The definition of $\oplus$

Question

I would like to understand why the books give two different concepts to $\oplus$ between vector spaces:

See:

Concept 1: $W=V_1\oplus V_2=\{(v_1,v_2) \mid v_1\in V_1, v_2\in V_2\}$.

Concept 2: $W=V_1\oplus V_2=\{v_1+v_2 \mid v_1\in V_1, v_2\in V_2\}$, where $V_1\cap V_2=\{0\}$.

Are they equivalent?

I'm thinking to prove this is an isomorphism $(v_1,v_2)\mapsto v_1+v_2$.

Thanks

Yes, they are equivalent. Remember that the second concept only makes sense when $V_1$ and $V_2$ are subspaces of $W$. Also, one has to be careful, as the first concept also applies for subspaces when $V_1 \cap V_2 \neq \emptyset$. Usually it is understood from context which of the concepts is meant. — Ulrik, Nov 16 '14 at 23:37
The first is known as the external direct sum. The second is the internal. They are isomorphic. — Dan Rust, Nov 16 '14 at 23:38
Possible duplicate of The direct sum $\oplus$ versus the cartesian product $\times$ — Watson, Mar 27 '16 at 15:29
@Watson It's not duplicate of that by any stretch of the imagination. — Matt Samuel, Mar 13 '17 at 10:21

score 4 · Answer 1 · edited May 03 '18 at 08:47

4

The two concepts are indistinguishable from the point of view of algebra.

The mathematical formulation of this is: $$ V_1\cap V_2 = \{0\}\implies \{(v_1 +v_2 |v_1 ∈V_1 ,v_2 ∈V_2 \} \cong\{ (v_1,v_2) |v_1 ∈V_1 ,v_2 ∈V_2 \} $$

I let you find the isomorphism.

edited May 03 '18 at 08:47

answered Nov 16 '14 at 23:40

mookid

28,236

score 4 · Answer 2 · edited Nov 17 '14 at 01:11

Take two subspaces $S,T$ of your vector space $V$. One can form the vector space $V'=S\times T$ that consists of pairs $(s,t)$ with coordinatewise addition and scalar multiplication. We can define a linear map $\eta:S\times T\to V$ that sends $(s,t)\to s+t$. This has image $S+T$ and kernel $\{(x,-x):x\in S\cap T\}\simeq S\cap T$ (prove this!) so it is an isomorphism of vector spaces if, and only if, $S+T=V$ and $S\cap T=0$. So, we have the two seemingly different concepts of "external" and "internal" direct sums, but essentially they are the same. We have a vector space $V$ and two subspaces $S,T$ such that $S\cap T=0$ and $S+T=V$. In the first case, $V=S\times T$, $S=\{(s,0):s\in S\}$ and $T=\{(0,t):t\in T\}$; in the second $V$ is an "arbitrary" vector space and $S,T$ satisfy the conditions mentioned above.

Essentially, this allows us to think of every vector of $V$ as decomposed into an $S$ component and a $T$ component, and uniquely so, and is very useful to understand vector spaces and their linear transformations. A great example is the Jordan canonical form or the Rational canonical forms, which are particular examples of a structure theorem for modules over PIDs. It allows, also, to inductively work our way out of a problem for a finite dimensional vector space by chopping out a one dimensional subspace for example, consider the proof that every orthogonal transformation is a composition of rotations and reflections, say.

$S\cap T$ is not a subset of $S\times T$. – mookid Nov 16 '14 at 23:50 — mookid, Nov 16 '14 at 23:50
@mookid Sure, that was just bad notation. Let me rephrase. – Pedro Nov 16 '14 at 23:51 — Pedro, Nov 16 '14 at 23:51

Matt Samuel · Answer 3 · 2014-11-17T00:14:18.733

There is a distinction to be made here. The first use, where we consider vectors of the form $(v_1,v_2)$, is called the external direct sum. We assume we have two vector spaces that are not necessarily subspaces of a particular vector space and we construct a new vector space from them. The second use, $v_1+v_2$ where $V_1$ and $V_2$ are subspaces of the same vector space such that $V_1\cap V_2=\{0\}$ is called the internal direct sum. Saying that a vector space is the internal direct sum of two subspaces is more specific than saying it is isomorphic to the direct sum of the two subspaces because we are given an actual decomposition on a vector space that was not defined as a direct sum in advance.

Don't fall into the trap of thinking that isomorphic structures are indistinguishable. A vector space can be written as an internal direct sum in many distinct ways, and which subspaces actually occur in the decomposition may be important because they may have some properties that are not dependent entirely on the vector space structure that will allow you to prove something.

For example, if we have an inner product on a vector space the vector space is the internal direct sum of a subspace and its orthogonal complement. The vector space is also the internal direct sum of the same subspace and a subspace different from the orthogonal complement, but that doesn't interest us as much when we are studying the inner product. The particular subspaces in the direct sum matter.

This comic illustrates the concept particularly well.

score -1 · Answer 4 · answered Nov 17 '14 at 00:31

In addition to MattS's excellent answer, we can add that the two notions, Cartesian products and external direct sum, differ when considering infinitely many vectors spaces.

We consider a family $(V_j)_{j\in J}$ of vector spaces, indexed by a set $J$. Then:

The product $\displaystyle{\prod_{j\in J}V_j}$ is, as a set, the Cartesian product of the $V_j$ with the following operations: if $v=(v_j)_{j\in J}$ and $w=(w_j)_{j\in J}$, $\lambda v=(\lambda v_j)_{j\in J}$ and $v+w=(v_j+w_j)_{j\in J}$. It is easily checked that these two operations give to $\displaystyle{\prod_{j\in J}V_j}$ a structure of vector spaces.
The external direct sum $\displaystyle{\bigoplus_{j\in J}V_j}$ is the subspace of $\displaystyle{\prod_{j\in J}V_j}$ of all vectors $v=(v_j)_{j\in J}$ such that all component $v_j$, except a finite number of them, are zero.

For a finite number of spaces $V_j$, these two spaces are the same.

These definition are not actually restricted to vector spaces but can be applied to many algebraic structures (groups, modules, rings,...), with the suitable operations.

Edit: after re-reading the OP and the comments, I feel this answer is more a complement to Pedro Arbizu's comment and Mookid's answer.

The OP didn't ask about products, rather about the seemingly different concepts of "inner" and "outer" coproduct of vector spaces! =D — Pedro, Nov 17 '14 at 00:41

The definition of $\oplus$

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