Finding characteristic and minimal polynomials and the Jordan normal form of $f$, knowing some relations for...












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Given a vector space $V$ of dimension $4$ and a base ${v_1,v_2,v_3,v_4}$, let $f$ be an endomorphism of $V$ such that $f^3=0$ and moreover $f(v_1)=f(v_2)=v_3$, $f(v_3)=kv_4$, and $f(v_4)inleft<v_1,v_4right>$, where $k$ is a real parameter.



I should find the characteristic polynomial $chi_f$, as well as the minimal one $m_f$, and the Jordan normal form of $f$, depending on $k$ and $f(v_4)$.



Now, here's my thoughts: $f^3=0$ means that either $m_f(t)=t^2$ or $m_f=t^3$; furthermore, $$0=f^3(v_1)=f^3(v_2)=f^2(v_3)=f(kv_4)=kcdot f(v_4) $$so I guess one should consider separately $k=0$, which lets $f(v_4)$ be a parameter, and $kne0$ which forces $f(v_4)=0$.



How should I continue?






linear-algebra vector-spaces linear-transformations jordan-normal-form minimal-polynomials







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    $begingroup$


    Given a vector space $V$ of dimension $4$ and a base ${v_1,v_2,v_3,v_4}$, let $f$ be an endomorphism of $V$ such that $f^3=0$ and moreover $f(v_1)=f(v_2)=v_3$, $f(v_3)=kv_4$, and $f(v_4)inleft<v_1,v_4right>$, where $k$ is a real parameter.



    I should find the characteristic polynomial $chi_f$, as well as the minimal one $m_f$, and the Jordan normal form of $f$, depending on $k$ and $f(v_4)$.



    Now, here's my thoughts: $f^3=0$ means that either $m_f(t)=t^2$ or $m_f=t^3$; furthermore, $$0=f^3(v_1)=f^3(v_2)=f^2(v_3)=f(kv_4)=kcdot f(v_4) $$so I guess one should consider separately $k=0$, which lets $f(v_4)$ be a parameter, and $kne0$ which forces $f(v_4)=0$.



    How should I continue?






    linear-algebra vector-spaces linear-transformations jordan-normal-form minimal-polynomials







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Given a vector space $V$ of dimension $4$ and a base ${v_1,v_2,v_3,v_4}$, let $f$ be an endomorphism of $V$ such that $f^3=0$ and moreover $f(v_1)=f(v_2)=v_3$, $f(v_3)=kv_4$, and $f(v_4)inleft<v_1,v_4right>$, where $k$ is a real parameter.



      I should find the characteristic polynomial $chi_f$, as well as the minimal one $m_f$, and the Jordan normal form of $f$, depending on $k$ and $f(v_4)$.



      Now, here's my thoughts: $f^3=0$ means that either $m_f(t)=t^2$ or $m_f=t^3$; furthermore, $$0=f^3(v_1)=f^3(v_2)=f^2(v_3)=f(kv_4)=kcdot f(v_4) $$so I guess one should consider separately $k=0$, which lets $f(v_4)$ be a parameter, and $kne0$ which forces $f(v_4)=0$.



      How should I continue?






      linear-algebra vector-spaces linear-transformations jordan-normal-form minimal-polynomials







      share|cite|improve this question











      $endgroup$




      Given a vector space $V$ of dimension $4$ and a base ${v_1,v_2,v_3,v_4}$, let $f$ be an endomorphism of $V$ such that $f^3=0$ and moreover $f(v_1)=f(v_2)=v_3$, $f(v_3)=kv_4$, and $f(v_4)inleft<v_1,v_4right>$, where $k$ is a real parameter.



      I should find the characteristic polynomial $chi_f$, as well as the minimal one $m_f$, and the Jordan normal form of $f$, depending on $k$ and $f(v_4)$.



      Now, here's my thoughts: $f^3=0$ means that either $m_f(t)=t^2$ or $m_f=t^3$; furthermore, $$0=f^3(v_1)=f^3(v_2)=f^2(v_3)=f(kv_4)=kcdot f(v_4) $$so I guess one should consider separately $k=0$, which lets $f(v_4)$ be a parameter, and $kne0$ which forces $f(v_4)=0$.



