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How to verify the rank-nullity theorem for the linear application $frac{d}{dx}in...
1
$begingroup$
I know that the rank-nullity theorem states that let $T: Vrightarrow W$ be a linear transformation, then:
$$ operatorname{Rank}(T)+ operatorname{Nullity}(T)=dim(V)$$
But since $frac{d}{dx}in operatorname{End}(mathbb{K}[x]_2)$ is also a linear transformation, I want to verify the rank-nullity theorem with this particular situation.
linear-algebra linear-transformations
edited Jan 17 at 20:55
Bernard
121k740116
asked Jan 17 at 20:52
KevinKevin
13311
$endgroup$
add a comment |
1
$begingroup$
I know that the rank-nullity theorem states that let $T: Vrightarrow W$ be a linear transformation, then:
$$ operatorname{Rank}(T)+ operatorname{Nullity}(T)=dim(V)$$
But since $frac{d}{dx}in operatorname{End}(mathbb{K}[x]_2)$ is also a linear transformation, I want to verify the rank-nullity theorem with this particular situation.
linear-algebra linear-transformations
edited Jan 17 at 20:55
Bernard
121k740116
asked Jan 17 at 20:52
KevinKevin
13311
$endgroup$
$begingroup$
What exactly do you mean by $Bbb K[x]_2$?
$endgroup$
– Robert Lewis
Jan 17 at 20:58
1
$begingroup$
Clarification question, is $mathbb{K}[x]_2$ the ring of polynomials of $mathbb{K}$ localized at the ideal $2$ or is $mathbb{K}$ the field with 2 elements?
$endgroup$
– Sheel Stueber
Jan 17 at 20:58
1
$begingroup$
$mathbb{K}[x]_2$ is the ring of polynomials of 𝕂 localized at the ideal 2 $(ax^2+bx+c)$
$endgroup$
– Kevin
Jan 17 at 21:01
$begingroup$
You can just take a proof of the abstract theorem and then substitute everything there with your example.
$endgroup$
– user635162
Jan 17 at 21:11
add a comment |
1
1
1
$begingroup$
I know that the rank-nullity theorem states that let $T: Vrightarrow W$ be a linear transformation, then:
$$ operatorname{Rank}(T)+ operatorname{Nullity}(T)=dim(V)$$
But since $frac{d}{dx}in operatorname{End}(mathbb{K}[x]_2)$ is also a linear transformation, I want to verify the rank-nullity theorem with this particular situation.
linear-algebra linear-transformations
edited Jan 17 at 20:55
Bernard
121k740116
asked Jan 17 at 20:52
KevinKevin
13311
$endgroup$
I know that the rank-nullity theorem states that let $T: Vrightarrow W$ be a linear transformation, then:
$$ operatorname{Rank}(T)+ operatorname{Nullity}(T)=dim(V)$$
But since $frac{d}{dx}in operatorname{End}(mathbb{K}[x]_2)$ is also a linear transformation, I want to verify the rank-nullity theorem with this particular situation.
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Jan 17 at 20:55
Bernard
121k740116
asked Jan 17 at 20:52
KevinKevin
13311
edited Jan 17 at 20:55
Bernard
121k740116
asked Jan 17 at 20:52
KevinKevin
13311
edited Jan 17 at 20:55
Bernard
121k740116
edited Jan 17 at 20:55
Bernard
121k740116
edited Jan 17 at 20:55
Bernard
121k740116
121k740116
asked Jan 17 at 20:52
KevinKevin
13311
asked Jan 17 at 20:52
KevinKevin
13311
asked Jan 17 at 20:52
KevinKevin
13311
13311
$begingroup$
What exactly do you mean by $Bbb K[x]_2$?
$endgroup$
– Robert Lewis
Jan 17 at 20:58
1
$begingroup$
Clarification question, is $mathbb{K}[x]_2$ the ring of polynomials of $mathbb{K}$ localized at the ideal $2$ or is $mathbb{K}$ the field with 2 elements?
$endgroup$
– Sheel Stueber
Jan 17 at 20:58
1
$begingroup$
$mathbb{K}[x]_2$ is the ring of polynomials of 𝕂 localized at the ideal 2 $(ax^2+bx+c)$
$endgroup$
– Kevin
Jan 17 at 21:01
$begingroup$
You can just take a proof of the abstract theorem and then substitute everything there with your example.
$endgroup$
– user635162
Jan 17 at 21:11
add a comment |
$begingroup$
What exactly do you mean by $Bbb K[x]_2$?
$endgroup$
– Robert Lewis
Jan 17 at 20:58
1
$begingroup$
Clarification question, is $mathbb{K}[x]_2$ the ring of polynomials of $mathbb{K}$ localized at the ideal $2$ or is $mathbb{K}$ the field with 2 elements?
$endgroup$
– Sheel Stueber
Jan 17 at 20:58
1
$begingroup$
$mathbb{K}[x]_2$ is the ring of polynomials of 𝕂 localized at the ideal 2 $(ax^2+bx+c)$
$endgroup$
– Kevin
Jan 17 at 21:01
$begingroup$
You can just take a proof of the abstract theorem and then substitute everything there with your example.
$endgroup$
– user635162
Jan 17 at 21:11
$begingroup$
What exactly do you mean by $Bbb K[x]_2$?
$endgroup$
– Robert Lewis
Jan 17 at 20:58
$begingroup$
What exactly do you mean by $Bbb K[x]_2$?
$endgroup$
– Robert Lewis
Jan 17 at 20:58
1
1
$begingroup$
Clarification question, is $mathbb{K}[x]_2$ the ring of polynomials of $mathbb{K}$ localized at the ideal $2$ or is $mathbb{K}$ the field with 2 elements?
$endgroup$
– Sheel Stueber
Jan 17 at 20:58
$begingroup$
Clarification question, is $mathbb{K}[x]_2$ the ring of polynomials of $mathbb{K}$ localized at the ideal $2$ or is $mathbb{K}$ the field with 2 elements?
$endgroup$
– Sheel Stueber
Jan 17 at 20:58
1
1
$begingroup$
$mathbb{K}[x]_2$ is the ring of polynomials of 𝕂 localized at the ideal 2 $(ax^2+bx+c)$
$endgroup$
– Kevin
Jan 17 at 21:01
$begingroup$
$mathbb{K}[x]_2$ is the ring of polynomials of 𝕂 localized at the ideal 2 $(ax^2+bx+c)$
$endgroup$
– Kevin
Jan 17 at 21:01
$begingroup$
You can just take a proof of the abstract theorem and then substitute everything there with your example.
$endgroup$
– user635162
Jan 17 at 21:11
$begingroup$
You can just take a proof of the abstract theorem and then substitute everything there with your example.
$endgroup$
– user635162
Jan 17 at 21:11
add a comment |
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$begingroup$
What exactly do you mean by $Bbb K[x]_2$?
$endgroup$
– Robert Lewis
Jan 17 at 20:58
1
$begingroup$
Clarification question, is $mathbb{K}[x]_2$ the ring of polynomials of $mathbb{K}$ localized at the ideal $2$ or is $mathbb{K}$ the field with 2 elements?
$endgroup$
– Sheel Stueber
Jan 17 at 20:58
1
$begingroup$
$mathbb{K}[x]_2$ is the ring of polynomials of 𝕂 localized at the ideal 2 $(ax^2+bx+c)$
$endgroup$
– Kevin
Jan 17 at 21:01
$begingroup$
You can just take a proof of the abstract theorem and then substitute everything there with your example.
$endgroup$
– user635162
Jan 17 at 21:11