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Let x = (3, −2).

Now I am supposed to find a vector that it is basis for $\mathbb{R}^2$.

What should I do first?

Also vector (1, 1) of the basis should be given in linear combination form.

MiMaKo
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  • Find a vector $y=(a,b)$ so that the pair of vectors $x,y$ is linearly independent. Do you know how to test whether two vectors are linearly independent or not? – Jimmy R. Oct 18 '17 at 10:28
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    I think so. If the determinant is 0, then it is linearly dependent. Otherwise it is independent, right? – MiMaKo Oct 18 '17 at 10:29
  • Yeah, correct. Once you find such a $y$ (there are many of them), you should find scalars (numbers) $\lambda_1,\lambda_2$ so that $(1,1)=\lambda_1 x+\lambda_2 y$. This is a system of two equations in two unknowns (the $\lambda$'s). – Jimmy R. Oct 18 '17 at 10:30
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    So, basically if the vector is (1, 0) then I got det. 2. So it is linearly independent. So for example (1,0) is a correct answer, right? – MiMaKo Oct 18 '17 at 10:39
  • Yeah, $(1,0)$ is a correct answer. And then $(1,1)=?(3,-2)+?(1,0)$. Solve for the two questionmarks and you are done. Hint: start from the second coordinate (the second $1$ in $(1,1)$). – Jimmy R. Oct 18 '17 at 10:40
  • The general method is to recall the basis extension theorem which provides an algorithm to extend a linearly independent set of vectors to a basis. Consider the matrix having the initial rows as the set of vectors and the rest of the rows being the standard canonical basis of $\Bbb R^n$. Applying Gauss-Jordan elimination should give you the vectors to include in the extension. – Prasun Biswas Oct 18 '17 at 10:45
  • So is it 1 and 1? -> As 13 - 12 = 1 and 11 + 10 = 1 – MiMaKo Oct 18 '17 at 10:45
  • Now, first coordinates go with first coordinates. $1=?3+?1$ and $1=?(-2)+?0$. The second is easier to solve, right? – Jimmy R. Oct 18 '17 at 10:48
  • Oh, I am so puzzled. I just can't get forward anymore. I got -0,5 but I don't know anymore what I am doing – MiMaKo Oct 18 '17 at 11:08

1 Answers1

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Show that $x$ and $y:=(1,1)$ are linearly independent. Then the calculation

$(1,1)= \alpha x+ \beta y$ is very(!) easy !

Fred
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  • 1(3-2) + 1(1+0)

    Is that correct in linear combination form?

    Honestly I don't know how linear combination form should look like...

     – MiMaKo Oct 18 '17 at 12:07
  • We have the equation $(1,1)= \alpha (3,-2)+ \beta (1,1)$. With only one eye one can see that $\alpha =0$ and $\beta=1$ – Fred Oct 18 '17 at 12:16