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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 $y \oplus z$?

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    $\begingroup$ 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 $y\oplus z=y+z$. $\endgroup$
    – Balloon
    Commented Apr 20, 2022 at 16:09

2 Answers 2

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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 $y \oplus z$.

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  • $\begingroup$ Thanks, but how does this differ from $y + z$? $\endgroup$
    – R. Rayl
    Commented Apr 20, 2022 at 15:56
  • $\begingroup$ 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 $y \oplus z$. $\endgroup$
    – carraig
    Commented Apr 20, 2022 at 15:59
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    $\begingroup$ 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 $X = Y \oplus Z$. $\endgroup$
    – hm1912
    Commented Apr 20, 2022 at 16:05
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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.

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