finding the dimension of eigenspace with characteristic and minimal polynomial












0












$begingroup$


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











$endgroup$

















    0












    $begingroup$


    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











    $endgroup$















      0












      0








      0





      $begingroup$


      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











      $endgroup$




      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 '18 at 9:27









      José Carlos Santos

      153k22123225




      153k22123225










      asked Jun 13 '17 at 9:46









      fateme jlfateme jl

      8710




      8710






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          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









          $endgroup$













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










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










          • $begingroup$
            @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.
            $endgroup$
            – DonAntonio
            Jun 13 '17 at 11:20








          • 1




            $begingroup$
            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.
            $endgroup$
            – DonAntonio
            Jun 13 '17 at 11:22












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



















          1












          $begingroup$

          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









          $endgroup$













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










          • $begingroup$
            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
            $endgroup$
            – Peter Melech
            Jun 16 '17 at 7:48



















          1












          $begingroup$

          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









          $endgroup$













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            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









            $endgroup$













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










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










            • $begingroup$
              @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.
              $endgroup$
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              $begingroup$
              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.
              $endgroup$
              – DonAntonio
              Jun 13 '17 at 11:22












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
















            1












            $begingroup$

            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









            $endgroup$













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










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










            • $begingroup$
              @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.
              $endgroup$
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              $begingroup$
              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.
              $endgroup$
              – DonAntonio
              Jun 13 '17 at 11:22












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














            1












            1








            1





            $begingroup$

            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









            $endgroup$



            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









            DonAntonioDonAntonio

            177k1492225




            177k1492225












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










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










            • $begingroup$
              @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.
              $endgroup$
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              $begingroup$
              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.
              $endgroup$
              – DonAntonio
              Jun 13 '17 at 11:22












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


















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










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










            • $begingroup$
              @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.
              $endgroup$
              – DonAntonio
              Jun 13 '17 at 11:20








            • 1




              $begingroup$
              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.
              $endgroup$
              – DonAntonio
              Jun 13 '17 at 11:22












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
















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




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












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




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












            $begingroup$
            @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.
            $endgroup$
            – DonAntonio
            Jun 13 '17 at 11:20






            $begingroup$
            @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.
            $endgroup$
            – DonAntonio
            Jun 13 '17 at 11:20






            1




            1




            $begingroup$
            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.
            $endgroup$
            – DonAntonio
            Jun 13 '17 at 11:22






            $begingroup$
            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.
            $endgroup$
            – DonAntonio
            Jun 13 '17 at 11:22














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




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











            1












            $begingroup$

            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









            $endgroup$













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










            • $begingroup$
              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
              $endgroup$
              – Peter Melech
              Jun 16 '17 at 7:48
















            1












            $begingroup$

            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









            $endgroup$













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










            • $begingroup$
              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
              $endgroup$
              – Peter Melech
              Jun 16 '17 at 7:48














            1












            1








            1





            $begingroup$

            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









            $endgroup$



            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 MelechPeter Melech

            2,562813




            2,562813












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










            • $begingroup$
              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
              $endgroup$
              – Peter Melech
              Jun 16 '17 at 7:48


















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










            • $begingroup$
              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
              $endgroup$
              – Peter Melech
              Jun 16 '17 at 7:48
















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




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












            $begingroup$
            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
            $endgroup$
            – Peter Melech
            Jun 16 '17 at 7:48




            $begingroup$
            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
            $endgroup$
            – Peter Melech
            Jun 16 '17 at 7:48











            1












            $begingroup$

            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









            $endgroup$


















              1












              $begingroup$

              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









              $endgroup$
















                1












                1








                1





                $begingroup$

                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









                $endgroup$



                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 $2$.







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                answered Jun 13 '17 at 10:08









                José Carlos SantosJosé Carlos Santos

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