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If $a\in S$ where $S$ is a subspace of $V$ and $b\in V$ \ $S$, does that necessarily mean that $a+b \not\in S$? I think it is true... Thanks.

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2 Answers 2

up vote 5 down vote accepted

Yes, it's true.

Because if $a\in S$ and $a+b\in S$, then $(a+b)-a = b\in S$. By contrapositive, if $a\in S$ and $b\notin S$, then $a+b\notin S$.

(Note, however, that the it is a common error for beginners to think that by a similar argument it would follow that "if $a\notin S$ and $b\notin S$, then $a+b\notin S$"; this is not true, since you can start with a $v\in S$, $x\notin S$, and take $a=x$ and $b=v-x$. Then $a,b\notin S$, but $a+b=v\in S$.)

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Thanks, esp for the note! – Dave Sep 10 '11 at 23:04

If $a\in S$ ($-a\in S$) and $a+b\in S$ then $a+b+(-a)\in S$ which is $b\in S$...

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