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How is direct sum of two vector subspaces different from the sum of two vector subspaces i.e. how is $X\oplus Y$ different from $X + Y$, where $X, Y$ are subspaces.

Stefan Hamcke
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Hashtag
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    For the direct sum, the intersection of $X$ and $Y$ must be trivial. That means every $v \in V$ (let $V$ be the ambient vector space) has a unique representation $v = x+y$ with $x\in X,, y \in Y$. For the not-necessarily-direct sum, such a represenattion need not be unique. – Daniel Fischer Sep 07 '13 at 12:48
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    It is the same with the additional constraint that $X\cap Y=0$ – Hagen von Eitzen Sep 07 '13 at 12:50
  • @DanielFischer can we say that sum of some vector subspaces is a necessary condition for the existence of direct sum of those two subspaces? – Fazzolini Nov 07 '15 at 12:07
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    @Fazzolini I don't understand what you're asking, sorry. Whenever we have a collection of subspaces of a vector space, the sum of these subspaces is defined. So the existence of the sum of subspaces isn't a condition at all. The sum of two subspaces is direct, if and only if the two subspaces have trivial intersection. – Daniel Fischer Nov 07 '15 at 12:15
  • @DanielFischer Suppose we have 3 subspaces in some space, namely, $U_1$, $U_2$ and $U_3$. Can we say that for some $V$ to be equal to direct sum of $U_1$, $U_2$ and $U_3$, the following two conditions should be met: (1) $V$ is the sum of $U_1$, $U_2$ and $U_3$ and (2) each element of $V$ can be written uniquely as $u_1 + u_2 + u_3$ where $u_1 \in U_1$, $u_2 \in U_2$ and $u_3 \in U_3$? – Fazzolini Nov 07 '15 at 14:04
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    @Fazzolini Yes, that is correct. Note that the uniqueness of the representation means $U_1 \cap (U_2 + U_3) = U_2 \cap (U_1 + U_3) = U_3 \cap (U_1 + U_2) = {0}$. – Daniel Fischer Nov 07 '15 at 14:11

1 Answers1

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For instance, in $\mathbb{R}^3$

$$ \langle e_1 \rangle + \langle e_2, e_3 \rangle = \langle e_1 \rangle \oplus \langle e_2, e_3 \rangle $$

is a direct sum. Whereas

$$ \langle e_1, e_2 \rangle + \langle e_2, e_3 \rangle $$

is not.

EDIT. $\langle e_1 \rangle $ stands for the vector subspace generated by vector $e_1 = (1,0,0)$. Maybe you write it $\mathrm{span}(e_1)$? The output of both sums is the same, namely the whole $\mathbb{R}^3$.

The point is that the first sum is a direct one because $\langle e_1 \rangle \cap \langle e_2, e_3\rangle = \left\{ (0,0,0)\right\}$, whereas $\langle e_1,e_2 \rangle \cap \langle e_2, e_3 \rangle = \langle e_2\rangle$. Hence the last one is not a direct sum.

Agustí Roig
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  • @Daniel Fischer Thanks. I have understood the comment by Daniel but example by Agusti is still not clear. – Hashtag Sep 07 '13 at 17:12
  • Why is it not clear? – Agustí Roig Sep 07 '13 at 23:26
  • @ Agusti Sorry for the delayed response.But could you clarify what are $ \langle e_1 \rangle $ and $ \langle e_1,e_2 \rangle $ - are they just standard basis vectors or are they vector spaces. In either cases what is the output of the two summations illustrated? – Hashtag Sep 14 '13 at 08:34
  • @Hashtag. I've edited my answer. – Agustí Roig Sep 14 '13 at 09:28
  • @ Agusti Thanks. Got the point you were trying to make. But if the output of both sums is identical in all cases then why do we need to make a classification between direct sum and sum. Does it benefit in any way if its a direct sum ..or are the properties of a direct sum different? – Hashtag Sep 15 '13 at 02:58
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    It's a piece of information you add to your sum: knowing that $X \oplus Y = X + Y$ allows you, for instance, to get a basis of $X+Y$ "for free": you just put together a basis of $X$ and a basis for $Y$. Otherwise, if the sum is NOT direct, the union of a basis of both $X$ and $Y$ is NOT a basis of $X + Y$. – Agustí Roig Sep 15 '13 at 04:43