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

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?

• how many do you need for basis? For, $\mathbb{R}^{n}$ you need a minimum of some numbers... – HumbleStudent Jan 8 '17 at 8:04
• @HumbleStudent, I know we need $3$, but that is not a dis-proof. That is the intuitive step, I want to completely disprove it – fasasa Jan 8 '17 at 8:06

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.

• Which theorem is this? Why must one vector be written as a combination of others in the basis? – fasasa Jan 8 '17 at 8:07
• @Gaandmit, see en.wikipedia.org/wiki/Standard_basis for detail of the definition of basis. – Lion Jan 8 '17 at 8:09
• We could also use $(1, 0, -1)$ right? – fasasa Jan 8 '17 at 8:17
• @Gaandmit Yes, it is correct. Actually, I just take the vector that perpendicular to the plane spanned by the given two vectors. – Lion Jan 8 '17 at 8:20

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)$\ne \mathbb{R}^3$.

• So to show that last statement, if we can find one example in $$\mathbb{R}^3$$ that shows that $span(B)$ does not work then we have the proof? – fasasa Jan 8 '17 at 8:09
• @Gaandmit right, just need to find a vector in R_3 that can't be written as a linear combo of the two vectors in B. – spaceisdarkgreen Jan 8 '17 at 8:10

Hint:

For e.g. show that $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ cannot be written using $B$.
• But $(0, 0, 0)$ can be written with all coefficients $=0$ – fasasa Jan 8 '17 at 8:11
• Oh, I meant $[1,1,1]$. Rectified. – 8hantanu Jan 8 '17 at 8:13