# Direct sum of two elements of subspaces of a vector space

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 $$y \oplus z$$?

• I don't know if this answers your question, but I like to think about this notation as follows, so I will let it as a comment. You can show that the condition of the definition implies that the linear map $Y\times Z\to X$ defined by $(y,z)\mapsto y+z$ is an isomorphism of vector spaces, where the operations $+$ and $\cdot$ in $Y\times Z$ are defined by $(y,z)+(y',z')=(y+y',z+z')$, etc. It is common to note the set $Y\times Z$ endowed with these operations by $Y\oplus Z$, and its elements $(y,z)$ by $y\oplus z$. Using implicitely the previous identification, we get that $y\oplus z=y+z$. Commented Apr 20, 2022 at 16:09

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 $$y \oplus z$$.

• Thanks, but how does this differ from $y + z$? Commented Apr 20, 2022 at 15:56
• The difference between $X=Y+Z$ and $X=Y \oplus Z$ is that in the latter case the expression of any element of $X$ as a sum of an element of $Y$ and $Z$ is unique. I believe the authors are emphasizing this with the notation $y \oplus z$. Commented Apr 20, 2022 at 15:59
• I think one should differentiate between $X = Y \oplus Z$ and $y \oplus z$. While the former is a statement, namely that the sum of the subspaces $Y$ and $Z$ is direct regarding $X$, the latter is a vector. So, if we let $v = y \oplus z$, then $v \in X$. Note that $y + z$ and $y \oplus z$ are the exact same vectors, the difference being only that in the latter notation one emphasises that the sum is direct, i.e. that $X = Y \oplus Z$. Commented Apr 20, 2022 at 16:05

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.