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I have doubts about whether

\begin{align*} \left\lbrace \begin{pmatrix} \frac{-1-i\sqrt{11}}{2} \\ 1 \end{pmatrix}, \begin{pmatrix} \frac{-1+i\sqrt{11}}{2} \\ 1 \end{pmatrix} \right\rbrace \end{align*}

is a basis for $\mathbb{C}^2$. I still do not work with bases for the field of complexes, that is why I have this doubt.

Gabriela
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    Over what field? Also, only have one set, so unclear what you mean by "which": it either is or isn't, there aren't any choices to be made. – Arturo Magidin Aug 19 '21 at 04:28
  • Base for $\mathbb{C}^2$ – Gabriela Aug 19 '21 at 04:31
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    They are two vectors in a 2-dimensional space. So just check for linear dependence by calculating the determinant and conclude. – Jyrki Lahtonen Aug 19 '21 at 04:41
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    Basic facts: In a vector space $V$ over a field $F$, a subset ${v_1,v_2} \subset V$ is linearly independent if and only if doesn't exists $a \in F$ such that $av_1 = v_2$. Also, if the space is finite-dimensional, any linearly independent subset of $V$ with $\dim(V)$ vectors is a basis. Does this help you? – azif00 Aug 19 '21 at 04:43
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    The determinant trick works if you want to show that these two form a basis of $\Bbb{C}^2$ over the field of complex numbers. The space $\Bbb{C}^2$ can also be viewed as a vector space over $\Bbb{R}$. Then it becomes 4-dimensional, and you need four vectors to have a basis over $\Bbb{R}$, such as $\binom 10$, $\binom i 0$, $\binom 01$, $\binom 0i$, to able to write a general vector as a linear combination: $$\binom{a+bi}{c+di}=a\binom10+b\binom i0+c\binom01+d\binom0i$$ with $a,b,c,d$ real. As opposed to $$\binom {z_1}{z_2}=z_1\binom 10+z_2\binom01$$ with $z_1,z_2$ complex. – Jyrki Lahtonen Aug 19 '21 at 04:47
  • (cont'd) I guess you are at a point in your studies when the distinction between vector spaces over different fields is still unfamiliar territory. – Jyrki Lahtonen Aug 19 '21 at 04:51
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    "Base for $\mathbb{C}^2$". Yes, you said that. I asked over what field. It makes a difference. It is two dimensional over $\mathbb{C}$,. but it is four dimensional over $\mathbb{R}$ and infinite dimensional over $\mathbb{Q}$. You need to specify the field.. – Arturo Magidin Aug 19 '21 at 04:59
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    @azif00: Your basic fact needs an addendum: it could be that there is no $a$ with $av_1=v_2$, but there is one with $v_1=av_2$ (if $v_1=0$ but $v_2\neq 0$)... – Arturo Magidin Aug 19 '21 at 05:00

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Firstly, Check the determinant of these vectors, to know about their independence .You can see they are obviously linearly independent. Now for the spanning try yourself.

Bjp Best
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