the problem is given below:
I can see that the set contain zero vector by saying:
⟨c,0,c⟩=⟨0,0,0⟩
c=0
but how to finde if the set is closed by addition of vectors and closed by multiplication of real-valued scalar ?
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1$\begingroup$ Every span is a subspace, no matter what it is a span of. (This is sometimes part of the definition of span, sometimes an early theorem). $\endgroup$– hmakholm left over MonicaApr 2, 2016 at 19:53
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1$\begingroup$ Both your thoughts and the answer by Emilio seem to assume that the problem says $$ H = \Biggl\{ \begin{bmatrix}c\\0\\c\end{bmatrix} \Bigg\vert\, x\in \mathbb R\Biggr\}$$ instead of what it actually says. If it actually says $$ H = \operatorname{span}\left\{\begin{bmatrix}c\\0\\c\end{bmatrix}\right\} \text{ with }c\in\mathbb R$$ then $H$ is something that depends on $c$ -- and you can't just set $c=0$ to prove that the $H$ you get when $c=2$ is a subspace ... $\endgroup$– hmakholm left over MonicaApr 2, 2016 at 19:55
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1$\begingroup$ x @AdiT: What is your definition of $\operatorname{span}$? $\endgroup$– hmakholm left over MonicaApr 2, 2016 at 20:02
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1$\begingroup$ x @AdiT: Then you can always get the zero vector as the empty linear combination -- or, if that unsettles you, as $0\cdot v_1$. $\endgroup$– hmakholm left over MonicaApr 2, 2016 at 20:05
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1$\begingroup$ x @AdiT: If you have two linear combinations, the distributive law will allow you to rearrange their sum so it end up being a single linear combination. $\endgroup$– hmakholm left over MonicaApr 2, 2016 at 20:20
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1 Answer
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Hint:
By definition the span of a set of vectors is the intersection of all the subspaces that contain the set, so it is a subspace.
If you want to prove that all vectors of the form $[c,0,c]^T$ are a subspace than take two vectors in $H$ as: $[a,0,a]^T$ and $[b,0,b]^T$, than:
$$ [a,0,a]^T+k[b,0,b]^T=[a+kb,0,a+kb]^T $$
answered Apr 2, 2016 at 19:50
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$\begingroup$ The comments of@Henning point out a wery important point so I've modified my answer. $\endgroup$ Apr 2, 2016 at 21:25