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The definition of direct sum that I learned is: $U$ is the direct sum of $U_1,U_2$ if $\forall u\in U,\exists !(u_1,u_2)\in U_1\times U_2:u=u_1+u_2$.

I want to understand what $U=U_1\oplus U_2$ means.

From the definition I know that every element of $U$ gets decomposed in a unique way on terms of $U_1,U_2$. However, does that also mean that every $u_1\in U_1,u_2\in U_2$ can be associated to a unique $u\in U$?

Thanks for helping! :D

Guilherme Salomé
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  • Two elements $u_1$ and $u_2$ can be uniquely associated to $u_1+u_2$... Is this what you meant in your last question? – coldnumber Aug 05 '15 at 20:25
  • @coldnumber Yep. I know that every element of $U$ can be uniquely associated with $U_1,U_2$. But can every element of $U_1+U_2$ be associated with an element of $U$ when we write $U=U_1\oplus U_2$. I guess this is a dumb question, but I was not understanding it clearly. – Guilherme Salomé Aug 05 '15 at 20:34

2 Answers2

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I think a characterization of the definition might be useful:

$$U_1 \oplus U_2 \iff U_1+U_2=U \text{ and } U_1 \cap U_2 = \{0\}$$

So let's see what each condition means:

$U_1+U_2=U$ means that given an element $u \in U$ you can always find elements $u_1 \in U_1$ and $u_2 \in U_2$ such that $u_1+u_2 = u$.

But the uniqueness of this way of writing $U$ comes from the condition that $U_1 \cap U_2=\{0\}$. If you have $U_1 +U_2=U$ and $U_1 \cap U_2 \neq \{0\}$, then there may be more than one choice for $u_1$ and $u_2$. Here's an example of a sum that is not a direct sum:

Take $U=\Bbb R^3$ and suppose $U_1=\langle e_1, e_2\rangle$ and $U_2=\langle e_2,e_3\rangle$. (Here $e_i$ is the vector with a $1$ in the $i$th coordinate and a $0$ in the other two.)

Then $U_1 \cap U_2 = \langle e_2\rangle$, and indeed elements of $U$ don't have a unique decomposition. For example, consider

$$(1,2,3) = e_1+2e_2+3e_3$$

This can be interpreted in many different ways; here are three of them:

$$(e_1)+(2e_2+3e_3) \text{ with $e_1 \in U_1$ and $2e_2+3e_3 \in U_2$} \\ (e_1+2e_2)+(3e_3) \text{ with $e_1+2e_2 \in U_1$ and $3e_3 \in U_2$} \\ (e_1+e_2)+(e_2+3e_3) \text{ with $e_1+e_2 \in U_1$ and $e_2+3e_3 \in U_2$}$$

Hence, in the context of your definition, in this example you could take $(u_1,u_2)$ to be any of $$(e_1, 2e_2+3e_3), (e_1+2e_2, 3e_3),(e_1+e_2, e_2+ 3e_3),$$ which means that $U$ is not the direct sum of $U_1$ and $U_2$.

coldnumber
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    Let me see if I understand: loosely speaking, the direct sum forms a bigger space within the two existing spaces, while the cartesian product forms a higher dimensional space? Also, it's not always the case that $U_2$ is the orthogonal complement of $U_1$, correct? You could have $U=\mathbb{R}^3$, $U_1 = \langle e_1, e_2 \rangle$, $U_2 = \langle (1,1,1) \rangle$. Then $U = U_1 \oplus U_2$. – Chester Aug 06 '15 at 00:08
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    What an succinct interpretation! Yes, each of the component subspaces is contained in the direct sum, which is always a subspace of the vector space. (That said, the direct sum and the cartesian product are isomorphic.) As your example shows, $U_1\oplus U_2 = U$ does not imply that $U_1 = U_2^\perp$, but the converse does hold: $U_1 = U_2^\perp \implies U_1 \oplus U_2 = U$. – coldnumber Aug 06 '15 at 00:34
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    Neat. Direct sum never really clicked with me until I read your answer. May I prod further? What is the advantage of having intersection containing only the null vector? Why not toss out the direct sum and use say, union of spans? Is direct sum a sort of generalization of independence between spaces? – Chester Aug 06 '15 at 00:45
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    Wow thanks :). If the intersection contains a nonzero vector, then the sum is not isomorphic to the cartesian product; the homomorphism sending an ordered pair $(x,y)$ to $x+y$ won't be injective. I think of it as a sort of decomposition, because if $U_1 \cap U_2 = {0}$ we're splitting $U$ into $U_1 \times U_2 \cong U_1 \oplus U_2$. Off the top of my head, I can recall one place where the direct sums are used: when splitting modules into their torsion and their torsion-free parts, but I'll take a look at my algebra book later today. I'm sure they pop up in some interesting places. – coldnumber Aug 06 '15 at 01:01
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We take vector spaces $U_1$ and $U_2$ and form $U_1 \times U_2$ by the natural component-wise addition and scalar multiplication: $$ (a,b)+(x,y) = (a+x,b+y) \qquad \& \qquad c(a,b) = (ca,cb)$$ we do have: $$ u=(u_1,u_2) = (u_1,0)+(0,u_2) $$ but, I wouldn't write $u=u_1+u_2$ here... unless I was identifying $u_1$ with $(u_1,0)$ and $u_2$ with $(0,u_2)$.

In contrast, I would write $U = U_1 \oplus U_2$ if both $U_1$ and $U_2$ were subspaces of $U$ for which every $u \in U$ is uniquely decomposed as $u=u_1+u_2$ for $u_1 \in U_1$ and $u_2 \in U_2$.

Of course, there is an isomorphism between these and often it is abused, wait, no, used.

James S. Cook
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