Calculate minimal polynomial of a matrix












4












$begingroup$


begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve






linear-algebra matrices minimal-polynomials







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    4












    $begingroup$


    begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
    I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve






    linear-algebra matrices minimal-polynomials







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      5



      $begingroup$


      begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
      I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve






      linear-algebra matrices minimal-polynomials







      share|cite|improve this question











      $endgroup$




      begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
      I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve






      linear-algebra matrices minimal-polynomials




      linear-algebra matrices minimal-polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited May 11 '17 at 4:19









      Parcly Taxel

      1




      1










      asked May 11 '17 at 4:08









      user439852user439852

      27113




      27113






















          3 Answers
          3






          active

          oldest

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          9












          $begingroup$

          These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.



          Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
          $$pmatrix{1\0\0\0}overset Amapsto
          pmatrix{0\1\0\1}overset Amapsto
          pmatrix{2\0\2\0}overset Amapsto
          pmatrix{0\4\0\4}
          $$
          with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.



          As you see, one can do entirely without the characteristic polynomial.






          share|cite|improve this answer











          $endgroup$





















            7












            $begingroup$

            All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$



            Hence $x(x^2-4)$ divides the minimal polynomial,



            Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.



            Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.



            Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.



            Another way to decide on the last part:
            The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
              $endgroup$
              – Pedro Tamaroff♦
              Dec 5 '18 at 15:57



















            1












            $begingroup$

            As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
              $endgroup$
              – Jose Brox
              Dec 5 '18 at 20:08











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            3 Answers
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            active

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            3 Answers
            3






            active

            oldest

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            active

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            active

            oldest

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            9












            $begingroup$

            These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.



            Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
            $$pmatrix{1\0\0\0}overset Amapsto
            pmatrix{0\1\0\1}overset Amapsto
            pmatrix{2\0\2\0}overset Amapsto
            pmatrix{0\4\0\4}
            $$
            with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.



            As you see, one can do entirely without the characteristic polynomial.






            share|cite|improve this answer











            $endgroup$


















              9












              $begingroup$

              These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.



              Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
              $$pmatrix{1\0\0\0}overset Amapsto
              pmatrix{0\1\0\1}overset Amapsto
              pmatrix{2\0\2\0}overset Amapsto
              pmatrix{0\4\0\4}
              $$
              with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.



              As you see, one can do entirely without the characteristic polynomial.






              share|cite|improve this answer











              $endgroup$
















                9












                9








                9





                $begingroup$

                These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.



                Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
                $$pmatrix{1\0\0\0}overset Amapsto
                pmatrix{0\1\0\1}overset Amapsto
                pmatrix{2\0\2\0}overset Amapsto
                pmatrix{0\4\0\4}
                $$
                with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.



                As you see, one can do entirely without the characteristic polynomial.






                share|cite|improve this answer











                $endgroup$



                These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.



                Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
                $$pmatrix{1\0\0\0}overset Amapsto
                pmatrix{0\1\0\1}overset Amapsto
                pmatrix{2\0\2\0}overset Amapsto
                pmatrix{0\4\0\4}
                $$
                with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.



                As you see, one can do entirely without the characteristic polynomial.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited May 11 '17 at 9:25

























                answered May 11 '17 at 4:42









                Marc van LeeuwenMarc van Leeuwen

                87.6k5110225




                87.6k5110225























                    7












                    $begingroup$

                    All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$



                    Hence $x(x^2-4)$ divides the minimal polynomial,



                    Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.



                    Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.



                    Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.



                    Another way to decide on the last part:
                    The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                      $endgroup$
                      – Pedro Tamaroff♦
                      Dec 5 '18 at 15:57
















                    7












                    $begingroup$

                    All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$



                    Hence $x(x^2-4)$ divides the minimal polynomial,



                    Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.



                    Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.



                    Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.



                    Another way to decide on the last part:
                    The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                      $endgroup$
                      – Pedro Tamaroff♦
                      Dec 5 '18 at 15:57














                    7












                    7








                    7





                    $begingroup$

                    All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$



                    Hence $x(x^2-4)$ divides the minimal polynomial,



                    Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.



                    Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.



                    Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.



                    Another way to decide on the last part:
                    The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.






                    share|cite|improve this answer











                    $endgroup$



                    All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$



                    Hence $x(x^2-4)$ divides the minimal polynomial,



                    Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.



                    Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.



                    Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.



                    Another way to decide on the last part:
                    The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited May 11 '17 at 4:43

























                    answered May 11 '17 at 4:33









                    Arpan1729Arpan1729

                    2,7871320




                    2,7871320








                    • 1




                      $begingroup$
                      Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                      $endgroup$
                      – Pedro Tamaroff♦
                      Dec 5 '18 at 15:57














                    • 1




                      $begingroup$
                      Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                      $endgroup$
                      – Pedro Tamaroff♦
                      Dec 5 '18 at 15:57








                    1




                    1




                    $begingroup$
                    Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                    $endgroup$
                    – Pedro Tamaroff♦
                    Dec 5 '18 at 15:57




                    $begingroup$
                    Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
                    $endgroup$
                    – Pedro Tamaroff♦
                    Dec 5 '18 at 15:57











                    1












                    $begingroup$

                    As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                      $endgroup$
                      – Jose Brox
                      Dec 5 '18 at 20:08
















                    1












                    $begingroup$

                    As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                      $endgroup$
                      – Jose Brox
                      Dec 5 '18 at 20:08














                    1












                    1








                    1





                    $begingroup$

                    As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)






                    share|cite|improve this answer









                    $endgroup$



                    As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 19 '18 at 5:22









                    user454200user454200

                    412




                    412












                    • $begingroup$
                      @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                      $endgroup$
                      – Jose Brox
                      Dec 5 '18 at 20:08


















                    • $begingroup$
                      @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                      $endgroup$
                      – Jose Brox
                      Dec 5 '18 at 20:08
















                    $begingroup$
                    @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                    $endgroup$
                    – Jose Brox
                    Dec 5 '18 at 20:08




                    $begingroup$
                    @ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
                    $endgroup$
                    – Jose Brox
                    Dec 5 '18 at 20:08


















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