Find matrices that commute with $operatorname{Diag}(1,1,-1)$.












6












$begingroup$



Let, $A = begin{bmatrix} 1 & 0 & 0 \ 0 & 1& 0 \ 0 & 0 & -1 end{bmatrix}$ . Then find $ S= {B in M_{3times3}(Bbb R) :AB=BA}$ .




My attempt:



Bare computation yielded ${Bin M_{3times3}(Bbb R) : B = begin{bmatrix} c_1 & c_2 & 0 \ c_3 & c_4& 0 \ 0 & 0 & c_5 end{bmatrix} , c_1,c_2,c_3,c_4,c_5 in Bbb R} subseteq S$ but I am having trouble showing the reverse inclusion (if it is true!).



I was also tempted to use eigenvalues and eigenspaces to reach some conclusion by looking at the diagonal form of $A$ (displaying its eigenvalues), but can't get my way through!



Thanks in advance for help.






linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations







share|cite|improve this question











$endgroup$

















    6












    $begingroup$



    Let, $A = begin{bmatrix} 1 & 0 & 0 \ 0 & 1& 0 \ 0 & 0 & -1 end{bmatrix}$ . Then find $ S= {B in M_{3times3}(Bbb R) :AB=BA}$ .




    My attempt:



    Bare computation yielded ${Bin M_{3times3}(Bbb R) : B = begin{bmatrix} c_1 & c_2 & 0 \ c_3 & c_4& 0 \ 0 & 0 & c_5 end{bmatrix} , c_1,c_2,c_3,c_4,c_5 in Bbb R} subseteq S$ but I am having trouble showing the reverse inclusion (if it is true!).



    I was also tempted to use eigenvalues and eigenspaces to reach some conclusion by looking at the diagonal form of $A$ (displaying its eigenvalues), but can't get my way through!



    Thanks in advance for help.






    linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations







    share|cite|improve this question











    $endgroup$















      6












      6








      6





      $begingroup$



      Let, $A = begin{bmatrix} 1 & 0 & 0 \ 0 & 1& 0 \ 0 & 0 & -1 end{bmatrix}$ . Then find $ S= {B in M_{3times3}(Bbb R) :AB=BA}$ .




      My attempt:



      Bare computation yielded ${Bin M_{3times3}(Bbb R) : B = begin{bmatrix} c_1 & c_2 & 0 \ c_3 & c_4& 0 \ 0 & 0 & c_5 end{bmatrix} , c_1,c_2,c_3,c_4,c_5 in Bbb R} subseteq S$ but I am having trouble showing the reverse inclusion (if it is true!).



      I was also tempted to use eigenvalues and eigenspaces to reach some conclusion by looking at the diagonal form of $A$ (displaying its eigenvalues), but can't get my way through!



      Thanks in advance for help.






      linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations







      share|cite|improve this question











      $endgroup$





      Let, $A = begin{bmatrix} 1 & 0 & 0 \ 0 & 1& 0 \ 0 & 0 & -1 end{bmatrix}$ . Then find $ S= {B in M_{3times3}(Bbb R) :AB=BA}$ .




      My attempt:



      Bare computation yielded ${Bin M_{3times3}(Bbb R) : B = begin{bmatrix} c_1 & c_2 & 0 \ c_3 & c_4& 0 \ 0 & 0 & c_5 end{bmatrix} , c_1,c_2,c_3,c_4,c_5 in Bbb R} subseteq S$ but I am having trouble showing the reverse inclusion (if it is true!).



      I was also tempted to use eigenvalues and eigenspaces to reach some conclusion by looking at the diagonal form of $A$ (displaying its eigenvalues), but can't get my way through!



