What does it mean for something to span $Z^n$? For example: do the vectors x,y, and z form a basis for $Z^2$?












0












$begingroup$


The question asks: do the vectors $[1,1,0], [0,1,1], [1,0,1]$ form a basis for $Z^2$?



I know how to check whether vectors form a basis normally: see if an arbitrary vector in $R^n$ can be written as a linear combination of the vectors (ex. $[1,1,0], [0,1,1], [1,0,1]$ forms a basis for $R^3$ because the matrix with those three as rows has a solution, meaning any vector can be written as a linear combination of them).



But how do I apply that idea to $Z$? I simplified the matrix under $Z$ and found that one of the three vectors was a combination of the other two, but I'm not sure how that really helps or how to apply that information.






linear-algebra modular-arithmetic







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  • 1




    $begingroup$
    $(1,1,0)notin Bbb{Z}^2$, but $(1,1,0)in Bbb{Z}^3$.
    $endgroup$
    – Dietrich Burde
    Dec 16 '18 at 19:25


















0












$begingroup$


The question asks: do the vectors $[1,1,0], [0,1,1], [1,0,1]$ form a basis for $Z^2$?



I know how to check whether vectors form a basis normally: see if an arbitrary vector in $R^n$ can be written as a linear combination of the vectors (ex. $[1,1,0], [0,1,1], [1,0,1]$ forms a basis for $R^3$ because the matrix with those three as rows has a solution, meaning any vector can be written as a linear combination of them).



But how do I apply that idea to $Z$? I simplified the matrix under $Z$ and found that one of the three vectors was a combination of the other two, but I'm not sure how that really helps or how to apply that information.






linear-algebra modular-arithmetic







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $(1,1,0)notin Bbb{Z}^2$, but $(1,1,0)in Bbb{Z}^3$.
    $endgroup$
    – Dietrich Burde
    Dec 16 '18 at 19:25
















0












0








0





$begingroup$


The question asks: do the vectors $[1,1,0], [0,1,1], [1,0,1]$ form a basis for $Z^2$?



I know how to check whether vectors form a basis normally: see if an arbitrary vector in $R^n$ can be written as a linear combination of the vectors (ex. $[1,1,0], [0,1,1], [1,0,1]$ forms a basis for $R^3$ because the matrix with those three as rows has a solution, meaning any vector can be written as a linear combination of them).



But how do I apply that idea to $Z$? I simplified the matrix under $Z$ and found that one of the three vectors was a combination of the other two, but I'm not sure how that really helps or how to apply that information.






linear-algebra modular-arithmetic







share|cite|improve this question









$endgroup$




The question asks: do the vectors $[1,1,0], [0,1,1], [1,0,1]$ form a basis for $Z^2$?



I know how to check whether vectors form a basis normally: see if an arbitrary vector in $R^n$ can be written as a linear combination of the vectors (ex. $[1,1,0], [0,1,1], [1,0,1]$ forms a basis for $R^3$ because the matrix with those three as rows has a solution, meaning any vector can be written as a linear combination of them).



But how do I apply that idea to $Z$? I simplified the matrix under $Z$ and found that one of the three vectors was a combination of the other two, but I'm not sure how that really helps or how to apply that information.






linear-algebra modular-arithmetic




linear-algebra modular-arithmetic






share|cite|improve this question













share|cite|improve this question











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asked Dec 16 '18 at 19:22









James RonaldJames Ronald

1807




1807








  • 1




    $begingroup$
    $(1,1,0)notin Bbb{Z}^2$, but $(1,1,0)in Bbb{Z}^3$.
    $endgroup$
    – Dietrich Burde
    Dec 16 '18 at 19:25
















  • 1




    $begingroup$
    $(1,1,0)notin Bbb{Z}^2$, but $(1,1,0)in Bbb{Z}^3$.
    $endgroup$
    – Dietrich Burde
    Dec 16 '18 at 19:25










1




1




$begingroup$
$(1,1,0)notin Bbb{Z}^2$, but $(1,1,0)in Bbb{Z}^3$.
$endgroup$
– Dietrich Burde
Dec 16 '18 at 19:25






$begingroup$
$(1,1,0)notin Bbb{Z}^2$, but $(1,1,0)in Bbb{Z}^3$.
$endgroup$
– Dietrich Burde
Dec 16 '18 at 19:25












1 Answer
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$begingroup$

The definition of a basis is the same as before. The coefficients are now in $Bbb{Z}$ instead of a field $K$. So ${e_1,ldots ,e_n}$ is a basis of $Bbb{Z}^n$ if $a_1e_1+cdots +a_ne_n=0$ implies that all $a_i=0$, and the integral linear combinations of $e_1,ldots ,e_n$ give $Bbb{Z}^n$. This is the case if and only if the associated matrix $A$ is in the group $GL_n(Bbb{Z})$. Note that $det(A)=pm 1$ then.






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    1 Answer
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    $begingroup$

    The definition of a basis is the same as before. The coefficients are now in $Bbb{Z}$ instead of a field $K$. So ${e_1,ldots ,e_n}$ is a basis of $Bbb{Z}^n$ if $a_1e_1+cdots +a_ne_n=0$ implies that all $a_i=0$, and the integral linear combinations of $e_1,ldots ,e_n$ give $Bbb{Z}^n$. This is the case if and only if the associated matrix $A$ is in the group $GL_n(Bbb{Z})$. Note that $det(A)=pm 1$ then.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The definition of a basis is the same as before. The coefficients are now in $Bbb{Z}$ instead of a field $K$. So ${e_1,ldots ,e_n}$ is a basis of $Bbb{Z}^n$ if $a_1e_1+cdots +a_ne_n=0$ implies that all $a_i=0$, and the integral linear combinations of $e_1,ldots ,e_n$ give $Bbb{Z}^n$. This is the case if and only if the associated matrix $A$ is in the group $GL_n(Bbb{Z})$. Note that $det(A)=pm 1$ then.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The definition of a basis is the same as before. The coefficients are now in $Bbb{Z}$ instead of a field $K$. So ${e_1,ldots ,e_n}$ is a basis of $Bbb{Z}^n$ if $a_1e_1+cdots +a_ne_n=0$ implies that all $a_i=0$, and the integral linear combinations of $e_1,ldots ,e_n$ give $Bbb{Z}^n$. This is the case if and only if the associated matrix $A$ is in the group $GL_n(Bbb{Z})$. Note that $det(A)=pm 1$ then.






        share|cite|improve this answer









        $endgroup$



        The definition of a basis is the same as before. The coefficients are now in $Bbb{Z}$ instead of a field $K$. So ${e_1,ldots ,e_n}$ is a basis of $Bbb{Z}^n$ if $a_1e_1+cdots +a_ne_n=0$ implies that all $a_i=0$, and the integral linear combinations of $e_1,ldots ,e_n$ give $Bbb{Z}^n$. This is the case if and only if the associated matrix $A$ is in the group $GL_n(Bbb{Z})$. Note that $det(A)=pm 1$ then.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 19:27









        Dietrich BurdeDietrich Burde

        80.6k647104




        80.6k647104






























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