Mixing
What is a basis of a vector space over the smallest finite field
$begingroup$
Consider a linear map $L:mathbb{F}_{2}^{2}rightarrow mathbb{F}_{2}^{3}$ where $mathbb{F}_{2}={0,1}$ (the finite field with two elements).
$L$ is such that when defined over $mathbb{R}$,i.e. $L:mathbb{R}^{2}rightarrow mathbb{R}^{3}$, it transforms the natural basis of $mathbb{R}^{2}$ like $L(1,0)=(1,1,3),L(0,1)=(-1,1,1)$.
How does $L$ transform the natural basis of $mathbb{F}_{2}^{2}$?
In order to answer this, I need to know what the natural basis of $mathbb{F}_{2}^{2}$ is. Lets assume it is the same as $mathbb{R}^{2}$ (although it is not clear to me why).
Then how do I find $L(1,0),L(0,1)$? Since $L(1,0)=(1,1,3)$ in $mathbb{R}$ do I convert the 3 to 1 and say $L(1,0)=(1,1,1)$?
Since $L(0,1)=(-1,1,1)$ in $mathbb{R}$ do I convert the $-1$ to $1$ and say $L(0,1)=(1,1,1)$?
linear-algebra abstract-algebra
asked Feb 1 at 2:25
user2175783user2175783
2007
$endgroup$
|
show 8 more comments
$begingroup$
Consider a linear map $L:mathbb{F}_{2}^{2}rightarrow mathbb{F}_{2}^{3}$ where $mathbb{F}_{2}={0,1}$ (the finite field with two elements).
$L$ is such that when defined over $mathbb{R}$,i.e. $L:mathbb{R}^{2}rightarrow mathbb{R}^{3}$, it transforms the natural basis of $mathbb{R}^{2}$ like $L(1,0)=(1,1,3),L(0,1)=(-1,1,1)$.
How does $L$ transform the natural basis of $mathbb{F}_{2}^{2}$?
In order to answer this, I need to know what the natural basis of $mathbb{F}_{2}^{2}$ is. Lets assume it is the same as $mathbb{R}^{2}$ (although it is not clear to me why).
Then how do I find $L(1,0),L(0,1)$? Since $L(1,0)=(1,1,3)$ in $mathbb{R}$ do I convert the 3 to 1 and say $L(1,0)=(1,1,1)$?
Since $L(0,1)=(-1,1,1)$ in $mathbb{R}$ do I convert the $-1$ to $1$ and say $L(0,1)=(1,1,1)$?
linear-algebra abstract-algebra
asked Feb 1 at 2:25
user2175783user2175783
2007
$endgroup$
1
$begingroup$
What does "when defined over $Bbb R$" mean? Given that $L$ maps $Bbb F_2^2$ to $Bbb F_2^3$ it does not map $Bbb R^2$ to $Bbb R^3$...
$endgroup$
– David C. Ullrich
Feb 1 at 13:40
1
$begingroup$
I read your post, so I know you mean "when $L:Bbb R^2toBbb R^3$". That makes no sense! Because given that $L:Bbb F_2^2toBbb F_2^3$ it is simply not true that $f:Bbb R^2toBbb R^3$.
$endgroup$
– David C. Ullrich
Feb 1 at 14:10
1
$begingroup$
@DavidC.Ullrich I am completely lost in what you are trying to say. The question gives $L:mathbb{R}^{2}rightarrow mathbb{R}^{3}$ and asks what is $L:mathbb{F}_{2}^{2}rightarrowmathbb{F}_{2}^{3}$, i.e. when we replace $mathbb{R}$ by $mathbb{F}_{2}$. Thats all. I am not a mathematician so I might be missing some subtlety that is obvious (but not to me).
$endgroup$
– user2175783
Feb 1 at 14:17
1
$begingroup$
The question as you wrote it does not give $L:Bbb R^2toBbb R^3$. If we do assume that $L:Bbb R^2toBbb R^3$ then asking what is $L:Bbb F_2^2toBbb F_2^3$ makes no sense - there's no such thhing as just "replacing $Bbb R$ by $Bbb F_2$" here.
