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Another task from test-exam:

True or false? $S$ is a linear subspace of a real vector space $V$. If $u,v \notin S$, then $u+v \notin S$

I think the statement is wrong because it's said that "$S$ is a linear subspace of a real vector space $V$". But then it's concluded that "if $u,v \notin S$, then $u+v \notin S$", which is a contradiction since $S$ is only a linear subspace of $V$ when $u+v \in S$.

Please tell me if I'm right and if not maybe you can explain shortly? This is from test-exam and we didn't get solutions.. But actually I'm convinced it's correct because I took some kind of definition from reading, well if I understood that correctly too..

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This is obviously false. Consider the plane $z=0$ in $\mathbb{R}^3$, which is a linear subspace of $\mathbb{R}^3$. The vectors $\hat{k}$ and $-\hat{k}$ are not in the plane, but their sum is.

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    Oh ok so really easy thank you, but I didn't see it myself :P But my attempt was right too? – 2017-02-28
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    (You can of course run the same argument for the line $y=0$ in $\mathbb{R}^2$, or for the trivial subspace of $\mathbb{R}$. I just happened to think of three dimensions first, because sometimes it's easier to find counterexamples in three dimensions than in two dimensions: that is a good strategy to have in your pocket for linear algebra.) – 2017-02-28
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    @cnmesr: no, your attempt is not even wrong, as Wolfgang Pauli allegedly said. Your attempt is incomprehensible. The definition of a linear subspace requires that it be closed under linear combinations: *if* $u, v\in S$ *then* $u+v\in S$. The definition says nothing about what happens when $u,v\notin S$. – 2017-02-28
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    Great, thanks :) – 2017-02-28
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    Sure already planned so but that annoying timer.. :p here we go – 2017-02-28
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This is false even in dimension $1$. Consider $V=\mathbb R$, $S=\{0\}$, $u=1$, $v=-1$.

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    Thank you very much too ! Glad I understood now. – 2017-02-28