finding the dimension of eigenspace with characteristic and minimal polynomial












0














I have a problem in understanding how author find the dimension of eigenspace of 3 and 1 in following example.
How can I calculate it?



For example, suppose you're given a $ 6 times 6 $
matrix and you calculate that its characteristic polynomial is $(t-3)^4(t-i)^2$, that
its minimal polynomial is $(t-3)^2 (t-i)^2$, that the 3-eigenspace is 3-dimensional,
and that the i-eigenspace is 1-dimensional.



Thanks for your help






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question





























    0














    I have a problem in understanding how author find the dimension of eigenspace of 3 and 1 in following example.
    How can I calculate it?



    For example, suppose you're given a $ 6 times 6 $
    matrix and you calculate that its characteristic polynomial is $(t-3)^4(t-i)^2$, that
    its minimal polynomial is $(t-3)^2 (t-i)^2$, that the 3-eigenspace is 3-dimensional,
    and that the i-eigenspace is 1-dimensional.



    Thanks for your help






    linear-algebra matrices eigenvalues-eigenvectors







    share|cite|improve this question



























      0












      0








      0







      I have a problem in understanding how author find the dimension of eigenspace of 3 and 1 in following example.
      How can I calculate it?



      For example, suppose you're given a $ 6 times 6 $
      matrix and you calculate that its characteristic polynomial is $(t-3)^4(t-i)^2$, that
      its minimal polynomial is $(t-3)^2 (t-i)^2$, that the 3-eigenspace is 3-dimensional,
      and that the i-eigenspace is 1-dimensional.



      Thanks for your help






      linear-algebra matrices eigenvalues-eigenvectors







      share|cite|improve this question















      I have a problem in understanding how author find the dimension of eigenspace of 3 and 1 in following example.
      How can I calculate it?



      For example, suppose you're given a $ 6 times 6 $
      matrix and you calculate that its characteristic polynomial is $(t-3)^4(t-i)^2$, that
      its minimal polynomial is $(t-3)^2 (t-i)^2$, that the 3-eigenspace is 3-dimensional,
      and that the i-eigenspace is 1-dimensional.



      Thanks for your help






      linear-algebra matrices eigenvalues-eigenvectors




      linear-algebra matrices eigenvalues-eigenvectors






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 at 9:27









      José Carlos Santos

      148k22117218




      148k22117218










      asked Jun 13 '17 at 9:46









      fateme jl

      8710




      8710






















          3 Answers
          3






          active

          oldest

          votes


















          1














          Let us take for example the eigenvalue $;lambda=3;$: that the exponent of its linear factor in the minimal polynomial is two means that two is the largest block in the Jordan Block (=JB) corresponding to $;3;$, which means that the JB, which clearly has size $;4times4;$ , has exactly three blocks: one of size two, two of size one. Something like the following, with the different colors marking the different block in $;JB_3;$ :



          $$JB_{lambda=3}=begin{pmatrix}
          color{red}3&color{red}1&0&0\
          color{red}0&color{red}3&0&0\
          0&0&color{green}3&0\
          0&0&0&color{blue}3end{pmatrix}$$



          Since the number of blocks in $;JB_lambda;$ gives us $;dim V_lambda;$ ,we're done.






          share|cite|improve this answer





















          • Isn't two Jordan blocks of size 2 a possibility?
            – Rab
            Jun 13 '17 at 10:05










          • Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
            – Rab
            Jun 13 '17 at 10:10










          • @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
            – DonAntonio
            Jun 13 '17 at 11:20








          • 1




            Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
            – DonAntonio
            Jun 13 '17 at 11:22












          • @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
            – Peter Melech
            Jun 16 '17 at 7:52



















          1














          That implies its Jordan normal form is:



          $begin{pmatrix}3&1&0&0&0&0\
          0&3&0&0&0&0\
          0&0&3&0&0&0\
          0&0&0&3&0&0\
          0&0&0&0&i&1\
          0&0&0&0&0&iend{pmatrix}$






          share|cite|improve this answer





















          • why we don't have two 2*2 blocks for eigenvalue 3?
            – fateme jl
            Jun 15 '17 at 12:05










          • Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
            – Peter Melech
            Jun 16 '17 at 7:48



















          1














          The Jordan form of your matrix must be of the type$$begin{pmatrix}3&*&0&0&0&0\0&3&*&0&0&0\0&0&3&*&0&0\0&0&0&3&0&0\0&0&0&0&i&*\0&0&0&0&0&iend{pmatrix},$$where each $*$ is either $0$ or $1$. But if, for instance, the $*$ from the fifth row is equal to $0$, then the exponent of $t-i$ in the minimal polynomial of the matrix must be $1$. Since it is $2$, that $*$ must be equal to $1$. Now, see what the other three $*$ must be so that the exponent of $t-3$ in the minimal polynomial of the matrix is $2$.






          share|cite|improve this answer





















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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

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            1














            Let us take for example the eigenvalue $;lambda=3;$: that the exponent of its linear factor in the minimal polynomial is two means that two is the largest block in the Jordan Block (=JB) corresponding to $;3;$, which means that the JB, which clearly has size $;4times4;$ , has exactly three blocks: one of size two, two of size one. Something like the following, with the different colors marking the different block in $;JB_3;$ :



            $$JB_{lambda=3}=begin{pmatrix}
            color{red}3&color{red}1&0&0\
            color{red}0&color{red}3&0&0\
            0&0&color{green}3&0\
            0&0&0&color{blue}3end{pmatrix}$$



            Since the number of blocks in $;JB_lambda;$ gives us $;dim V_lambda;$ ,we're done.






            share|cite|improve this answer





















            • Isn't two Jordan blocks of size 2 a possibility?
              – Rab
              Jun 13 '17 at 10:05










            • Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
              – Rab
              Jun 13 '17 at 10:10










            • @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
              – DonAntonio
              Jun 13 '17 at 11:22












            • @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
              – Peter Melech
              Jun 16 '17 at 7:52
















            1














            Let us take for example the eigenvalue $;lambda=3;$: that the exponent of its linear factor in the minimal polynomial is two means that two is the largest block in the Jordan Block (=JB) corresponding to $;3;$, which means that the JB, which clearly has size $;4times4;$ , has exactly three blocks: one of size two, two of size one. Something like the following, with the different colors marking the different block in $;JB_3;$ :



            $$JB_{lambda=3}=begin{pmatrix}
            color{red}3&color{red}1&0&0\
            color{red}0&color{red}3&0&0\
            0&0&color{green}3&0\
            0&0&0&color{blue}3end{pmatrix}$$



            Since the number of blocks in $;JB_lambda;$ gives us $;dim V_lambda;$ ,we're done.






            share|cite|improve this answer





















            • Isn't two Jordan blocks of size 2 a possibility?
              – Rab
              Jun 13 '17 at 10:05










            • Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
              – Rab
              Jun 13 '17 at 10:10










            • @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
              – DonAntonio
              Jun 13 '17 at 11:22












            • @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
              – Peter Melech
              Jun 16 '17 at 7:52














            1












            1








            1






            Let us take for example the eigenvalue $;lambda=3;$: that the exponent of its linear factor in the minimal polynomial is two means that two is the largest block in the Jordan Block (=JB) corresponding to $;3;$, which means that the JB, which clearly has size $;4times4;$ , has exactly three blocks: one of size two, two of size one. Something like the following, with the different colors marking the different block in $;JB_3;$ :



            $$JB_{lambda=3}=begin{pmatrix}
            color{red}3&color{red}1&0&0\
            color{red}0&color{red}3&0&0\
            0&0&color{green}3&0\
            0&0&0&color{blue}3end{pmatrix}$$



            Since the number of blocks in $;JB_lambda;$ gives us $;dim V_lambda;$ ,we're done.






