Is $\left\{\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}0\\0\\0\end{bmatrix}\right\}$ a vector space? this post says


Since U is only two vectors, it is clear that U is not a vector space.

if I add zero-vector into U
$$
U^{'} = 
\{
\begin{bmatrix}
1\\
0\\
0\\
\end{bmatrix}
,
\begin{bmatrix}
0\\
1\\
0\\
\end{bmatrix}
,
\begin{bmatrix}
0\\
0\\
0\\
\end{bmatrix}
\}
\subset R^3
$$
Is $(U^{'}, +)$ a vector space?
 A: Still not.
It isn't closed under addition (or scalar multiplication).  Either one of these would preclude it from being a vector space. 
That is (to consider the addition part): $\begin {pmatrix}1\\0\\0\end{pmatrix}+\begin {pmatrix} 0\\1\\0\end{pmatrix}=\begin {pmatrix}1\\1\\0\end {pmatrix}\not\in U'$.
A: No, $$U'=\left\{\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}0\\0\\0\end{bmatrix}\right\}$$ is not a vector space. And it is not, because this set contains the element $$\begin{bmatrix}1\\0\\0\end{bmatrix}\in U'$$ but it does not contain $$2\cdot\begin{bmatrix}1\\0\\0\end{bmatrix}\notin U'$$
which means it does not satisfy all conditions a vector space should satisfy.
A: No finite subset $F$ of $\mathbb R^3$ other than $\bigl\{\begin{bmatrix}0&0&0\end{bmatrix}^T\bigr\}$ is a vector space, because if $v\in F$ is a non-null vector, then $nv\notin F$ if $n\in\mathbb N$ is large enough.
A: No. Except for the $0$-dimensional space $\lbrace 0 \rbrace$ there is no finite vector space over $\mathbb{R}$.
What you wrote down is not even a vector, but a matrix.
Assuming that you just wanted to add the zero vector, it is still not a vector space as for example $(2,0,0) \not \in U'$, but $(1,0,0) \in U'$. This means that $U'$ Is not closed under scalar multiplication.