      Thanks in advance for help.






      linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations




      linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 23 at 9:20









      José Carlos Santos

      167k22132235




      167k22132235










      asked Jun 23 '18 at 10:22









      CoherentCoherent

      1,147623




      1,147623






















          2 Answers
          2






          active

          oldest

          votes


















          8












          $begingroup$

          Yes, the reverse inclusion is true. Note that if$$B=begin{pmatrix}b_{11}&b_{12}&b_{13}\b_{21}&b_{22}&b_{23}\b_{31}&b_{32}&b_{33}end{pmatrix},$$then$$AB-BA=begin{pmatrix}0&0&2b_{13}\0&0&2b_{23}\-2b_{31}&-2b_{32}&0end{pmatrix}.$$Therefore, $AB=BA$ if and only if $b_{13}=b_{23}=b_{31}=b_{32}=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you give some idea with the Eigen spaces and Eigen values.
            $endgroup$
            – Coherent
            Jun 23 '18 at 10:30










          • $begingroup$
            The block nature should make it obvious what the eigenspaces are..
            $endgroup$
            – Cameron Williams
            Jun 23 '18 at 10:34










          • $begingroup$
            I meant some arguments by Eigen spaces and Eigen values
            $endgroup$
            – Coherent
            Jun 23 '18 at 10:46










          • $begingroup$
            @ThatIs Right now, I can't think of any.
            $endgroup$
            – José Carlos Santos
            Jun 23 '18 at 10:47










          • $begingroup$
            @JoséCarlosSantos I've tried to give an argument via Eigen spaces and all.
            $endgroup$
            – Coherent
            Oct 7 '18 at 16:31



















          0












          $begingroup$

          Opposed to the direct computation, this is an approach via eigenspaces.



          Let $M,N$ be any two matrices such that $MN=NM$. Let $c$ be an eigenvalue of $M$ and let's call $E_c$ to be the eigenspace associated to the eigen value $c$ .



          Then for any $v in E_c$ , $M(Nv)=(MN)v=(NM)v=N(Mv)=N(cv)=c(Nv)$ . So we get that, $N(E_c) subset E_c$ .



          Now back to our given problem. $A$ clearly has two eigenvalues : 1 and -1 with the former having multiplicity 2 and the latter having multiplicity 1.



          A routine computation shows that, $E_1 = span{(1,0,0),(0,1,0)}$ and $E_{-1} = span{(0,0,1)}$.



          So we get that, $B(1,0,0) in E_1, B(0,1,0) in E_1, B(0,0,1) in E_{-1} $ i.e. $exists c_1, dots ,c_5in Bbb R$ such that ,$$B(1,0,0)=c_1(1,0,0)+c_2 (0,1,0) + 0(0,0,1)$$ $$B(0,1,0)=c_3(1,0,0)+c_4 (0,1,0)+0(0,0,1)$$ $$B(0,0,1)=0(1,0,0)+0 (0,1,0)+c_5(0,0,1)$$



          And hence, in the standard basis, we get, $$B=begin{bmatrix} c_1 & c_2 & 0 \ c_3 & c_4 & 0 \ 0 & 0 & c_5 end{bmatrix}$$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
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            active

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            8












            $begingroup$

            Yes, the reverse inclusion is true. Note that if$$B=begin{pmatrix}b_{11}&b_{12}&b_{13}\b_{21}&b_{22}&b_{23}\b_{31}&b_{32}&b_{33}end{pmatrix},$$then$$AB-BA=begin{pmatrix}0&0&2b_{13}\0&0&2b_{23}\-2b_{31}&-2b_{32}&0end{pmatrix}.$$Therefore, $AB=BA$ if and only if $b_{13}=b_{23}=b_{31}=b_{32}=0$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Can you give some idea with the Eigen spaces and Eigen values.
              $endgroup$
              – Coherent
              Jun 23 '18 at 10:30










            • $begingroup$
              The block nature should make it obvious what the eigenspaces are..
              $endgroup$
              – Cameron Williams
              Jun 23 '18 at 10:34










            • $begingroup$
              I meant some arguments by Eigen spaces and Eigen values
              $endgroup$
              – Coherent
              Jun 23 '18 at 10:46










            • $begingroup$
              @ThatIs Right now, I can't think of any.
              $endgroup$
              – José Carlos Santos
              Jun 23 '18 at 10:47










