1
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $a+b=v\in S$.)

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.