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Show that the list $(1,2), (3,5)$ is a basis for $\mathbb R^2$.

In order to show it is a basis, I have show that the list is linearly independent and that the list spans $\mathbb R^2$.

So I understand how to show linear independence (you simply set up a system of equations). However, I am uncertain as to how to show that it spans $\mathbb R^2$. How can you be certain that just by using $(1,2)$ and $(3,5)$ you will be able to produce all $\mathbb R^2$?

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  • $\begingroup$ It depend on what is $F^2$. For $\mathbb R^2$, notice that $span((1,2),(3,5))=span(e_1,e_2)=\mathbb R^2$ where $e_1=(1,0)$ and $e_2=(0,1)$. Now, $\mathbb R^2=span(e_1,e_2)$ mean that if $x\in\mathbb R^2$, there is $\alpha ,\beta \in\mathbb R$ s.t. $x=\alpha e_1+\beta e_2$. $\endgroup$ – Surb Jan 28 '16 at 10:10
  • $\begingroup$ How do you know that span((1,2),(3,5)) = span((1,0), (0,1))? $\endgroup$ – El Fufu Jan 28 '16 at 10:15
  • $\begingroup$ because $(1,2)=e_1+2e_2$ and $(3,5)=3e_1+5e_2$ therefore $span((1,2),(3,5))\subset span(e_1,e_2)$. You can prove as well the other inclusion. $\endgroup$ – Surb Jan 28 '16 at 10:17
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To show that $(1,2),\ (3,5)$ span $\mathbb R^2$, you need to show that given $(a,b)\in \mathbb R^2$, we can find $\alpha,\beta\in\mathbb R$ such that $$(a,b) =\alpha(1,2)+\beta(3,5)$$ or equivalently, we need to solve the simultaneous equations $$a = \alpha + 3\beta\\ b=2\alpha + 5\beta.$$ One can manually show that these simultaneous equations always have solutions for any $(a,b)$, either by solving as usual, or by using the fact that the matrix $$\begin{pmatrix}1&3\\2&5\end{pmatrix}$$is invertible.

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The determinant of $$\begin{pmatrix}1&3\\2&5\end{pmatrix}$$ is nonzero over any field $F$. Hence the vectors are a basis for $F^2$. In particular this is true for $F=\mathbb{R}$.

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  • $\begingroup$ That was my edit. Given that the OP had asked about $\mathbb R^2$ in the last line and not specified what $\mathbb F$ was, I felt that the question would be cleaner if it was just about $\mathbb R$. Feel free to roll back if you disagree. $\endgroup$ – Mathmo123 Jan 28 '16 at 10:22
  • $\begingroup$ @Mathmo123 You are right; in this context the OP probably means $\mathbb{R}^2$. $\endgroup$ – Dietrich Burde Jan 28 '16 at 10:26
  • $\begingroup$ I actually did mean F^2, where F is supposed to be R (real-values) and C(complex-values). $\endgroup$ – El Fufu Jan 28 '16 at 10:30
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A linearly independent list with the same cardinality as a basis is a spanning set, so a basis. (Assuming finite dimensional spaces, of course.)

This is because every linearly independent list can be extended to a basis; so if yours wasn't a basis, you'd end up with a basis of $R^2$ with more than two elements.

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  • $\begingroup$ Whilst this answers the question, given that the OP has stated that they don't really understand what spanning means, I'm not sure how useful this is. $\endgroup$ – Mathmo123 Jan 28 '16 at 10:11
  • $\begingroup$ @Mathmo123 The OP says to be uncertain as to “show that it spans $F^2$”, not about the meaning of “spanning”. $\endgroup$ – egreg Jan 28 '16 at 10:12
  • $\begingroup$ Nonetheless, the result you are quoting is a very powerful result. It's overkill to use it in the 2-dimensional case. If the OP doesn't yet understand how to manually show that $(1,2),\ (3,5)$ generate $\mathbb R^2$, then I doubt they would understand the proof of this theorem. $\endgroup$ – Mathmo123 Jan 28 '16 at 10:16
  • $\begingroup$ @Mathmo123 Give a man a fish… $\endgroup$ – egreg Jan 28 '16 at 10:18
  • $\begingroup$ Just out of curiosity ( I am not native english speaking), is the fish referring to this proverb ? $\endgroup$ – Dietrich Burde Jan 28 '16 at 10:48
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There's an easier way to show that it is a basis of $R^2$.

There's a sentence that states that if you can show a a group of vectors are linearly independent, and the dimension of the group is equal to the dimension of the space, then that group is a basis for the space.

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  • $\begingroup$ You have to be careful with the term "group", as it means something else in group theory. Also you should try to make your statement more precise. $\endgroup$ – Dietrich Burde Jan 28 '16 at 10:43

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