$AB=0$ solving for matrices in MatLab











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I have the equation $AB = 0$, where $A$ and $B$ are matrices. And I know a matrix $B$ and want to find a matrix $A$.



How can I do it in MatLab?










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  • Are there any hypothesis on $B$ ?
    – nicomezi
    Nov 13 at 11:16












  • I know only that matrix $A$ may be in view as $A=(I_k P)$
    – Multifora
    Nov 13 at 11:23












  • Type help null into the command window.
    – user1551
    Nov 13 at 13:09















up vote
2
down vote

favorite












I have the equation $AB = 0$, where $A$ and $B$ are matrices. And I know a matrix $B$ and want to find a matrix $A$.



How can I do it in MatLab?










share|cite|improve this question
























  • Are there any hypothesis on $B$ ?
    – nicomezi
    Nov 13 at 11:16












  • I know only that matrix $A$ may be in view as $A=(I_k P)$
    – Multifora
    Nov 13 at 11:23












  • Type help null into the command window.
    – user1551
    Nov 13 at 13:09













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I have the equation $AB = 0$, where $A$ and $B$ are matrices. And I know a matrix $B$ and want to find a matrix $A$.



How can I do it in MatLab?










share|cite|improve this question















I have the equation $AB = 0$, where $A$ and $B$ are matrices. And I know a matrix $B$ and want to find a matrix $A$.



How can I do it in MatLab?







matrices matlab matrix-equations






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share|cite|improve this question













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edited Nov 13 at 11:13









amWhy

191k27223437




191k27223437










asked Nov 13 at 11:06









Multifora

162




162












  • Are there any hypothesis on $B$ ?
    – nicomezi
    Nov 13 at 11:16












  • I know only that matrix $A$ may be in view as $A=(I_k P)$
    – Multifora
    Nov 13 at 11:23












  • Type help null into the command window.
    – user1551
    Nov 13 at 13:09


















  • Are there any hypothesis on $B$ ?
    – nicomezi
    Nov 13 at 11:16












  • I know only that matrix $A$ may be in view as $A=(I_k P)$
    – Multifora
    Nov 13 at 11:23












  • Type help null into the command window.
    – user1551
    Nov 13 at 13:09
















Are there any hypothesis on $B$ ?
– nicomezi
Nov 13 at 11:16






Are there any hypothesis on $B$ ?
– nicomezi
Nov 13 at 11:16














I know only that matrix $A$ may be in view as $A=(I_k P)$
– Multifora
Nov 13 at 11:23






I know only that matrix $A$ may be in view as $A=(I_k P)$
– Multifora
Nov 13 at 11:23














Type help null into the command window.
– user1551
Nov 13 at 13:09




Type help null into the command window.
– user1551
Nov 13 at 13:09










1 Answer
1






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up vote
0
down vote













Let us assume that all matrices involved are square ($n times n$).



You have first to convert your issue by transposition :



$B^T A^T = 0$



which amounts to say :



$$B^T [V_1 |V_2|...|V_n] = [0 | 0 | ... | 0] iff forall i, B^T V_i=0$$



which means that all $V_i$ should belong to the kernel of $B^T$, giving a manner to build such matrices.



Particular cases :




  • a) The kernel of $B^T$ is reduced to ${0}$ : only the null matrix is possible for $A$.


  • b) The kernel of $B^T$ has dimension 1, with basis $V$, then $A^T=[a_1V|a_2V|...|a_nV]$.



Let us take an example for case (b):



  B=[1 2 3
2 2 4
3 5 8];% matrix with rank<3, here rank(B)=2, thus dim(ker(B'))=1
K=null(B'); % the columns of K are a basis of the Kernel
V=K(:,1); % in fact, in this case, V and K are identical
a=rand(1,3);
A=[a(1)*V,a(2)*V,a(3)*V]';% note the transposition operator
A*B,%test : must be zero





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    1 Answer
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    1 Answer
    1






    active

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    active

    oldest

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    active

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    up vote
    0
    down vote













    Let us assume that all matrices involved are square ($n times n$).



    You have first to convert your issue by transposition :



    $B^T A^T = 0$



    which amounts to say :



    $$B^T [V_1 |V_2|...|V_n] = [0 | 0 | ... | 0] iff forall i, B^T V_i=0$$



    which means that all $V_i$ should belong to the kernel of $B^T$, giving a manner to build such matrices.