      How should I continue?






      linear-algebra vector-spaces linear-transformations jordan-normal-form minimal-polynomials




      linear-algebra vector-spaces linear-transformations jordan-normal-form minimal-polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 29 at 14:12









      Marc van Leeuwen

      87.8k5111225




      87.8k5111225










      asked Jan 20 at 23:41









      LearnerLearner

      17510




      17510






















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          $begingroup$

          Note that $f$ cannot be a Jordan block can by of size 4, since $f^3=0$. Also $f^3=0$ forces all eigenvalues of $f$ to be zero, so the characteristic polynomial is $chi_f(lambda)=lambda^4$.



          Assume first that $kne0$. As you say, $f(v_4)=0$. Then $v_1,v_3,v_4$ form a basis for a $3times 3$ Jordan block. So the Jordan form of $f$ is
          $$
          begin{bmatrix} 0&1&0&0\ 0&0&1&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
          $$
          Take now the case $k=0$. If $bne0$, then
          $$
          x=tfrac ab,v_1+tfrac a{b^2},v_3+v_4
          $$
          satisfies $f(x)=bx$, so $b$ is an eigenvalue. But this is not possible, as it contradicts $f^3=0$. Thus $b=0$, and so $f(v_4)=av_1$ for some $a$. So, if $ane0$, then $v_3,v_1,tfrac1a,v_4$ is a basis for a $3times 3$ Jordan block. If $a=0$, then $f(v_4)=0$. Then $v_1,v_3$ give you a $2times 2$ Jordan block, and you can complete the basis with $v_4$ and $v_1-v_2$, both in the kernel of $f$.



          In summary: the deciding factors are $k$ and $a$ in $f(v_4)=av_1+bv_4$.




          • If $k=0$ and $a=0$, the Jordan form is
            $$
            begin{bmatrix} 0&1&0&0\ 0&0&0&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
            $$


          • If $k=0$ and $ane0$, or if $kne0$, the Jordan form is
            $$
            begin{bmatrix} 0&1&0&0\ 0&0&1&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
            $$







          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            Note that $f$ cannot be a Jordan block can by of size 4, since $f^3=0$. Also $f^3=0$ forces all eigenvalues of $f$ to be zero, so the characteristic polynomial is $chi_f(lambda)=lambda^4$.



            Assume first that $kne0$. As you say, $f(v_4)=0$. Then $v_1,v_3,v_4$ form a basis for a $3times 3$ Jordan block. So the Jordan form of $f$ is
            $$
            begin{bmatrix} 0&1&0&0\ 0&0&1&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
            $$
            Take now the case $k=0$. If $bne0$, then
            $$
            x=tfrac ab,v_1+tfrac a{b^2},v_3+v_4
            $$
            satisfies $f(x)=bx$, so $b$ is an eigenvalue. But this is not possible, as it contradicts $f^3=0$. Thus $b=0$, and so $f(v_4)=av_1$ for some $a$. So, if $ane0$, then $v_3,v_1,tfrac1a,v_4$ is a basis for a $3times 3$ Jordan block. If $a=0$, then $f(v_4)=0$. Then $v_1,v_3$ give you a $2times 2$ Jordan block, and you can complete the basis with $v_4$ and $v_1-v_2$, both in the kernel of $f$.



            In summary: the deciding factors are $k$ and $a$ in $f(v_4)=av_1+bv_4$.




            • If $k=0$ and $a=0$, the Jordan form is
              $$
              begin{bmatrix} 0&1&0&0\ 0&0&0&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
              $$


            • If $k=0$ and $ane0$, or if $kne0$, the Jordan form is
              $$
              begin{bmatrix} 0&1&0&0\ 0&0&1&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
              $$







            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Note that $f$ cannot be a Jordan block can by of size 4, since $f^3=0$. Also $f^3=0$ forces all eigenvalues of $f$ to be zero, so the characteristic polynomial is $chi_f(lambda)=lambda^4$.



              Assume first that $kne0$. As you say, $f(v_4)=0$. Then $v_1,v_3,v_4$ form a basis for a $3times 3$ Jordan block. So the Jordan form of $f$ is
              $$
              begin{bmatrix} 0&1&0&0\ 0&0&1&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
              $$
              Take now the case $k=0$. If $bne0$, then
              $$
              x=tfrac ab,v_1+tfrac a{b^2},v_3+v_4
              $$
              satisfies $f(x)=bx$, so $b$ is an eigenvalue. But this is not possible, as it contradicts $f^3=0$. Thus $b=0$, and so $f(v_4)=av_1$ for some $a$. So, if $ane0$, then $v_3,v_1,tfrac1a,v_4$ is a basis for a $3times 3$ Jordan block. If $a=0$, then $f(v_4)=0$. Then $v_1,v_3$ give you a $2times 2$ Jordan block, and you can complete the basis with $v_4$ and $v_1-v_2$, both in the kernel of $f$.