            • $begingroup$
              @JoséCarlosSantos I've tried to give an argument via Eigen spaces and all.
              $endgroup$
              – Coherent
              Oct 7 '18 at 16:31
















            8












            $begingroup$

            Yes, the reverse inclusion is true. Note that if$$B=begin{pmatrix}b_{11}&b_{12}&b_{13}\b_{21}&b_{22}&b_{23}\b_{31}&b_{32}&b_{33}end{pmatrix},$$then$$AB-BA=begin{pmatrix}0&0&2b_{13}\0&0&2b_{23}\-2b_{31}&-2b_{32}&0end{pmatrix}.$$Therefore, $AB=BA$ if and only if $b_{13}=b_{23}=b_{31}=b_{32}=0$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Can you give some idea with the Eigen spaces and Eigen values.
              $endgroup$
              – Coherent
              Jun 23 '18 at 10:30










            • $begingroup$
              The block nature should make it obvious what the eigenspaces are..
              $endgroup$
              – Cameron Williams
              Jun 23 '18 at 10:34










            • $begingroup$
              I meant some arguments by Eigen spaces and Eigen values
              $endgroup$
              – Coherent
              Jun 23 '18 at 10:46










            • $begingroup$
              @ThatIs Right now, I can't think of any.
              $endgroup$
              – José Carlos Santos
              Jun 23 '18 at 10:47










            • $begingroup$
              @JoséCarlosSantos I've tried to give an argument via Eigen spaces and all.
              $endgroup$
              – Coherent
              Oct 7 '18 at 16:31














            8












            8








            8





            $begingroup$

            Yes, the reverse inclusion is true. Note that if$$B=begin{pmatrix}b_{11}&b_{12}&b_{13}\b_{21}&b_{22}&b_{23}\b_{31}&b_{32}&b_{33}end{pmatrix},$$then$$AB-BA=begin{pmatrix}0&0&2b_{13}\0&0&2b_{23}\-2b_{31}&-2b_{32}&0end{pmatrix}.$$Therefore, $AB=BA$ if and only if $b_{13}=b_{23}=b_{31}=b_{32}=0$.






            share|cite|improve this answer









            $endgroup$



            Yes, the reverse inclusion is true. Note that if$$B=begin{pmatrix}b_{11}&b_{12}&b_{13}\b_{21}&b_{22}&b_{23}\b_{31}&b_{32}&b_{33}end{pmatrix},$$then$$AB-BA=begin{pmatrix}0&0&2b_{13}\0&0&2b_{23}\-2b_{31}&-2b_{32}&0end{pmatrix}.$$Therefore, $AB=BA$ if and only if $b_{13}=b_{23}=b_{31}=b_{32}=0$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jun 23 '18 at 10:29









            José Carlos SantosJosé Carlos Santos

            167k22132235




            167k22132235












            • $begingroup$
              Can you give some idea with the Eigen spaces and Eigen values.
              $endgroup$
              – Coherent
              Jun 23 '18 at 10:30










            • $begingroup$
              The block nature should make it obvious what the eigenspaces are..
              $endgroup$
              – Cameron Williams
              Jun 23 '18 at 10:34










            • $begingroup$
              I meant some arguments by Eigen spaces and Eigen values
              $endgroup$
              – Coherent
              Jun 23 '18 at 10:46










            • $begingroup$
              @ThatIs Right now, I can't think of any.
              $endgroup$
              – José Carlos Santos
              Jun 23 '18 at 10:47










            • $begingroup$
              @JoséCarlosSantos I've tried to give an argument via Eigen spaces and all.
              $endgroup$
              – Coherent
              Oct 7 '18 at 16:31


















            • $begingroup$
              Can you give some idea with the Eigen spaces and Eigen values.
              $endgroup$
              – Coherent
              Jun 23 '18 at 10:30










            • $begingroup$
              The block nature should make it obvious what the eigenspaces are..
              $endgroup$
              – Cameron Williams
              Jun 23 '18 at 10:34










            • $begingroup$
              I meant some arguments by Eigen spaces and Eigen values
              $endgroup$
              – Coherent
              Jun 23 '18 at 10:46