$endgroup$
– David C. Ullrich
Feb 1 at 14:20
1
$begingroup$
@DavidC.Ullrich The question shows how $L$ transforms a basis of $mathbb{R}^{2}$ and hence any vector in $mathbb{R}^{2}$. I imagine this is enough to answer how $L$ transforms the basis of $mathbb{F}^{2}_{2}$ but I dont know how.
$endgroup$
– user2175783
Feb 1 at 14:26
|
show 8 more comments
$begingroup$
Consider a linear map $L:mathbb{F}_{2}^{2}rightarrow mathbb{F}_{2}^{3}$ where $mathbb{F}_{2}={0,1}$ (the finite field with two elements).
$L$ is such that when defined over $mathbb{R}$,i.e. $L:mathbb{R}^{2}rightarrow mathbb{R}^{3}$, it transforms the natural basis of $mathbb{R}^{2}$ like $L(1,0)=(1,1,3),L(0,1)=(-1,1,1)$.
How does $L$ transform the natural basis of $mathbb{F}_{2}^{2}$?
In order to answer this, I need to know what the natural basis of $mathbb{F}_{2}^{2}$ is. Lets assume it is the same as $mathbb{R}^{2}$ (although it is not clear to me why).
Then how do I find $L(1,0),L(0,1)$? Since $L(1,0)=(1,1,3)$ in $mathbb{R}$ do I convert the 3 to 1 and say $L(1,0)=(1,1,1)$?
Since $L(0,1)=(-1,1,1)$ in $mathbb{R}$ do I convert the $-1$ to $1$ and say $L(0,1)=(1,1,1)$?
linear-algebra abstract-algebra
asked Feb 1 at 2:25
user2175783user2175783
2007
$endgroup$
Consider a linear map $L:mathbb{F}_{2}^{2}rightarrow mathbb{F}_{2}^{3}$ where $mathbb{F}_{2}={0,1}$ (the finite field with two elements).
$L$ is such that when defined over $mathbb{R}$,i.e. $L:mathbb{R}^{2}rightarrow mathbb{R}^{3}$, it transforms the natural basis of $mathbb{R}^{2}$ like $L(1,0)=(1,1,3),L(0,1)=(-1,1,1)$.
How does $L$ transform the natural basis of $mathbb{F}_{2}^{2}$?
In order to answer this, I need to know what the natural basis of $mathbb{F}_{2}^{2}$ is. Lets assume it is the same as $mathbb{R}^{2}$ (although it is not clear to me why).
Then how do I find $L(1,0),L(0,1)$? Since $L(1,0)=(1,1,3)$ in $mathbb{R}$ do I convert the 3 to 1 and say $L(1,0)=(1,1,1)$?
Since $L(0,1)=(-1,1,1)$ in $mathbb{R}$ do I convert the $-1$ to $1$ and say $L(0,1)=(1,1,1)$?
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Feb 1 at 2:25
user2175783user2175783
2007
asked Feb 1 at 2:25
user2175783user2175783
2007
edited Feb 1 at 13:02
user2175783
asked Feb 1 at 2:25
user2175783user2175783
2007
asked Feb 1 at 2:25
user2175783user2175783
2007
asked Feb 1 at 2:25
user2175783user2175783
2007
2007
1
$begingroup$
What does "when defined over $Bbb R$" mean? Given that $L$ maps $Bbb F_2^2$ to $Bbb F_2^3$ it does not map $Bbb R^2$ to $Bbb R^3$...
$endgroup$
– David C. Ullrich
Feb 1 at 13:40
1
$begingroup$
I read your post, so I know you mean "when $L:Bbb R^2toBbb R^3$". That makes no sense! Because given that $L:Bbb F_2^2toBbb F_2^3$ it is simply not true that $f:Bbb R^2toBbb R^3$.
$endgroup$
– David C. Ullrich
Feb 1 at 14:10
1
$begingroup$
@DavidC.Ullrich I am completely lost in what you are trying to say. The question gives $L:mathbb{R}^{2}rightarrow mathbb{R}^{3}$ and asks what is $L:mathbb{F}_{2}^{2}rightarrowmathbb{F}_{2}^{3}$, i.e. when we replace $mathbb{R}$ by $mathbb{F}_{2}$. Thats all. I am not a mathematician so I might be missing some subtlety that is obvious (but not to me).
