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







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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


















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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 20:55









Bernard

121k740116




121k740116










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




















  • $begingroup$
    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












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