            • $begingroup$
              @ThatIs Right now, I can't think of any.
              $endgroup$
              – José Carlos Santos
              Jun 23 '18 at 10:47










            • $begingroup$
              @JoséCarlosSantos I've tried to give an argument via Eigen spaces and all.
              $endgroup$
              – Coherent
              Oct 7 '18 at 16:31
















            $begingroup$
            Can you give some idea with the Eigen spaces and Eigen values.
            $endgroup$
            – Coherent
            Jun 23 '18 at 10:30




            $begingroup$
            Can you give some idea with the Eigen spaces and Eigen values.
            $endgroup$
            – Coherent
            Jun 23 '18 at 10:30












            $begingroup$
            The block nature should make it obvious what the eigenspaces are..
            $endgroup$
            – Cameron Williams
            Jun 23 '18 at 10:34




            $begingroup$
            The block nature should make it obvious what the eigenspaces are..
            $endgroup$
            – Cameron Williams
            Jun 23 '18 at 10:34












            $begingroup$
            I meant some arguments by Eigen spaces and Eigen values
            $endgroup$
            – Coherent
            Jun 23 '18 at 10:46




            $begingroup$
            I meant some arguments by Eigen spaces and Eigen values
            $endgroup$
            – Coherent
            Jun 23 '18 at 10:46












            $begingroup$
            @ThatIs Right now, I can't think of any.
            $endgroup$
            – José Carlos Santos
            Jun 23 '18 at 10:47




            $begingroup$
            @ThatIs Right now, I can't think of any.
            $endgroup$
            – José Carlos Santos
            Jun 23 '18 at 10:47












            $begingroup$
            @JoséCarlosSantos I've tried to give an argument via Eigen spaces and all.
            $endgroup$
            – Coherent
            Oct 7 '18 at 16:31




            $begingroup$
            @JoséCarlosSantos I've tried to give an argument via Eigen spaces and all.
            $endgroup$
            – Coherent
            Oct 7 '18 at 16:31











            0












            $begingroup$

            Opposed to the direct computation, this is an approach via eigenspaces.



            Let $M,N$ be any two matrices such that $MN=NM$. Let $c$ be an eigenvalue of $M$ and let's call $E_c$ to be the eigenspace associated to the eigen value $c$ .



            Then for any $v in E_c$ , $M(Nv)=(MN)v=(NM)v=N(Mv)=N(cv)=c(Nv)$ . So we get that, $N(E_c) subset E_c$ .



            Now back to our given problem. $A$ clearly has two eigenvalues : 1 and -1 with the former having multiplicity 2 and the latter having multiplicity 1.



            A routine computation shows that, $E_1 = span{(1,0,0),(0,1,0)}$ and $E_{-1} = span{(0,0,1)}$.



            So we get that, $B(1,0,0) in E_1, B(0,1,0) in E_1, B(0,0,1) in E_{-1} $ i.e. $exists c_1, dots ,c_5in Bbb R$ such that ,$$B(1,0,0)=c_1(1,0,0)+c_2 (0,1,0) + 0(0,0,1)$$ $$B(0,1,0)=c_3(1,0,0)+c_4 (0,1,0)+0(0,0,1)$$ $$B(0,0,1)=0(1,0,0)+0 (0,1,0)+c_5(0,0,1)$$



            And hence, in the standard basis, we get, $$B=begin{bmatrix} c_1 & c_2 & 0 \ c_3 & c_4 & 0 \ 0 & 0 & c_5 end{bmatrix}$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Opposed to the direct computation, this is an approach via eigenspaces.



              Let $M,N$ be any two matrices such that $MN=NM$. Let $c$ be an eigenvalue of $M$ and let's call $E_c$ to be the eigenspace associated to the eigen value $c$ .



              Then for any $v in E_c$ , $M(Nv)=(MN)v=(NM)v=N(Mv)=N(cv)=c(Nv)$ . So we get that, $N(E_c) subset E_c$ .



              Now back to our given problem. $A$ clearly has two eigenvalues : 1 and -1 with the former having multiplicity 2 and the latter having multiplicity 1.



