Find a Matrix A on the ring of integers modulo 3 so that KerA=ImB.











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B={{1,1,1},{0,1,2},{2,1,0},{0,2,2}}



I understand that each vector from then span of column vectors of B is a solution for Ax=o and that matrix A should have four columns. However I don't know how many rows it should have and how to find it.






linear-algebra matrices modular-arithmetic







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    B={{1,1,1},{0,1,2},{2,1,0},{0,2,2}}



    I understand that each vector from then span of column vectors of B is a solution for Ax=o and that matrix A should have four columns. However I don't know how many rows it should have and how to find it.






    linear-algebra matrices modular-arithmetic







    share|cite|improve this question
























      up vote
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      down vote

      favorite









      up vote
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      down vote

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      B={{1,1,1},{0,1,2},{2,1,0},{0,2,2}}



      I understand that each vector from then span of column vectors of B is a solution for Ax=o and that matrix A should have four columns. However I don't know how many rows it should have and how to find it.






      linear-algebra matrices modular-arithmetic







      share|cite|improve this question













      B={{1,1,1},{0,1,2},{2,1,0},{0,2,2}}



      I understand that each vector from then span of column vectors of B is a solution for Ax=o and that matrix A should have four columns. However I don't know how many rows it should have and how to find it.






      linear-algebra matrices modular-arithmetic




      linear-algebra matrices modular-arithmetic






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Nov 17 at 10:02









      Oleksandr

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          Suppose $A$ has $r$ rows. As $mathbb Z_3$ is a field, by rank-nullity theorem,
          $$
          4 =text{dimension of domain of $A$}=operatorname{rank}(A)+operatorname{nullity}(A)=operatorname{rank}(A)+operatorname{rank}(B).
          $$
          Hence $r:=operatorname{rank}(A)=4-operatorname{rank}(B)$ and $A$ must have at least $r$ rows.






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            Suppose $A$ has $r$ rows. As $mathbb Z_3$ is a field, by rank-nullity theorem,
            $$
            4 =text{dimension of domain of $A$}=operatorname{rank}(A)+operatorname{nullity}(A)=operatorname{rank}(A)+operatorname{rank}(B).
            $$
            Hence $r:=operatorname{rank}(A)=4-operatorname{rank}(B)$ and $A$ must have at least $r$ rows.






            share|cite|improve this answer

























              up vote
              1
              down vote













              Suppose $A$ has $r$ rows. As $mathbb Z_3$ is a field, by rank-nullity theorem,
              $$
              4 =text{dimension of domain of $A$}=operatorname{rank}(A)+operatorname{nullity}(A)=operatorname{rank}(A)+operatorname{rank}(B).
              $$
              Hence $r:=operatorname{rank}(A)=4-operatorname{rank}(B)$ and $A$ must have at least $r$ rows.






              share|cite|improve this answer























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                1
                down vote










                up vote
                1
                down vote









                Suppose $A$ has $r$ rows. As $mathbb Z_3$ is a field, by rank-nullity theorem,
                $$
                4 =text{dimension of domain of $A$}=operatorname{rank}(A)+operatorname{nullity}(A)=operatorname{rank}(A)+operatorname{rank}(B).
                $$
                Hence $r:=operatorname{rank}(A)=4-operatorname{rank}(B)$ and $A$ must have at least $r$ rows.






                share|cite|improve this answer












                Suppose $A$ has $r$ rows. As $mathbb Z_3$ is a field, by rank-nullity theorem,
                $$
                4 =text{dimension of domain of $A$}=operatorname{rank}(A)+operatorname{nullity}(A)=operatorname{rank}(A)+operatorname{rank}(B).
                $$
                Hence $r:=operatorname{rank}(A)=4-operatorname{rank}(B)$ and $A$ must have at least $r$ rows.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 17 at 10:53









                user1551

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