Proving that the set of vectors is not basis for R^3

Question

let $B = \begin{Bmatrix} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix} \end{Bmatrix} $, show that $B$ is not a basis for $\mathbb{R}^3$.

From the definition of a basis, we must have $\text{span} \space \{ B\} = S \subseteq \mathbb{R}^n$ and that $B$ is linearly independent.

Fact: It is true that $B$ is a linearly independent vector set, so we must disprove the first part of the definition.

So our goal is to disprove that $\text{span} \space \{B\} \ne S = \mathbb{R}^3$?

So in our case it is true that $S = \mathbb{R}^3$ right?

Answer 1

It suffices to find a vector in $\mathbb{R}^3$ such that cannot be represented as a linear combination of the given basis.

To this end, let us take $a=(-1,3,1)^\top$ and this vector will do the job.

Answer 2

I'm not positive about your notation/definition, but for $B$ to be a basis for $\mathbb{R}^3$, must have Span(B) = $\mathbb{R}^3$ and $B$ linearly independent. Since the vectors are LI, yes, your goal is to show Span(B)$ \ne \mathbb{R}^3$.

Answer 3

Hint:

Proof by contradiction:

For e.g. show that $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ cannot be written using $B$.