General formula to find number of Bases in $(mathbb{Z}/2mathbb{Z})^n$












3












$begingroup$


If we consider $(mathbb{Z}/2mathbb{Z})^2$ (n=2) as a vector space, we get the Bases:



B1 = {$begin{pmatrix}1\0end{pmatrix},begin{pmatrix}0\1end{pmatrix}$}
B2 = {$begin{pmatrix}1\0end{pmatrix},begin{pmatrix}1\1end{pmatrix}$}
B3 = {$begin{pmatrix}0\1end{pmatrix},begin{pmatrix}1\1end{pmatrix}$}



So we get 3 Bases



For n=3 I came up with the following calculation: $${7*6*4over 3!}= 28$$
There are $2^3$ possible vectors in $(mathbb{Z}/2mathbb{Z})^3$ and the Bases must have 3
linearly independent vectors.



A Basis does not include the $0$ vector so 7*(number of possible vectors left (6))*(number of vectors left that aren't a linear combination of the first two (4)) and we divide that by 3! since we don't care about the order in which the vectors are chosen.



So we get 28 Bases



My questions are: Is this correct so far? and does $$prod_{k=0}^{n-1}{(2^n-2^k)over (k+1)}$$



work as a possible general formula for any n? (I don't know how to show this)



It would be much appreciated if you could help me show this a little bit more rigorously.






linear-algebra cyclic-groups







share|cite|improve this question









$endgroup$












  • $begingroup$
    You might find this relevant: en.wikipedia.org/wiki/General_linear_group
    $endgroup$
    – Randall
    Dec 13 '18 at 4:20






  • 1




    $begingroup$
    take a look at this.
    $endgroup$
    – dezdichado
    Dec 13 '18 at 4:54
















3












$begingroup$


If we consider $(mathbb{Z}/2mathbb{Z})^2$ (n=2) as a vector space, we get the Bases:



B1 = {$begin{pmatrix}1\0end{pmatrix},begin{pmatrix}0\1end{pmatrix}$}
B2 = {$begin{pmatrix}1\0end{pmatrix},begin{pmatrix}1\1end{pmatrix}$}
B3 = {$begin{pmatrix}0\1end{pmatrix},begin{pmatrix}1\1end{pmatrix}$}



So we get 3 Bases



For n=3 I came up with the following calculation: $${7*6*4over 3!}= 28$$
There are $2^3$ possible vectors in $(mathbb{Z}/2mathbb{Z})^3$ and the Bases must have 3
linearly independent vectors.



A Basis does not include the $0$ vector so 7*(number of possible vectors left (6))*(number of vectors left that aren't a linear combination of the first two (4)) and we divide that by 3! since we don't care about the order in which the vectors are chosen.



So we get 28 Bases



My questions are: Is this correct so far? and does $$prod_{k=0}^{n-1}{(2^n-2^k)over (k+1)}$$



work as a possible general formula for any n? (I don't know how to show this)



It would be much appreciated if you could help me show this a little bit more rigorously.






linear-algebra cyclic-groups







share|cite|improve this question









$endgroup$












  • $begingroup$
    You might find this relevant: en.wikipedia.org/wiki/General_linear_group
    $endgroup$
    – Randall
    Dec 13 '18 at 4:20






  • 1




    $begingroup$
    take a look at this.
    $endgroup$
    – dezdichado
    Dec 13 '18 at 4:54














3












3








3





$begingroup$


If we consider $(mathbb{Z}/2mathbb{Z})^2$ (n=2) as a vector space, we get the Bases:



B1 = {$begin{pmatrix}1\0end{pmatrix},begin{pmatrix}0\1end{pmatrix}$}
B2 = {$begin{pmatrix}1\0end{pmatrix},begin{pmatrix}1\1end{pmatrix}$}
B3 = {$begin{pmatrix}0\1end{pmatrix},begin{pmatrix}1\1end{pmatrix}$}



So we get 3 Bases



For n=3 I came up with the following calculation: $${7*6*4over 3!}= 28$$
There are $2^3$ possible vectors in $(mathbb{Z}/2mathbb{Z})^3$ and the Bases must have 3
linearly independent vectors.



A Basis does not include the $0$ vector so 7*(number of possible vectors left (6))*(number of vectors left that aren't a linear combination of the first two (4)) and we divide that by 3! since we don't care about the order in which the vectors are chosen.



So we get 28 Bases



My questions are: Is this correct so far? and does $$prod_{k=0}^{n-1}{(2^n-2^k)over (k+1)}$$



work as a possible general formula for any n? (I don't know how to show this)



It would be much appreciated if you could help me show this a little bit more rigorously.






linear-algebra cyclic-groups







share|cite|improve this question









$endgroup$




If we consider $(mathbb{Z}/2mathbb{Z})^2$ (n=2) as a vector space, we get the Bases:



B1 = {$begin{pmatrix}1\0end{pmatrix},begin{pmatrix}0\1end{pmatrix}$}
B2 = {$begin{pmatrix}1\0end{pmatrix},begin{pmatrix}1\1end{pmatrix}$}
B3 = {$begin{pmatrix}0\1end{pmatrix},begin{pmatrix}1\1end{pmatrix}$}



So we get 3 Bases



For n=3 I came up with the following calculation: $${7*6*4over 3!}= 28$$
There are $2^3$ possible vectors in $(mathbb{Z}/2mathbb{Z})^3$ and the Bases must have 3
linearly independent vectors.



A Basis does not include the $0$ vector so 7*(number of possible vectors left (6))*(number of vectors left that aren't a linear combination of the first two (4)) and we divide that by 3! since we don't care about the order in which the vectors are chosen.



So we get 28 Bases



My questions are: Is this correct so far? and does $$prod_{k=0}^{n-1}{(2^n-2^k)over (k+1)}$$



work as a possible general formula for any n? (I don't know how to show this)



It would be much appreciated if you could help me show this a little bit more rigorously.






linear-algebra cyclic-groups




linear-algebra cyclic-groups






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 4:18









Nick T. WolffNick T. Wolff

161




161












  • $begingroup$
    You might find this relevant: en.wikipedia.org/wiki/General_linear_group
    $endgroup$
    – Randall
    Dec 13 '18 at 4:20






  • 1




    $begingroup$
    take a look at this.
    $endgroup$
    – dezdichado
    Dec 13 '18 at 4:54


















  • $begingroup$
    You might find this relevant: en.wikipedia.org/wiki/General_linear_group
    $endgroup$
    – Randall
    Dec 13 '18 at 4:20






  • 1




    $begingroup$
    take a look at this.
    $endgroup$
    – dezdichado
    Dec 13 '18 at 4:54
















$begingroup$
You might find this relevant: en.wikipedia.org/wiki/General_linear_group
$endgroup$
– Randall
Dec 13 '18 at 4:20




$begingroup$
You might find this relevant: en.wikipedia.org/wiki/General_linear_group
$endgroup$
– Randall
Dec 13 '18 at 4:20




1




1




$begingroup$
take a look at this.
$endgroup$
– dezdichado
Dec 13 '18 at 4:54




$begingroup$
take a look at this.
$endgroup$
– dezdichado
Dec 13 '18 at 4:54










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