              In summary: the deciding factors are $k$ and $a$ in $f(v_4)=av_1+bv_4$.




              • If $k=0$ and $a=0$, the Jordan form is
                $$
                begin{bmatrix} 0&1&0&0\ 0&0&0&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
                $$


              • If $k=0$ and $ane0$, or if $kne0$, the Jordan form is
                $$
                begin{bmatrix} 0&1&0&0\ 0&0&1&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
                $$







              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Note that $f$ cannot be a Jordan block can by of size 4, since $f^3=0$. Also $f^3=0$ forces all eigenvalues of $f$ to be zero, so the characteristic polynomial is $chi_f(lambda)=lambda^4$.



                Assume first that $kne0$. As you say, $f(v_4)=0$. Then $v_1,v_3,v_4$ form a basis for a $3times 3$ Jordan block. So the Jordan form of $f$ is
                $$
                begin{bmatrix} 0&1&0&0\ 0&0&1&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
                $$
                Take now the case $k=0$. If $bne0$, then
                $$
                x=tfrac ab,v_1+tfrac a{b^2},v_3+v_4
                $$
                satisfies $f(x)=bx$, so $b$ is an eigenvalue. But this is not possible, as it contradicts $f^3=0$. Thus $b=0$, and so $f(v_4)=av_1$ for some $a$. So, if $ane0$, then $v_3,v_1,tfrac1a,v_4$ is a basis for a $3times 3$ Jordan block. If $a=0$, then $f(v_4)=0$. Then $v_1,v_3$ give you a $2times 2$ Jordan block, and you can complete the basis with $v_4$ and $v_1-v_2$, both in the kernel of $f$.



                In summary: the deciding factors are $k$ and $a$ in $f(v_4)=av_1+bv_4$.




                • If $k=0$ and $a=0$, the Jordan form is
                  $$
                  begin{bmatrix} 0&1&0&0\ 0&0&0&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
                  $$


                • If $k=0$ and $ane0$, or if $kne0$, the Jordan form is
                  $$
                  begin{bmatrix} 0&1&0&0\ 0&0&1&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
                  $$







                share|cite|improve this answer









                $endgroup$



                Note that $f$ cannot be a Jordan block can by of size 4, since $f^3=0$. Also $f^3=0$ forces all eigenvalues of $f$ to be zero, so the characteristic polynomial is $chi_f(lambda)=lambda^4$.



                Assume first that $kne0$. As you say, $f(v_4)=0$. Then $v_1,v_3,v_4$ form a basis for a $3times 3$ Jordan block. So the Jordan form of $f$ is
                $$
                begin{bmatrix} 0&1&0&0\ 0&0&1&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
                $$
                Take now the case $k=0$. If $bne0$, then
                $$
                x=tfrac ab,v_1+tfrac a{b^2},v_3+v_4
                $$
                satisfies $f(x)=bx$, so $b$ is an eigenvalue. But this is not possible, as it contradicts $f^3=0$. Thus $b=0$, and so $f(v_4)=av_1$ for some $a$. So, if $ane0$, then $v_3,v_1,tfrac1a,v_4$ is a basis for a $3times 3$ Jordan block. If $a=0$, then $f(v_4)=0$. Then $v_1,v_3$ give you a $2times 2$ Jordan block, and you can complete the basis with $v_4$ and $v_1-v_2$, both in the kernel of $f$.



                In summary: the deciding factors are $k$ and $a$ in $f(v_4)=av_1+bv_4$.




                • If $k=0$ and $a=0$, the Jordan form is
                  $$
                  begin{bmatrix} 0&1&0&0\ 0&0&0&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
                  $$


                • If $k=0$ and $ane0$, or if $kne0$, the Jordan form is
                  $$
                  begin{bmatrix} 0&1&0&0\ 0&0&1&0\ 0&0&0&0\ 0&0&0&0end{bmatrix}.
                  $$








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                answered Jan 21 at 2:37









                Martin ArgeramiMartin Argerami

                128k1183183




                128k1183183






























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