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From what I have read the definition of the direct sum of the vector spaces $V_1$, $V_2$ is the set $V_1\times V_2$ with the operations of addition and scalar multiplication defined as follows (Knapp, 2006): \[(u_1,u_2)+(v_1,v_2)=(u_1+v_1,u_2+v_2)\] \[c(v_1,v_2)=(cv_1,cv_2)\] But then today I came across the fact that if $V$ has linear subspaces $V_1$, $V_2$ and every $v\in V$ can be written uniquly as: $$v=v_1+v_2$$ for $v_i \in V_i$ then $V=V_1\oplus V_2$. I cannot see how this makes sense given the definition above. Is the above definition wrong? If not how can $V$ have the form $(v_1,v_2)$ when $v_1,v_2 \in V$?

Xam
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Quantum spaghettification
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3 Answers3

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The sum of two subspaces of a vector space $V_1,V_2\subseteq V$ (denoted as $V_1+V_2$ is defined as

$$V_1 + V_2 = \{v_1 + v_2| v_1\in V_1\land v_2\in V_2\}$$

and it is a vector subspace of $V$.

The direct sum of two vector spaces $V_1, V_2$, (denoted as $V_1\oplus V_2$) is a vector space $V_1\times V_2$ with operations defined as you wrote.

The connection between a direct and ordinary sum is that if $V_1, V_2\subset V$ and $V_1$ and $V_2$ are linearly independent, then $V_1+V_2$ is isomorphic to $V_1\oplus V_2$.

5xum
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The $+$ is defined on pairs of subspaces of a given vector space and $\times$ on pairs of vector spaces (not considered as subspaces of any sorts) and we write $V_1 \oplus V_2$ in place of $V_1 + V_2$, if $V_1 \cap V_2 = \{0\}$.

These two things are "the same" in the sense, that if $V_1,V_2$ are subvectorspaces of $V$ with $V_1 \cap V_2 = \{0\}$, then $V_1 \oplus V_2$ is isomorphic to $V_1 \times V_2$ which is not hard to prove.

Stefan Perko
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  • Do you mean then $V_1\oplus V_2$ is isomorphic to $V_1 +V_2$ instead of $V_1 \times V_2$? – Quantum spaghettification Aug 02 '17 at 14:07
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    @Quantumspaghettification They are the same. The notation $V_1 \oplus V_2$ in place of $V_1 + V_2$, if $V_1 \cap V_2 = {0}$, is just very popular (I guess to be brief? It's the same idea as disjoint unions of subsets, c.f. https://math.stackexchange.com/questions/1631396/what-is-the-difference-between-disjoint-union-and-union). EDIT: sorry, I misread your comment. I mean: $V_1 \oplus V_2 = V_1 + V_2 \cong V_1 \times V_2$ ($\cong$ meaning "is isomorphic to") – Stefan Perko Aug 02 '17 at 14:11
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I have just found this pdf: http://www.math.ncku.edu.tw/~fjmliou/advcal/sumvspace.pdf which summarizes direct sums, and explains what is going on here. I will here outline what it says. Thanks also to the other answers which helped clarify things.

This is to do with the difference between an internal and external direct product:

The External Direct Sum

Let $V$ and $W$ be two vector spaces over $F$ then we the external direct product $V\oplus_e W$ is defined as I have given it in the question. i.e. the vector space with underlying set $V\times W$ and operators defined by: \[(u_1,u_2)+(v_1,v_2)=(u_1+v_1,u_2+v_2)\] \[c(v_1,v_2)=(cv_1,cv_2)\]

The Internal Direct Sum

For $V_1$ and $V_2$ subspaces of $V$ which satisfy:

  • for each $v\in V$ there exists $v_1 \in V_1$ and $v_2 \in V_2$ such that $v=v_1+v_2$.
  • $V_1 \cap V_2 = \{0\}$

in this case we write $V=V_1 \oplus_i V_2$ and $V$ is the inner direct product of $V_1$ and $V_2$.

Link between them

There is a linear isomorphism such that $V_1\oplus_e V_2$ is isomorphic to $V_1 \oplus_i V_2$

Knight white
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Quantum spaghettification
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