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October 18th, 2017, 07:27 AM   #1
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bases of the reals^2

question: find 3 different bases of R^2

My answer: I thought for R^2 there are only two standard base vectors e1,e2 where e1=(0,1) e2=(1,0) which spans every vector in R^2 and can be written as a unique linear combination. So what is the 3rd bases of R^2 or could the question be wrong?? Thanks
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October 18th, 2017, 07:45 AM   #2
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Hi !

A basis of $\displaystyle {\mathbb R}^2$ is a set of two vectors that span all the vectors of $\displaystyle {\mathbb R}^2$. The famous basis is what you mentioned, which is called the canonical basis
$\displaystyle B=(e_1,e_2)$ where $\displaystyle e_1=(1,0)$ and $\displaystyle e_2=(0,1)$. But this is only a single basis - although it has two vectors.

Did you get it ?

Now for answering your question, we can use this property:
Let $\displaystyle w=(w_1,w_2)$ and $\displaystyle v=(v_1,v_2)$, then $\displaystyle (w,v)$ is a basis of $\displaystyle {\mathbb R}^2$ iff the determinant of $\displaystyle (w,v)$ is not null.
That means:
$\displaystyle
\begin{vmatrix}
w_1 &v_1 \\
w_2 &v_2
\end{vmatrix}\ne 0
$.

So, all that you need is finding 3 pairs of vectors that satisfy that condition.

Sorry for bad English!
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Last edited by skipjack; October 19th, 2017 at 04:13 PM.
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October 18th, 2017, 09:36 AM   #3
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Note that finding such vectors is straightforward in $\mathbb R^2$. Any pair of non-collinear vectors will do.
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Last edited by skipjack; October 19th, 2017 at 04:15 PM.
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October 18th, 2017, 12:13 PM   #4
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Ahhh, I see. I didn't realise that b=(e1,e2) only forms one single basis in R^2,

so would the second basis vectors be, for example, w=(1,1) v=(1,-1)

and the third bases vectors be w=(2,2) v=(2,-2) ??

Would this work since the det(w*v) is not equal to 0?

Thanks again!

Last edited by skipjack; October 19th, 2017 at 04:16 PM.
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October 18th, 2017, 12:48 PM   #5
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Yes, both are bases, but your teacher might not like you quoting those two pairs as they are just scalar multiples of each other. I'd find a third pair.

Not that your pair don't have to be at right-angles, just not collinear.
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Last edited by skipjack; October 19th, 2017 at 04:16 PM.
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October 18th, 2017, 08:15 PM   #6
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By definition, a basis for a space is any spanning set of vectors which is linearly independent. I would also hesitate to ${\bf define}$ a basis in terms of determinants. The "right" way to look for a basis is to iteratively build a set of vectors by adding vectors to your set which don't lie in the span of the current vectors in your set.

The problem with thinking of things in terms of determinants is

1. It is not a useful computational tool.

2. It does not give any insight into what a basis is.

3. Determinants only make sense for square matrices. However, non-square matrices show up often in math and questions about their important subspaces (kernel and image) are common. In this case, determinants aren't even defined, but the concept of a basis is.

For example, the following 3 vectors
$(1,1,1,0), (1,1,0,1),(1,0,1,1)$ span a 3-dimensional subspace of $\mathbb{R}^4$, hence they are a basis for this subspace. But you can't compute their determinant.
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Last edited by skipjack; October 19th, 2017 at 04:18 PM.
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