    Particular cases :




    • a) The kernel of $B^T$ is reduced to ${0}$ : only the null matrix is possible for $A$.


    • b) The kernel of $B^T$ has dimension 1, with basis $V$, then $A^T=[a_1V|a_2V|...|a_nV]$.



    Let us take an example for case (b):



      B=[1 2 3
    2 2 4
    3 5 8];% matrix with rank<3, here rank(B)=2, thus dim(ker(B'))=1
    K=null(B'); % the columns of K are a basis of the Kernel
    V=K(:,1); % in fact, in this case, V and K are identical
    a=rand(1,3);
    A=[a(1)*V,a(2)*V,a(3)*V]';% note the transposition operator
    A*B,%test : must be zero





    share|cite|improve this answer



























      up vote
      0
      down vote













      Let us assume that all matrices involved are square ($n times n$).



      You have first to convert your issue by transposition :



      $B^T A^T = 0$



      which amounts to say :



      $$B^T [V_1 |V_2|...|V_n] = [0 | 0 | ... | 0] iff forall i, B^T V_i=0$$



      which means that all $V_i$ should belong to the kernel of $B^T$, giving a manner to build such matrices.



      Particular cases :




      • a) The kernel of $B^T$ is reduced to ${0}$ : only the null matrix is possible for $A$.


      • b) The kernel of $B^T$ has dimension 1, with basis $V$, then $A^T=[a_1V|a_2V|...|a_nV]$.



      Let us take an example for case (b):



        B=[1 2 3
      2 2 4
      3 5 8];% matrix with rank<3, here rank(B)=2, thus dim(ker(B'))=1
      K=null(B'); % the columns of K are a basis of the Kernel
      V=K(:,1); % in fact, in this case, V and K are identical
      a=rand(1,3);
      A=[a(1)*V,a(2)*V,a(3)*V]';% note the transposition operator
      A*B,%test : must be zero





      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        Let us assume that all matrices involved are square ($n times n$).



        You have first to convert your issue by transposition :



        $B^T A^T = 0$



        which amounts to say :



        $$B^T [V_1 |V_2|...|V_n] = [0 | 0 | ... | 0] iff forall i, B^T V_i=0$$



        which means that all $V_i$ should belong to the kernel of $B^T$, giving a manner to build such matrices.



        Particular cases :




        • a) The kernel of $B^T$ is reduced to ${0}$ : only the null matrix is possible for $A$.


        • b) The kernel of $B^T$ has dimension 1, with basis $V$, then $A^T=[a_1V|a_2V|...|a_nV]$.



        Let us take an example for case (b):



          B=[1 2 3
        2 2 4
        3 5 8];% matrix with rank<3, here rank(B)=2, thus dim(ker(B'))=1
        K=null(B'); % the columns of K are a basis of the Kernel
        V=K(:,1); % in fact, in this case, V and K are identical
        a=rand(1,3);
        A=[a(1)*V,a(2)*V,a(3)*V]';% note the transposition operator
        A*B,%test : must be zero





        share|cite|improve this answer














        Let us assume that all matrices involved are square ($n times n$).



        You have first to convert your issue by transposition :



        $B^T A^T = 0$



        which amounts to say :



        $$B^T [V_1 |V_2|...|V_n] = [0 | 0 | ... | 0] iff forall i, B^T V_i=0$$



        which means that all $V_i$ should belong to the kernel of $B^T$, giving a manner to build such matrices.



        Particular cases :




        • a) The kernel of $B^T$ is reduced to ${0}$ : only the null matrix is possible for $A$.


        • b) The kernel of $B^T$ has dimension 1, with basis $V$, then $A^T=[a_1V|a_2V|...|a_nV]$.



        Let us take an example for case (b):



          B=[1 2 3
        2 2 4
        3 5 8];% matrix with rank<3, here rank(B)=2, thus dim(ker(B'))=1
        K=null(B'); % the columns of K are a basis of the Kernel
        V=K(:,1); % in fact, in this case, V and K are identical
        a=rand(1,3);
        A=[a(1)*V,a(2)*V,a(3)*V]';% note the transposition operator
        A*B,%test : must be zero






        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 15 at 13:19

























        answered Nov 15 at 13:08









        Jean Marie

        28k41848




        28k41848






























             

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