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What is the difference between sum of two vectors and direct sum of two vector subspaces?

My textbook is confusing about it. Any help would be appreciated.

Thanks in advance!

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    $\begingroup$ I think the phrase "direct sum" goes with "vector spaces", not with vectors. $\endgroup$ – hardmath Feb 24 '15 at 14:52
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Direct sum is a term for subspaces, while sum is defined for vectors. We can take the sum of subspaces, but then their intersection need not be $\{0\}$.

Example: Let $u=(0,1),v=(1,0),w=(1,0)$. Then

  • $u+v=(1,1)$ (sum of vectors),
  • $\operatorname{span}(v)+\operatorname{span}(w)=\operatorname{span}(v)$, so the sum is not direct,
  • $\operatorname{span}(u)\oplus\operatorname{span}(v)=\Bbb R^2$, here the sum is direct because $\operatorname{span}(u)\cap\operatorname{span}(v)=\{0\}$,
  • $u\oplus v $ makes no sense in this context.

Note that the direct sum of subspaces of a vector space is not the same thing as the direct sum of some vector spaces.

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    $\begingroup$ Having a sum equal to the whole space is not a requirement for a direct sum of subspaces. It is a requirement however for complementary subspaces (and another part of their definition is that their sum must be direct). $\endgroup$ – Marc van Leeuwen Feb 24 '15 at 15:21
  • $\begingroup$ @MarcvanLeeuwen right, actually it has to be the whole space $W$ with $W = U+V$ and $U\cap V =\{0\}$. $\endgroup$ – Surb Feb 24 '15 at 15:30
  • $\begingroup$ How does $\operatorname{span}(v)+\operatorname{span}(w)=\operatorname{span}(v)$? $\endgroup$ – user398843 Feb 21 at 1:44
  • $\begingroup$ @user398843 well $v=w$ so they span the same line. More generally if $V$ is a real vector space and $v=\alpha w\in V$ for some $\alpha \neq 0$, then $\text{span}(v)=\{\beta v \mid \beta \in \Bbb R\} = \{\beta/\alpha w \mid \beta \in \Bbb R\} = \{\beta w \mid \beta \in \Bbb R\} =\text{span}(w).$ Of course this is not restricted to real vector spaces but is simpler to discuss in a comment. $\endgroup$ – Surb Feb 21 at 9:28
  • $\begingroup$ Sorry, my bad. I read it as $\operatorname{span}(u)+\operatorname{span}(w)=\operatorname{span}(v)$. $\endgroup$ – user398843 Feb 21 at 15:25
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In Axler's Linear Algebra Done Right, he defines the sum of subspaces $U + V$ as

$\{u + v : u \in U, v \in V \}$.

He then says that $W = U \oplus V$ if

(1) $W = U + V$, and

(2) The representation of each $w$ as $u + v$ is unique.

This is a different way of presenting these definitions than most texts, but it's equivalent to other definitions of direct sum.

In anyone's book, the sum and direct sum of subspaces are always defined; and the sum of vectors is always defined; but there's no such thing as a direct sum of vectors.

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    $\begingroup$ Nor is there such a thing a the direct sum of subspaces that have a nonzero intersection. $\endgroup$ – Marc van Leeuwen Feb 24 '15 at 15:23
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You need to keep in mind these two definitions:

Definition 1: sum of subsets

Suppose $U_1,...,U_m$ are subsets of $V$. The sum of $U_1,...,U_m$ denoted by $U_1+...+U_m$ is the set of all possible sums of elements $U_1,...,U_m$. More precisely,

$$U_1 + ... + U_m = \{u_1+...+u_m: u_1\in U_1,...,u_m\in U_m\}$$

Definition 2: direct sum

Suppose $U_1,...,U_m$ are subspaces of $V$.

The sum $U_1+...+U_m$ is called a direct sum if and only if each element of $U_1+...+U_m$ can be written in one and only one way as a sum of $u_1+...+u_m$ where each $u_j$ is in $U_j$. The notation for such a sum of subsets is $U_1\oplus ... \oplus U_m$.

There's a shortcut to identify direct sums:

Suppose $U_1,...,U_m$ are subspaces of $V$. Then $U_1+...+U_m$ is a direct sum if and only if the only way to write $0$ as a sum $u_1+...+u_m$ is by taking each $u_j$ equal to $0$.

Also,

Suppose $U$ and $W$ are subspaces of $V$. Then $U+W$ is a direct sum if and only if $U\cap W =\{0\}$.

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