              A routine computation shows that, $E_1 = span{(1,0,0),(0,1,0)}$ and $E_{-1} = span{(0,0,1)}$.



              So we get that, $B(1,0,0) in E_1, B(0,1,0) in E_1, B(0,0,1) in E_{-1} $ i.e. $exists c_1, dots ,c_5in Bbb R$ such that ,$$B(1,0,0)=c_1(1,0,0)+c_2 (0,1,0) + 0(0,0,1)$$ $$B(0,1,0)=c_3(1,0,0)+c_4 (0,1,0)+0(0,0,1)$$ $$B(0,0,1)=0(1,0,0)+0 (0,1,0)+c_5(0,0,1)$$



              And hence, in the standard basis, we get, $$B=begin{bmatrix} c_1 & c_2 & 0 \ c_3 & c_4 & 0 \ 0 & 0 & c_5 end{bmatrix}$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Opposed to the direct computation, this is an approach via eigenspaces.



                Let $M,N$ be any two matrices such that $MN=NM$. Let $c$ be an eigenvalue of $M$ and let's call $E_c$ to be the eigenspace associated to the eigen value $c$ .



                Then for any $v in E_c$ , $M(Nv)=(MN)v=(NM)v=N(Mv)=N(cv)=c(Nv)$ . So we get that, $N(E_c) subset E_c$ .



                Now back to our given problem. $A$ clearly has two eigenvalues : 1 and -1 with the former having multiplicity 2 and the latter having multiplicity 1.



                A routine computation shows that, $E_1 = span{(1,0,0),(0,1,0)}$ and $E_{-1} = span{(0,0,1)}$.



                So we get that, $B(1,0,0) in E_1, B(0,1,0) in E_1, B(0,0,1) in E_{-1} $ i.e. $exists c_1, dots ,c_5in Bbb R$ such that ,$$B(1,0,0)=c_1(1,0,0)+c_2 (0,1,0) + 0(0,0,1)$$ $$B(0,1,0)=c_3(1,0,0)+c_4 (0,1,0)+0(0,0,1)$$ $$B(0,0,1)=0(1,0,0)+0 (0,1,0)+c_5(0,0,1)$$



                And hence, in the standard basis, we get, $$B=begin{bmatrix} c_1 & c_2 & 0 \ c_3 & c_4 & 0 \ 0 & 0 & c_5 end{bmatrix}$$






                share|cite|improve this answer









                $endgroup$



                Opposed to the direct computation, this is an approach via eigenspaces.



                Let $M,N$ be any two matrices such that $MN=NM$. Let $c$ be an eigenvalue of $M$ and let's call $E_c$ to be the eigenspace associated to the eigen value $c$ .



                Then for any $v in E_c$ , $M(Nv)=(MN)v=(NM)v=N(Mv)=N(cv)=c(Nv)$ . So we get that, $N(E_c) subset E_c$ .



                Now back to our given problem. $A$ clearly has two eigenvalues : 1 and -1 with the former having multiplicity 2 and the latter having multiplicity 1.



                A routine computation shows that, $E_1 = span{(1,0,0),(0,1,0)}$ and $E_{-1} = span{(0,0,1)}$.



                So we get that, $B(1,0,0) in E_1, B(0,1,0) in E_1, B(0,0,1) in E_{-1} $ i.e. $exists c_1, dots ,c_5in Bbb R$ such that ,$$B(1,0,0)=c_1(1,0,0)+c_2 (0,1,0) + 0(0,0,1)$$ $$B(0,1,0)=c_3(1,0,0)+c_4 (0,1,0)+0(0,0,1)$$ $$B(0,0,1)=0(1,0,0)+0 (0,1,0)+c_5(0,0,1)$$



                And hence, in the standard basis, we get, $$B=begin{bmatrix} c_1 & c_2 & 0 \ c_3 & c_4 & 0 \ 0 & 0 & c_5 end{bmatrix}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Oct 7 '18 at 16:29









                CoherentCoherent

                1,147623




                1,147623






























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                    in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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                    1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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