends on context. There are two notions of direct sum: the inner direct sum and the outer direct sum.

With the inner direct sum, we have some large vector space and two subspaces. We must assume that the intersection of the two subspaces is zero in order to form the inner direct sum.

With the outer direct sum, we take two vector spaces that have nothing to do with each other and slam them together. I.e. you get a vector in the outer direct sum by appending vectors from your two vector spaces.

It turns out that these two notions are isomorphic. That is, if you take the inner direct sum of two subspaces in a big vector space, that is isomorphic to the vector space obtained by taking the outer direct sum of those subspaces as vector spaces in their own right.

In your example, we are viewing the two copies of $\mathbb{R}$ as distinct vector spaces that have nothing to do with each other and then taking the outer direct sum of them.

To think about it in terms of the inner direct sum, think of one of the copies of $\mathbb{R}$ as the x-axis and the other copy of $\mathbb{R}$ as the y-axis. Then their intersection is zero, and taking the inner direct sum yields $\mathbb{R}^2$.

Ken Duna
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  • Comments say we are taking the internal direct sum and you are saying we are taking the outer direct sum. Who is right? Thanks! Or are you saying that we can view these two copies as distinct and "smash" them together and it would be exactly the same as thinking of it as an inner sum? I guess that's what isomorphism means. – David South Mar 21 '16 at 18:01
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    Written as is in your question, it would have to be the outer direct sum. Since as you pointed out, the intersection of those two thing is not zero. However if you think of those $\mathbb{R}$'s as sitting inside the plane, then you can think of the statement as taking the inner direct sum. This is what Xiao does in his comment. – Ken Duna Mar 21 '16 at 18:04
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    "Or are you saying that we can view these two copies as distinct and "smash" them together and it would be exactly the same as thinking of it as an inner sum?" That is exactly right. – Ken Duna Mar 21 '16 at 18:05
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    @DavidSouth the comment that said you refer to an internal direct some meant the first part (zero-intersection). – quid Mar 21 '16 at 18:06