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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?

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

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

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  • $\begingroup$ 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? $\endgroup$ – fasasa Jan 8 '17 at 8:09
  • $\begingroup$ @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. $\endgroup$ – spaceisdarkgreen Jan 8 '17 at 8:10
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Hint:

Proof by contradiction:

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

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  • $\begingroup$ But $(0, 0, 0)$ can be written with all coefficients $=0$ $\endgroup$ – fasasa Jan 8 '17 at 8:11
  • $\begingroup$ Oh, I meant $[1,1,1]$. Rectified. $\endgroup$ – 8hantanu Jan 8 '17 at 8:13

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