            share|cite|improve this answer












            Let us take for example the eigenvalue $;lambda=3;$: that the exponent of its linear factor in the minimal polynomial is two means that two is the largest block in the Jordan Block (=JB) corresponding to $;3;$, which means that the JB, which clearly has size $;4times4;$ , has exactly three blocks: one of size two, two of size one. Something like the following, with the different colors marking the different block in $;JB_3;$ :



            $$JB_{lambda=3}=begin{pmatrix}
            color{red}3&color{red}1&0&0\
            color{red}0&color{red}3&0&0\
            0&0&color{green}3&0\
            0&0&0&color{blue}3end{pmatrix}$$



            Since the number of blocks in $;JB_lambda;$ gives us $;dim V_lambda;$ ,we're done.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jun 13 '17 at 10:02









            DonAntonio

            176k1491225




            176k1491225












            • Isn't two Jordan blocks of size 2 a possibility?
              – Rab
              Jun 13 '17 at 10:05










            • Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
              – Rab
              Jun 13 '17 at 10:10










            • @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
              – DonAntonio
              Jun 13 '17 at 11:22












            • @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
              – Peter Melech
              Jun 16 '17 at 7:52


















            • Isn't two Jordan blocks of size 2 a possibility?
              – Rab
              Jun 13 '17 at 10:05










            • Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
              – Rab
              Jun 13 '17 at 10:10










            • @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
              – DonAntonio
              Jun 13 '17 at 11:22












            • @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
              – Peter Melech
              Jun 16 '17 at 7:52
















            Isn't two Jordan blocks of size 2 a possibility?
            – Rab
            Jun 13 '17 at 10:05




            Isn't two Jordan blocks of size 2 a possibility?
            – Rab
            Jun 13 '17 at 10:05












            Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
            – Rab
            Jun 13 '17 at 10:10




            Cause the minimal polynomial exponent is the largest JB, but I could be wrong, I am still studying it in uni.
            – Rab
            Jun 13 '17 at 10:10












            @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
            – DonAntonio
            Jun 13 '17 at 11:20






            @RabMakh You're completely correct, but in this case we're given that $;dim V_3=3;$ , so there can't be only two blocks but three.
            – DonAntonio
            Jun 13 '17 at 11:20






            1




            1




            Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
            – DonAntonio
            Jun 13 '17 at 11:22






            Hmm...wait, re-reading your question, the eigenspace's dimension is not given, right? If so, then there exist two possibilities, indeed: what I wrote and two $';2times2;$ blocks, I believe.
            – DonAntonio
            Jun 13 '17 at 11:22














            @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
            – Peter Melech
            Jun 16 '17 at 7:52




            @DonAntonio "3-eigenspace is 3-dimensional", it is given ...
            – Peter Melech
            Jun 16 '17 at 7:52











            1














            That implies its Jordan normal form is:



            $begin{pmatrix}3&1&0&0&0&0\
            0&3&0&0&0&0\
            0&0&3&0&0&0\
            0&0&0&3&0&0\
            0&0&0&0&i&1\
            0&0&0&0&0&iend{pmatrix}$






            share|cite|improve this answer





















            • why we don't have two 2*2 blocks for eigenvalue 3?
              – fateme jl
              Jun 15 '17 at 12:05










            • Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
              – Peter Melech
              Jun 16 '17 at 7:48
















            1














            That implies its Jordan normal form is:



            $begin{pmatrix}3&1&0&0&0&0\
            0&3&0&0&0&0\
            0&0&3&0&0&0\
            0&0&0&3&0&0\
            0&0&0&0&i&1\
            0&0&0&0&0&iend{pmatrix}$






            share|cite|improve this answer





















            • why we don't have two 2*2 blocks for eigenvalue 3?
              – fateme jl
              Jun 15 '17 at 12:05










            • Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
              – Peter Melech
              Jun 16 '17 at 7:48














            1












            1








            1






            That implies its Jordan normal form is:



            $begin{pmatrix}3&1&0&0&0&0\
            0&3&0&0&0&0\
            0&0&3&0&0&0\
            0&0&0&3&0&0\
            0&0&0&0&i&1\
            0&0&0&0&0&iend{pmatrix}$






            share|cite|improve this answer












            That implies its Jordan normal form is:



