# Determine that H is a subspace of R^3

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 ?

• 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). Apr 2, 2016 at 19:53
• 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 ... Apr 2, 2016 at 19:55
• x @AdiT: What is your definition of $\operatorname{span}$? Apr 2, 2016 at 20:02
• x @AdiT: Then you can always get the zero vector as the empty linear combination -- or, if that unsettles you, as $0\cdot v_1$. Apr 2, 2016 at 20:05
• 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. Apr 2, 2016 at 20:20

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:
$$[a,0,a]^T+k[b,0,b]^T=[a+kb,0,a+kb]^T$$