$endgroup$
– user2175783
Feb 1 at 14:17
1
$begingroup$
The question as you wrote it does not give $L:Bbb R^2toBbb R^3$. If we do assume that $L:Bbb R^2toBbb R^3$ then asking what is $L:Bbb F_2^2toBbb F_2^3$ makes no sense - there's no such thhing as just "replacing $Bbb R$ by $Bbb F_2$" here.
$endgroup$
– David C. Ullrich
Feb 1 at 14:20
1
$begingroup$
@DavidC.Ullrich The question shows how $L$ transforms a basis of $mathbb{R}^{2}$ and hence any vector in $mathbb{R}^{2}$. I imagine this is enough to answer how $L$ transforms the basis of $mathbb{F}^{2}_{2}$ but I dont know how.
$endgroup$
– user2175783
Feb 1 at 14:26
|
show 8 more comments
1
$begingroup$
What does "when defined over $Bbb R$" mean? Given that $L$ maps $Bbb F_2^2$ to $Bbb F_2^3$ it does not map $Bbb R^2$ to $Bbb R^3$...
$endgroup$
– David C. Ullrich
Feb 1 at 13:40
1
$begingroup$
I read your post, so I know you mean "when $L:Bbb R^2toBbb R^3$". That makes no sense! Because given that $L:Bbb F_2^2toBbb F_2^3$ it is simply not true that $f:Bbb R^2toBbb R^3$.
$endgroup$
– David C. Ullrich
Feb 1 at 14:10
1
$begingroup$
@DavidC.Ullrich I am completely lost in what you are trying to say. The question gives $L:mathbb{R}^{2}rightarrow mathbb{R}^{3}$ and asks what is $L:mathbb{F}_{2}^{2}rightarrowmathbb{F}_{2}^{3}$, i.e. when we replace $mathbb{R}$ by $mathbb{F}_{2}$. Thats all. I am not a mathematician so I might be missing some subtlety that is obvious (but not to me).
$endgroup$
– user2175783
Feb 1 at 14:17
1
$begingroup$
The question as you wrote it does not give $L:Bbb R^2toBbb R^3$. If we do assume that $L:Bbb R^2toBbb R^3$ then asking what is $L:Bbb F_2^2toBbb F_2^3$ makes no sense - there's no such thhing as just "replacing $Bbb R$ by $Bbb F_2$" here.
$endgroup$
– David C. Ullrich
Feb 1 at 14:20
1
$begingroup$
@DavidC.Ullrich The question shows how $L$ transforms a basis of $mathbb{R}^{2}$ and hence any vector in $mathbb{R}^{2}$. I imagine this is enough to answer how $L$ transforms the basis of $mathbb{F}^{2}_{2}$ but I dont know how.
$endgroup$
– user2175783
Feb 1 at 14:26
1
1
$begingroup$
What does "when defined over $Bbb R$" mean? Given that $L$ maps $Bbb F_2^2$ to $Bbb F_2^3$ it does not map $Bbb R^2$ to $Bbb R^3$...
$endgroup$
– David C. Ullrich
Feb 1 at 13:40
$begingroup$
What does "when defined over $Bbb R$" mean? Given that $L$ maps $Bbb F_2^2$ to $Bbb F_2^3$ it does not map $Bbb R^2$ to $Bbb R^3$...
$endgroup$
– David C. Ullrich
Feb 1 at 13:40
1
1
$begingroup$
I read your post, so I know you mean "when $L:Bbb R^2toBbb R^3$". That makes no sense! Because given that $L:Bbb F_2^2toBbb F_2^3$ it is simply not true that $f:Bbb R^2toBbb R^3$.
$endgroup$
– David C. Ullrich
Feb 1 at 14:10
$begingroup$
I read your post, so I know you mean "when $L:Bbb R^2toBbb R^3$". That makes no sense! Because given that $L:Bbb F_2^2toBbb F_2^3$ it is simply not true that $f:Bbb R^2toBbb R^3$.
$endgroup$
– David C. Ullrich
Feb 1 at 14:10
1
1
$begingroup$
@DavidC.Ullrich I am completely lost in what you are trying to say. The question gives $L:mathbb{R}^{2}rightarrow mathbb{R}^{3}$ and asks what is $L:mathbb{F}_{2}^{2}rightarrowmathbb{F}_{2}^{3}$, i.e. when we replace $mathbb{R}$ by $mathbb{F}_{2}$. Thats all. I am not a mathematician so I might be missing some subtlety that is obvious (but not to me).
