Suppose U is a subspace of V. Then U+U = U+{0}=U, right? So the operation of addition of vector spaces does not have unique additive identities.

*typo in title

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If U is a vector space (doesn't need to be a subspace of a larger vector space V) then U + [any subspace of U] = U. This is easily derived from considering the basis vectors of U and any of it's possible subspaces.

mathwonk
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i thought i had found someone else addicted to subspaces, like me.

matt grime
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That depends on how you define addition of vector spaces.

Personally I would use direct sum, when there is a well defined identity, the zero vector space.

You cannot add a subspace of U to U whilst considering it as a subspace. You can add a vector space isomorphic to a subspace of U (which is just any subspace of smaller dimension in fact) and get something strictly different from U, barring adding the zero vector space).

Vector addition is defined as the set of all possible sums of vectors from the two vector spaces.

There's something I don't understand about direct sums. It's not a binary operator or anything, so V direct sum W could possibly produce nothing at all. It's like saying 'A disjoint union B'. All we can say is either the statement is true or false. So how does one 'use' direct sums?

matt grime
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The addition you define is that which operates on subspaces of some given vector space. It isn't 'vector addition', it is addition of subspaces, perhaps even 'addition of vector subspaces'.

Anyway, I think you have a misconception about the (entirely different) subject of direct sum.

Why is direct sum not a binary operator on the category of vector spaces (or any other abelian category)? I'd be interested to hear why you think this, since it certainly is a binary operator, actually it can be an n-ary operator, or possibly even an A-operator for any integer n, or cardinal A.

Direct sum must take place in some ambient category, and it is never the case that $V\oplus W$ is zero when V and W are non-zero in the category of vector spaces.

The direct sum of A and B is the smallest vector space containing both A and B as subspaces intersecting in 0 only.

It is the vector space analogue of disjoint union (coproduct), and I am perplexed by what "All we can say is either the statement is true or false" has to do with anything.

So how do you 'take' the disjoint union of a set with itself?

matt grime
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it is two copies of the object, each having the canonical injection from one copy of the set. that is the disjoint part of disjoint union.
we are after all only dealing with things upto isomorphism. thus it is just a set with twice as many elements and a pair of maps.

How do you take the direct sum of a vector space with itself?

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mathwonk
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to construct the disjoint union of two copies of the set S, you paint one copy red and the other copy blue.

more precisely,

consider the cartesian product Sx{red} and Sx{blue}. and take the union. Say you direct sum a vector space with itself. Then doesn't it lose the fact that each vector in the direct sum can be written as a linear combination of vectors in a unique way?

mathwonk
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no because redv + blue w is different from redw + blue v.

matt grime
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You do understand that you've been using direct sums of vector spaces all the time? R^2, R^3, etc is the same thing as $\mathbb{R}\oplus\mathbb{R}$ etc.

So $$\mathbb{R}\oplus\mathbb{R} = \mathbb{R}\times\mathbb{R}$$?

matt grime
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Of course. In an abelian category with finite coproducts (direct sum) and products they agree. Note this is not true for sets. Coproduct (disjoint union) and product definitely do not agree.

Then why the charade with direct sums? Why not just study Cartesian products?

matt grime
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What Charade? Direct sum and direct product are different, except in very rare cases.

If X_i are vector spaces then

$$\prod X_i$$

and

$$\coprod X_i \cong \oplus X_i$$

are definitely different if the index runs over any infinite set.

In set terms you don't believe (disjoint) union or product are the same do you?

Incidentally, you can think of the former as the set of 'tuples' indexed by i, and the latter the subset of them that are zero in all but finitely many entries.

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I see. Thanks for the clarification.

matt grime