Show that $M_1$ is not a generating set of $\mathbb R^3$.

Question

I have a set of vectors, $M_1$ which is defined as the following: $$M_1:=[\begin{pmatrix}1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix}0 \\ 1 \\ 1 \end{pmatrix}]$$ I have to show that $M_1$ isn't a generating set of $\mathbb R^3$, even though it's linearly independent. My initial idea was that, because $$\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}≠a\cdot\begin{pmatrix}1 \\ 0 \\ 1 \end{pmatrix}+ b \cdot\begin{pmatrix}0 \\ 1 \\ 1 \end{pmatrix}$$ therefore $M_1$ isn't a generating set of $\mathbb R^3$. However I would like to know if there is any other way to show that $M_1$ is not a generating set of $\mathbb R^3$.

Answer 1

It is not possible that two vectors generate $\mathbb{R}^{3}$, since $dim(\mathbb{R}^{3})=3$ and the dimension of the span of two vectors can be $0$, $1$ or $2$, depending on the vectors you have. In order to find a "$\mathbb{R}^{3}$ generator", you'll have to find three linearly independent vectors.

Answer 2

$M_1$ contains only two vectors. So $\mathrm{Vect}(M_1)$ is at most of dimension $2$. Then it cannot be $\mathbb{R}^3$.

Answer 3

If $M_1$ were a generating set for $\mathbb R^3$ then $\operatorname{dim}\mathbb R^3\le2$, which is false.