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This is a notation clarification question: If $S$ and $T$ are subspaces of a vector space $V$, is $S+T$ equivalent to $S\cup T$? Thanks.

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3 Answers 3

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No; for example, if $V=\mathbb{R}^2$ is the plane, and $S=\{(a,0)\in V\mid a\in\mathbb{R}\}$ and $T=\{(0,b)\in V\mid b\in\mathbb{R}\}$ are the $x$-axis and $y$-axis (which are subspaces of $V$), respectively, then $$S+T=\{(a,0)+(0,b)\in V\mid a,b\in\mathbb{R}\}=\{(a,b)\in V\mid a,b\in \mathbb{R}\}=V$$ i.e. $S+T$ is the whole plane, but $$S\cup T=\{(a,0)\in V\mid a\in\mathbb{R}\}\cup\{(0,b)\in V\mid b\in \mathbb{R}\}=\{(a,b)\in V\mid \text{ either }a=0\text{ or }b=0\}$$ is just the union of the $x$-axis and $y$-axis, which is not even a subspace of $V$ (for example, it is not closed under addition: $(1,0)\in S\cup T$ and $(0,1)\in S\cup T$, but $(1,1)\notin S\cup T$).

However, $S+T$ is the subspace spanned by $S\cup T$, i.e. $S+T=\text{span}(S\cup T)$.

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no, $S+T=\{ s+t \colon s \in S, t \in T\}$, $S \cup T = \{ u \colon u \in S \vee u \in T\}$.

Eg. let $A$, $B$ nonempty subsets of linear space $X$, then $\mbox{Lin}(A \cup B) = \mbox{Lin}(A) + \mbox{Lin}(B)$, but $\mbox{Lin}(A +B) \subset \mbox{Lin}(A) + \mbox{Lin}(B)$. Take $A=\{(1,0)\}$, $B=\{(0,1)\}$ in $\mathbb{R}^2$.

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If by equivalent you mean equal, then the union of two subspaces is a subspace if and only if one of the subspaces is contained in the other. If by equivalent you mean that they generate the same subspace, then yes, $S+T = \langle S , T \rangle = \langle S \cup T \rangle$.

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