$endgroup$
– user2175783
Feb 1 at 14:17
$begingroup$
@DavidC.Ullrich I am completely lost in what you are trying to say. The question gives $L:mathbb{R}^{2}rightarrow mathbb{R}^{3}$ and asks what is $L:mathbb{F}_{2}^{2}rightarrowmathbb{F}_{2}^{3}$, i.e. when we replace $mathbb{R}$ by $mathbb{F}_{2}$. Thats all. I am not a mathematician so I might be missing some subtlety that is obvious (but not to me).
$endgroup$
– user2175783
Feb 1 at 14:17
1
1
$begingroup$
The question as you wrote it does not give $L:Bbb R^2toBbb R^3$. If we do assume that $L:Bbb R^2toBbb R^3$ then asking what is $L:Bbb F_2^2toBbb F_2^3$ makes no sense - there's no such thhing as just "replacing $Bbb R$ by $Bbb F_2$" here.
$endgroup$
– David C. Ullrich
Feb 1 at 14:20
$begingroup$
The question as you wrote it does not give $L:Bbb R^2toBbb R^3$. If we do assume that $L:Bbb R^2toBbb R^3$ then asking what is $L:Bbb F_2^2toBbb F_2^3$ makes no sense - there's no such thhing as just "replacing $Bbb R$ by $Bbb F_2$" here.
$endgroup$
– David C. Ullrich
Feb 1 at 14:20
1
1
$begingroup$
@DavidC.Ullrich The question shows how $L$ transforms a basis of $mathbb{R}^{2}$ and hence any vector in $mathbb{R}^{2}$. I imagine this is enough to answer how $L$ transforms the basis of $mathbb{F}^{2}_{2}$ but I dont know how.
$endgroup$
– user2175783
Feb 1 at 14:26
$begingroup$
@DavidC.Ullrich The question shows how $L$ transforms a basis of $mathbb{R}^{2}$ and hence any vector in $mathbb{R}^{2}$. I imagine this is enough to answer how $L$ transforms the basis of $mathbb{F}^{2}_{2}$ but I dont know how.
$endgroup$
– user2175783
Feb 1 at 14:26
|
show 8 more comments
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1
$begingroup$
What does "when defined over $Bbb R$" mean? Given that $L$ maps $Bbb F_2^2$ to $Bbb F_2^3$ it does not map $Bbb R^2$ to $Bbb R^3$...
$endgroup$
– David C. Ullrich
Feb 1 at 13:40
1
$begingroup$
I read your post, so I know you mean "when $L:Bbb R^2toBbb R^3$". That makes no sense! Because given that $L:Bbb F_2^2toBbb F_2^3$ it is simply not true that $f:Bbb R^2toBbb R^3$.
$endgroup$
– David C. Ullrich
Feb 1 at 14:10
1
$begingroup$
@DavidC.Ullrich I am completely lost in what you are trying to say. The question gives $L:mathbb{R}^{2}rightarrow mathbb{R}^{3}$ and asks what is $L:mathbb{F}_{2}^{2}rightarrowmathbb{F}_{2}^{3}$, i.e. when we replace $mathbb{R}$ by $mathbb{F}_{2}$. Thats all. I am not a mathematician so I might be missing some subtlety that is obvious (but not to me).
$endgroup$
– user2175783
Feb 1 at 14:17
1
$begingroup$
The question as you wrote it does not give $L:Bbb R^2toBbb R^3$. If we do assume that $L:Bbb R^2toBbb R^3$ then asking what is $L:Bbb F_2^2toBbb F_2^3$ makes no sense - there's no such thhing as just "replacing $Bbb R$ by $Bbb F_2$" here.
$endgroup$
– David C. Ullrich
Feb 1 at 14:20
1
$begingroup$
@DavidC.Ullrich The question shows how $L$ transforms a basis of $mathbb{R}^{2}$ and hence any vector in $mathbb{R}^{2}$. I imagine this is enough to answer how $L$ transforms the basis of $mathbb{F}^{2}_{2}$ but I dont know how.
$endgroup$
– user2175783
Feb 1 at 14:26