            $begin{pmatrix}3&1&0&0&0&0\
            0&3&0&0&0&0\
            0&0&3&0&0&0\
            0&0&0&3&0&0\
            0&0&0&0&i&1\
            0&0&0&0&0&iend{pmatrix}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jun 13 '17 at 10:06









            Peter Melech

            2,529813




            2,529813












            • why we don't have two 2*2 blocks for eigenvalue 3?
              – fateme jl
              Jun 15 '17 at 12:05










            • Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
              – Peter Melech
              Jun 16 '17 at 7:48


















            • why we don't have two 2*2 blocks for eigenvalue 3?
              – fateme jl
              Jun 15 '17 at 12:05










            • Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
              – Peter Melech
              Jun 16 '17 at 7:48
















            why we don't have two 2*2 blocks for eigenvalue 3?
            – fateme jl
            Jun 15 '17 at 12:05




            why we don't have two 2*2 blocks for eigenvalue 3?
            – fateme jl
            Jun 15 '17 at 12:05












            Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
            – Peter Melech
            Jun 16 '17 at 7:48




            Because the dimension of the eigenspace is 3, there must be three Jordan blocks, each one containing one entry corresponding to an eigenvector, because of the exponent 2 in the minimal polynomial the first block is 2*2, the remaining blocks must be 1*1
            – Peter Melech
            Jun 16 '17 at 7:48











            1














            The Jordan form of your matrix must be of the type$$begin{pmatrix}3&*&0&0&0&0\0&3&*&0&0&0\0&0&3&*&0&0\0&0&0&3&0&0\0&0&0&0&i&*\0&0&0&0&0&iend{pmatrix},$$where each $*$ is either $0$ or $1$. But if, for instance, the $*$ from the fifth row is equal to $0$, then the exponent of $t-i$ in the minimal polynomial of the matrix must be $1$. Since it is $2$, that $*$ must be equal to $1$. Now, see what the other three $*$ must be so that the exponent of $t-3$ in the minimal polynomial of the matrix is $2$.






            share|cite|improve this answer


























              1














              The Jordan form of your matrix must be of the type$$begin{pmatrix}3&*&0&0&0&0\0&3&*&0&0&0\0&0&3&*&0&0\0&0&0&3&0&0\0&0&0&0&i&*\0&0&0&0&0&iend{pmatrix},$$where each $*$ is either $0$ or $1$. But if, for instance, the $*$ from the fifth row is equal to $0$, then the exponent of $t-i$ in the minimal polynomial of the matrix must be $1$. Since it is $2$, that $*$ must be equal to $1$. Now, see what the other three $*$ must be so that the exponent of $t-3$ in the minimal polynomial of the matrix is $2$.






              share|cite|improve this answer
























                1












                1








                1






                The Jordan form of your matrix must be of the type$$begin{pmatrix}3&*&0&0&0&0\0&3&*&0&0&0\0&0&3&*&0&0\0&0&0&3&0&0\0&0&0&0&i&*\0&0&0&0&0&iend{pmatrix},$$where each $*$ is either $0$ or $1$. But if, for instance, the $*$ from the fifth row is equal to $0$, then the exponent of $t-i$ in the minimal polynomial of the matrix must be $1$. Since it is $2$, that $*$ must be equal to $1$. Now, see what the other three $*$ must be so that the exponent of $t-3$ in the minimal polynomial of the matrix is $2$.






                share|cite|improve this answer












                The Jordan form of your matrix must be of the type$$begin{pmatrix}3&*&0&0&0&0\0&3&*&0&0&0\0&0&3&*&0&0\0&0&0&3&0&0\0&0&0&0&i&*\0&0&0&0&0&iend{pmatrix},$$where each $*$ is either $0$ or $1$. But if, for instance, the $*$ from the fifth row is equal to $0$, then the exponent of $t-i$ in the minimal polynomial of the matrix must be $1$. Since it is $2$, that $*$ must be equal to $1$. Now, see what the other three $*$ must be so that the exponent of $t-3$ in the minimal polynomial of the matrix is $2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jun 13 '17 at 10:08









                José Carlos Santos

                148k22117218




                148k22117218






























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                    Cabo Verde

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                      Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                    Gyllenstierna

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