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Let $E$ be a $mathbb{K}$-vector space of arbitrary dimension, $ainmathbb{K}setminus{0}$ and $uinmathcal{L}(E)$ an endomorphism of $E$ satisfying the following equation $$a^2u-3au^2+u^3=0_{mathcal{L}(E)}.$$ Show that $E=ker(u)bigoplus Im(u)$.






linear-algebra







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$endgroup$

















    0












    $begingroup$


    Let $E$ be a $mathbb{K}$-vector space of arbitrary dimension, $ainmathbb{K}setminus{0}$ and $uinmathcal{L}(E)$ an endomorphism of $E$ satisfying the following equation $$a^2u-3au^2+u^3=0_{mathcal{L}(E)}.$$ Show that $E=ker(u)bigoplus Im(u)$.






    linear-algebra







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $E$ be a $mathbb{K}$-vector space of arbitrary dimension, $ainmathbb{K}setminus{0}$ and $uinmathcal{L}(E)$ an endomorphism of $E$ satisfying the following equation $$a^2u-3au^2+u^3=0_{mathcal{L}(E)}.$$ Show that $E=ker(u)bigoplus Im(u)$.






      linear-algebra







      share|cite|improve this question









      $endgroup$




      Let $E$ be a $mathbb{K}$-vector space of arbitrary dimension, $ainmathbb{K}setminus{0}$ and $uinmathcal{L}(E)$ an endomorphism of $E$ satisfying the following equation $$a^2u-3au^2+u^3=0_{mathcal{L}(E)}.$$ Show that $E=ker(u)bigoplus Im(u)$.






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Mar 16 at 17:11









      Eliakim MachadoEliakim Machado

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          1 Answer
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          1












          $begingroup$

          Proving that $mathbf{operatorname{Ker}(u)capoperatorname{Im}(u)={0}}$:



          If $yin operatorname{Im}(u)$, then $y=u(x)$ for some $xin E$. Assume that $yinoperatorname{Ker}(u)$. This is, $0=u(y)=u^2(x)$. Then $0=(a^2-3au^2+u^3)(x)=a^2u(x)=a^2y$. Therefore $operatorname{Ker}(u)capoperatorname{Im}(u)={0}$.



          Proving that $mathbf{Esubseteq operatorname{Im}(u)oplusoperatorname{Ker}(u)}$:



          Consider the polynomials $P(t)=t^2-3at+a^2$ and $Q(t)=t$. They are relatively prime. Therefore, there are polynomials $M,N$ such that



          $$P(t)M(t)+Q(t)N(t)=1$$



          Evaluating this equation in $u$ and applying it to an arbitrary element $xin E$, we get $$P(u)M(u)(x)+Q(u)N(u)(x)=(x)$$



          Therefore, $x$ is the sum of an element of $operatorname{Im}(u)$, which is $Q(u)N(u)(x)=uN(u)(x)$ and an element of $operatorname{Ker}(u)$, which is the other term $P(u)M(u)(x)=(a^2-3au+u^2)M(u)(x)$.






          share|cite|improve this answer











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            $begingroup$

            Proving that $mathbf{operatorname{Ker}(u)capoperatorname{Im}(u)={0}}$:



            If $yin operatorname{Im}(u)$, then $y=u(x)$ for some $xin E$. Assume that $yinoperatorname{Ker}(u)$. This is, $0=u(y)=u^2(x)$. Then $0=(a^2-3au^2+u^3)(x)=a^2u(x)=a^2y$. Therefore $operatorname{Ker}(u)capoperatorname{Im}(u)={0}$.



            Proving that $mathbf{Esubseteq operatorname{Im}(u)oplusoperatorname{Ker}(u)}$:



            Consider the polynomials $P(t)=t^2-3at+a^2$ and $Q(t)=t$. They are relatively prime. Therefore, there are polynomials $M,N$ such that



            $$P(t)M(t)+Q(t)N(t)=1$$



            Evaluating this equation in $u$ and applying it to an arbitrary element $xin E$, we get $$P(u)M(u)(x)+Q(u)N(u)(x)=(x)$$



