What does it mean for something to span $Z^n$? For example: do the vectors x,y, and z form a basis for $Z^2$?
$begingroup$
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
asked Dec 16 '18 at 19:22
James RonaldJames Ronald
1807
$endgroup$
add a comment |
$begingroup$
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
asked Dec 16 '18 at 19:22
James RonaldJames Ronald
1807
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 16 '18 at 19:22
James RonaldJames Ronald
1807
$endgroup$
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
asked Dec 16 '18 at 19:22
James RonaldJames Ronald
1807
asked Dec 16 '18 at 19:22
James RonaldJames Ronald
1807
asked Dec 16 '18 at 19:22
James RonaldJames Ronald
1807
asked Dec 16 '18 at 19:22
James RonaldJames Ronald
1807
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 16 '18 at 19:27
Dietrich BurdeDietrich Burde
80.6k647104
$endgroup$
add a comment |
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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.
answered Dec 16 '18 at 19:27
Dietrich BurdeDietrich Burde
80.6k647104
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 16 '18 at 19:27
Dietrich BurdeDietrich Burde
80.6k647104
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 16 '18 at 19:27
Dietrich BurdeDietrich Burde
80.6k647104
$endgroup$
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.
answered Dec 16 '18 at 19:27
Dietrich BurdeDietrich Burde
80.6k647104
answered Dec 16 '18 at 19:27
Dietrich BurdeDietrich Burde
80.6k647104
answered Dec 16 '18 at 19:27
Dietrich BurdeDietrich Burde
80.6k647104
answered Dec 16 '18 at 19:27
Dietrich BurdeDietrich Burde
80.6k647104
80.6k647104
add a comment |
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$begingroup$
$(1,1,0)notin Bbb{Z}^2$, but $(1,1,0)in Bbb{Z}^3$.
$endgroup$
– Dietrich Burde
Dec 16 '18 at 19:25