Determine dimension of set of matrices given that each matrix has same eigenvalue 0 and eigenvector












1












$begingroup$


Let V be the set of all $A_{3×3}$ matrices, where all of them have eigenvalue 0 and corresponding eigenvector $y=(1,2,3)^T$.



Assuming we have showed that V is a vector space, determine the dimension of V.




From the definition we know that:
$x$ and $lambda$ are eigenvector and eigenvalue iff $(A-lambda I)x=0$ .



So for each matrix $A_i$ in our set$$
(A_i-0)
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
=0
$$
$$
A_i
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
=0
$$
$$
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
in N(A_i)
$$
From this we find that $Nullity(A_i)geq1 Rightarrow $



$Rank(A_i) = n-Nullity(A_i)=3-1=2 Rightarrow $



$Rank(A_i)=Dim(C(A_i)) leq 2$



However, I am not sure how can I proceed from here?



1) How can determine the dimension of the matrix $A_i$ for sure?



2) How is that connected to the dimension of entire vector space formed by such matrices?






linear-algebra vector-spaces eigenvalues-eigenvectors







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let V be the set of all $A_{3×3}$ matrices, where all of them have eigenvalue 0 and corresponding eigenvector $y=(1,2,3)^T$.



    Assuming we have showed that V is a vector space, determine the dimension of V.




    From the definition we know that:
    $x$ and $lambda$ are eigenvector and eigenvalue iff $(A-lambda I)x=0$ .



    So for each matrix $A_i$ in our set$$
    (A_i-0)
    begin{pmatrix}
    1 \
    2 \
    3 \
    end{pmatrix}
    =0
    $$
    $$
    A_i
    begin{pmatrix}
    1 \
    2 \
    3 \
    end{pmatrix}
    =0
    $$
    $$
    begin{pmatrix}
    1 \
    2 \
    3 \
    end{pmatrix}
    in N(A_i)
    $$
    From this we find that $Nullity(A_i)geq1 Rightarrow $



    $Rank(A_i) = n-Nullity(A_i)=3-1=2 Rightarrow $



    $Rank(A_i)=Dim(C(A_i)) leq 2$



    However, I am not sure how can I proceed from here?



    1) How can determine the dimension of the matrix $A_i$ for sure?



    2) How is that connected to the dimension of entire vector space formed by such matrices?






    linear-algebra vector-spaces eigenvalues-eigenvectors







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let V be the set of all $A_{3×3}$ matrices, where all of them have eigenvalue 0 and corresponding eigenvector $y=(1,2,3)^T$.



      Assuming we have showed that V is a vector space, determine the dimension of V.




      From the definition we know that:
      $x$ and $lambda$ are eigenvector and eigenvalue iff $(A-lambda I)x=0$ .



      So for each matrix $A_i$ in our set$$
      (A_i-0)
      begin{pmatrix}
      1 \
      2 \
      3 \
      end{pmatrix}
      =0
      $$
      $$
      A_i
      begin{pmatrix}
      1 \
      2 \
      3 \
      end{pmatrix}
      =0
      $$
      $$
      begin{pmatrix}
      1 \
      2 \
      3 \
      end{pmatrix}
      in N(A_i)
      $$
      From this we find that $Nullity(A_i)geq1 Rightarrow $



      $Rank(A_i) = n-Nullity(A_i)=3-1=2 Rightarrow $



      $Rank(A_i)=Dim(C(A_i)) leq 2$



      However, I am not sure how can I proceed from here?



      1) How can determine the dimension of the matrix $A_i$ for sure?



      2) How is that connected to the dimension of entire vector space formed by such matrices?






      linear-algebra vector-spaces eigenvalues-eigenvectors







      share|cite|improve this question









      $endgroup$




      Let V be the set of all $A_{3×3}$ matrices, where all of them have eigenvalue 0 and corresponding eigenvector $y=(1,2,3)^T$.



      Assuming we have showed that V is a vector space, determine the dimension of V.




      From the definition we know that:
      $x$ and $lambda$ are eigenvector and eigenvalue iff $(A-lambda I)x=0$ .



      So for each matrix $A_i$ in our set$$
      (A_i-0)
      begin{pmatrix}
      1 \
      2 \
      3 \
      end{pmatrix}
      =0
      $$
      $$
      A_i
      begin{pmatrix}
      1 \
      2 \
      3 \
      end{pmatrix}
      =0
      $$
      $$
      begin{pmatrix}
      1 \
      2 \
      3 \
      end{pmatrix}
      in N(A_i)
      $$
      From this we find that $Nullity(A_i)geq1 Rightarrow $



      $Rank(A_i) = n-Nullity(A_i)=3-1=2 Rightarrow $



      $Rank(A_i)=Dim(C(A_i)) leq 2$



      However, I am not sure how can I proceed from here?



      1) How can determine the dimension of the matrix $A_i$ for sure?



      2) How is that connected to the dimension of entire vector space formed by such matrices?






      linear-algebra vector-spaces eigenvalues-eigenvectors




      linear-algebra vector-spaces eigenvalues-eigenvectors






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 15 at 19:38









      Jill WhiteJill White

      133




      133






















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