            Therefore, $x$ is the sum of an element of $operatorname{Im}(u)$, which is $Q(u)N(u)(x)=uN(u)(x)$ and an element of $operatorname{Ker}(u)$, which is the other term $P(u)M(u)(x)=(a^2-3au+u^2)M(u)(x)$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Proving that $mathbf{operatorname{Ker}(u)capoperatorname{Im}(u)={0}}$:



              If $yin operatorname{Im}(u)$, then $y=u(x)$ for some $xin E$. Assume that $yinoperatorname{Ker}(u)$. This is, $0=u(y)=u^2(x)$. Then $0=(a^2-3au^2+u^3)(x)=a^2u(x)=a^2y$. Therefore $operatorname{Ker}(u)capoperatorname{Im}(u)={0}$.



              Proving that $mathbf{Esubseteq operatorname{Im}(u)oplusoperatorname{Ker}(u)}$:



              Consider the polynomials $P(t)=t^2-3at+a^2$ and $Q(t)=t$. They are relatively prime. Therefore, there are polynomials $M,N$ such that



              $$P(t)M(t)+Q(t)N(t)=1$$



              Evaluating this equation in $u$ and applying it to an arbitrary element $xin E$, we get $$P(u)M(u)(x)+Q(u)N(u)(x)=(x)$$



              Therefore, $x$ is the sum of an element of $operatorname{Im}(u)$, which is $Q(u)N(u)(x)=uN(u)(x)$ and an element of $operatorname{Ker}(u)$, which is the other term $P(u)M(u)(x)=(a^2-3au+u^2)M(u)(x)$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Proving that $mathbf{operatorname{Ker}(u)capoperatorname{Im}(u)={0}}$:



                If $yin operatorname{Im}(u)$, then $y=u(x)$ for some $xin E$. Assume that $yinoperatorname{Ker}(u)$. This is, $0=u(y)=u^2(x)$. Then $0=(a^2-3au^2+u^3)(x)=a^2u(x)=a^2y$. Therefore $operatorname{Ker}(u)capoperatorname{Im}(u)={0}$.



                Proving that $mathbf{Esubseteq operatorname{Im}(u)oplusoperatorname{Ker}(u)}$:



                Consider the polynomials $P(t)=t^2-3at+a^2$ and $Q(t)=t$. They are relatively prime. Therefore, there are polynomials $M,N$ such that



                $$P(t)M(t)+Q(t)N(t)=1$$



                Evaluating this equation in $u$ and applying it to an arbitrary element $xin E$, we get $$P(u)M(u)(x)+Q(u)N(u)(x)=(x)$$



                Therefore, $x$ is the sum of an element of $operatorname{Im}(u)$, which is $Q(u)N(u)(x)=uN(u)(x)$ and an element of $operatorname{Ker}(u)$, which is the other term $P(u)M(u)(x)=(a^2-3au+u^2)M(u)(x)$.






                share|cite|improve this answer











                $endgroup$



                Proving that $mathbf{operatorname{Ker}(u)capoperatorname{Im}(u)={0}}$:



                If $yin operatorname{Im}(u)$, then $y=u(x)$ for some $xin E$. Assume that $yinoperatorname{Ker}(u)$. This is, $0=u(y)=u^2(x)$. Then $0=(a^2-3au^2+u^3)(x)=a^2u(x)=a^2y$. Therefore $operatorname{Ker}(u)capoperatorname{Im}(u)={0}$.



                Proving that $mathbf{Esubseteq operatorname{Im}(u)oplusoperatorname{Ker}(u)}$:



                Consider the polynomials $P(t)=t^2-3at+a^2$ and $Q(t)=t$. They are relatively prime. Therefore, there are polynomials $M,N$ such that



                $$P(t)M(t)+Q(t)N(t)=1$$



                Evaluating this equation in $u$ and applying it to an arbitrary element $xin E$, we get $$P(u)M(u)(x)+Q(u)N(u)(x)=(x)$$



                Therefore, $x$ is the sum of an element of $operatorname{Im}(u)$, which is $Q(u)N(u)(x)=uN(u)(x)$ and an element of $operatorname{Ker}(u)$, which is the other term $P(u)M(u)(x)=(a^2-3au+u^2)M(u)(x)$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 16 at 17:38

























                answered Mar 16 at 17:31









                user647486user647486

                582110




                582110






























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