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Could someone provide examples of sum and direct sum of vector subspaces?

Maybe with geometric vectors, row vectors, polynomials, anything...

Difference between sum and direct sum

The way I get it is that every direct sum is also a sum.

What is an example (if not too trivial even better) of a sum which is not a direct sum?

I have a book with a few definitions and theorems but not a single example related to these two concepts.

Note that by sum I don't mean union in set theory sense but sum of two subspaces as defined here:

https://math.stackexchange.com/a/1163346/116591

peter.petrov
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  • Consider a one dimensional space, and two subspaces, each equal to the whole space. Then the sum of those two subspaces is again the whole space, but the direct sum is two dimensional. – lulu Sep 05 '21 at 15:59
  • @march the sum would be direct regardless of whether the two vectors are orthogonal. The definition of a direct sum does not require a scalar product (which is needed to define orthogonality) – Al.G. Sep 05 '21 at 16:18
  • @lulu Why is the direct sum two dimensional?! I think it will not exist in this example... – peter.petrov Sep 05 '21 at 16:28
  • Why would it not exist? It consists of pairs $(\vec v_1, \vec v_2)$ where both of the $\vec v_i$ are in the space. – lulu Sep 05 '21 at 16:30
  • @lulu No pairs here. You are talking about something else, it seems, not direct sum and sum of subspaces. – peter.petrov Sep 05 '21 at 22:00
  • I don't understand. here is a routine discussion of direct sums. Note that the definition involves ordered pairs. – lulu Sep 05 '21 at 22:03
  • @lulu Yeah, we're talking of two different definitions. This PDF here claims they are the essentially the same?! https://math.vanderbilt.edu/rolenl/LinearAlgebraNotes13.pdf I specified in my post which definition of "direct sum" I have in mind. Anyway, I got the examples I needed, thanks. – peter.petrov Sep 06 '21 at 09:42

2 Answers2

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The point of a direct sum is that each of the summands is independent of the others in the sense that if $V=\bigoplus_{k=1}^n U_k$, then every vector in $V$ can be uniquely expressed as a sum of exactly one vector per direct summand (in contrast to a non-direct sum, where the expression exists but is not unique). For instance, if we have two subspaces of $\mathbb R^3$ where $U_1$ is spanned by $(1,0,0)$ and $U_2$ is spanned by $(0,1,0)$ and $(0,0,1)$, then there is only one way to express $(a,b,c)$ as a sum of a vector from $U_1$ and a vector from $U_2$: it must be $(a,0,0)+(0,b,c)$. So $U_1+U_2$ is direct, or in short: $V=U_1\oplus U_2$.

A very simple example of a sum which is not direct: $V+V=V$ for all vector spaces $V$. Any sum of two vectors in $V$ is again a vector in $V$, from which the equality follows. This sum is only direct if $V=0$. Otherwise there are obviously many ways to express a vector in $V$ as a sum of two other vectors, also from $V$. A less trivial example: Take a 3d vector space $V$ with basis $\{b_1,b_2,b_3\}$, and the subspaces $U_1$ spanned by $\{b_1,b_2\}$ and $U_2$ spanned by $\{b_2,b_3\}$. Then every vector in $V$ can be written as a sum of one vector from $U_1$ and one from $U_2$, but not uniquely so. For instance, $b_2$ can be written as $2b_2-b_2$, or as $b_2+0$, or many other combinations.

Vercassivelaunos
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If the vector subspaces $V_1, \ldots ,V_n$ have the property that

$$ V_k\cap \left(\sum_{j\neq k}V_j\right)=\{0\} $$

for each $k$ then it sum is named as a direct sum and denoted by $\bigoplus_{k=1}^n V_k$. It have the property that if $v\in \bigoplus_{k=1}^n V_k$ then there exists a unique decomposition $v=\sum_{k=1}^n v_k$ such that $v_k\in V_k$ for each $k$. The typical example of direct sum is that when each $V_k=\operatorname{span}(v_k)$ and the list $v_1,\ldots ,v_n$ is linearly independent in $V$, by example the sum of vector subspaces of $\mathbb{R}^2$ given by

$$ \operatorname{span}((1,0))+\operatorname{span}((0,1)) $$

is direct. However the sum of vector subspaces of $\mathbb{R}^3$ given by $$ \operatorname{span}((1,0,0))+\operatorname{span}((0,1,0))+\operatorname{span}((1,1,0)) $$ is not direct, as

$$ \operatorname{span}((1,1,0))\cap \left(\operatorname{span}((1,0,0))+\operatorname{span}((0,1,0))\right)=\operatorname{span}((1,1,0))\neq \{0\} $$