Direct sum of two elements of subspaces of a vector space

Question

Help me to understand what the authors of this paper (p. 3) mean by the direct sum of two elements in a vector space.

Let $X$ be a vector space with subspaces $Y$ and $Z$

Definition: X is a direct sum of $Y$ and $Z$, denoted $X = Y\oplus Z$, if $X = \{y+z : y \in Y, z \in Z\}$ and $Y \cap Z = \{0\}$.

The authors then say that "we write $y \oplus z$ to denote the direct sum of elements $y \in Y$, $z \in Z$ of subspaces $Y$ and $Z$, respectively, of $X$".

What is meant by the direct sum of elements $y \oplus z$?

Answer 1

Every element in $X$ can be expressed uniquely as the sum of an element of $Y$ and an element of $Z$, this is what it means for $X$ to be the direct sum of $Y$ and $Z$. The authors denote this as $y \oplus z$.

Answer 2

An example shall better explain this, I believe:

$$\Bbb R^3:=V_1\oplus V_2\;,\;\;\text{with}\;\;V_1:=Span\{(1,0,0)\}\,,\;\;V_2:= Span\{(0,1,0), (0,0,1)\}$$

and we indeed have that for all $\;(x,y,z)\in\Bbb R^3\;$ :

$$\;(x,y,z)=x(1,0,0)+y(0,1,0)+z(0,0,1)=u_1\oplus u_2\;,\;\;u_k\in V_k$$

and this expression is unique. This follows at once from the fact that the vectors that spann $\;V_1, V_2\;$ are a basis of $\;\Bbb R^3\;$, and in fact this condition is equivalent to a direct sum: we have that $\;V=W\oplus U\;$ iff the set-theoretical union of basis of $\;U, W\;$ is a basis of $\;V\;$, and this happens in the finite dimensional case iff $\;\dim V=\dim W +\dim U\;$ .

Now take

$$\;V_1=Span (1,0,0), (0,1,0)\}\;,\;\;V_2=Span\{(1,1,1), (0,0,1)\}$$

As before, for any $\;(x,y,z)\in\Bbb R^3\;$ :

$$(x,y,z)=x(1,0,0)+y(0,1,0)+z(0,0,1)\in V_1+V_2\,, \text{and}\;x(1,0,0)+y(0,1,0)\in V_1,z(0,0,1)\in V_2$$

but also

$$(x,y,z)=(y-x)(0,1,0)+x(1,1,1)+(z-x)(0,0,1),\,$$

with

$$(y-x)(0,1,0)\in V_1,\,x(1,1,1)+(z-x)(0,0,1)\in V_2$$

Thus this last is just an expression of $\;\Bbb R^3\;$ as asum of two of its subspaces but not a direct sum.