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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

Elena
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  • Yes, It's a basis written in a particular order. – Ethan Bolker Mar 15 '17 at 14:11
  • Hey would it be possible to show me why it would just really help when I try other questions – Elena Mar 15 '17 at 14:19
  • All I did was read the words "basis" and "ordered" and see that they described what you wrote down. Reading the words (then in this case going back to the definitions) is always a good problem solving strategy. – Ethan Bolker Mar 15 '17 at 14:23
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    As written $S$ has two elements: $e_1-e_2$ and $e_3$. is that what you intended? – lulu Mar 15 '17 at 14:31
  • OOPS I didn't read carefully. @lulu 's comment is crucial. My answer is right if the $-$ sign is a typo, Hurkyl 's is right if the $-$ is what you meant. So the moral is still "read carefully". – Ethan Bolker Mar 15 '17 at 14:36
  • Yes I just meant for S=[e1-e2,e3] I was just confused about it only having 2 elements and didn't know how to prove linear independence – Elena Mar 15 '17 at 14:40
  • The edit by @Dave changes the meaning, since ${ e_1 - e_2, e_3 }$ surely means a set, which is unordered. I've changed that back to square brackets. –  Mar 15 '17 at 14:52

2 Answers2

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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 $\mathbb{R}^3$.

  • How about for T=[2e1-e2 , e2+e3 , 2e1 +e3] I got this to be an ordered basis of R^3 because the list of vectors it forms are linearly independent, am I right in thinking so? – Elena Mar 15 '17 at 14:43
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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.

Frieder
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