Is this an ordered basis

Question

I was just having trouble with this question.

If $S=[e_1-e_2,e_3]$ where $e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)$

Would the list $S$ be an ordered basis of $\Bbb R^3$

Help would really be appreciated on this one guys

Answer 1

As

• $S$ has two elements
• $\mathbb{R}^3$ is three dimensional
• The number of elements in a basis for a vector space is the dimension of the space

$S$ can't possibly be a basis for $\mathbb{R}^3$.

Answer 2

With given

$$S = \left\{ {{e_1} - {e_2},{e_3}} \right\}$$

we try to find a representation

$$u \cdot \left( {{e_1} - {e_2}} \right) + v \cdot {e_3} = \left( {{x_1},{x_2},{x_3}} \right)$$

for any $\left( {{x_1},{x_2},{x_3}} \right) \in {\mathbb{R}^3}$.

That is

$$u \cdot \left( {1, - 1,0} \right) + v \cdot \left( {0,0,1} \right) = \left( {{x_1},{x_2},{x_3}} \right)$$

or

$$\left( {u, - u,v} \right) = \left( {{x_1},{x_2},{x_3}} \right)$$

Comparing, we see

$$\begin{gathered} u = {x_1} \hfill \\ u = - {x_2} \hfill \\ v = {x_3} \hfill \\ \end{gathered}$$

Which means

$${x_2} = - {x_1}$$

So, only vectors with shape

$$\left( {{x_1}, - {x_1},{x_3}} \right) \in {\mathbb{R}^3}$$

can be represented by $S$.

And there are a lot more vectors of ${\mathbb{R}^3}$, which haven't this kind of shape.

$S$ can't be an ordered base.