Jordan Normal Form: Two times the same basis vector?!












0












$begingroup$


I have 3 dimensional matrix $$A =
left(begin{array}{c} 2 & 1 & 0 \ -1 & 0 & 1 \ 1 & 3 & 1end{array}right)$$
and want to find a Jordan Form for it and a basis for the Jordan Form. My procedure: I calculated the characteristic polynomial $chi_A(lambda) = -(2-lambda)^2(1+lambda)$ and found the roots $lambda_1 = 2$ with algebraic multiplicity $mu_1 = 2$ and $lambda_2 = -1$ with algebraic multiplicity $mu_2 = -1$, respectively. Then, for $lambda_1$, I found that a basis for the kernel of $A - 2 I$ is the vector $left(begin{array}{c} 1 \ 0 \ 1end{array}right).$ Clearly, this has subspace has dimension $gamma_{11} = 1$ which is less than $mu_1 = 2$, so I have to continue and calculate the kernel of $(A - 2I)^2$. A basis for this space is given by $left(begin{array}{c} 1 \ 0 \ 1end{array}right), left(begin{array}{c} 1 \ 1 \ 0end{array}right).$ Since now the geometric multiplicity equals the algebraic multiplicity, I am finished with calculating kernels. Now I have to pick some vector $w_{12}$ in the kernel of $(A-2I)^2$ which is not in $(A-2 I)$. An obvious choice is $w_{12} = left(begin{array}{c} 1 \ 1 \ 0end{array}right)$. Then: $$w_{11} = (A - 2I) = left(begin{array}{c}1 \ -3 \ 4end{array}right).$$
Now turning to $lambda_2$, a basis for the kernel is $left(begin{array}{c} 1 \ -3 \ 4end{array}right)$.



But then I get stuck because I have two times the exact same vector in my basis which of course is not enough to span a 3 dimensional space. I cannot see what I did wrong or where my mistake comes from. What do I do in such a situation?






linear-algebra jordan-normal-form







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  • $begingroup$
    You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
    $endgroup$
    – amd
    Dec 13 '18 at 0:49
















0












$begingroup$


I have 3 dimensional matrix $$A =
left(begin{array}{c} 2 & 1 & 0 \ -1 & 0 & 1 \ 1 & 3 & 1end{array}right)$$
and want to find a Jordan Form for it and a basis for the Jordan Form. My procedure: I calculated the characteristic polynomial $chi_A(lambda) = -(2-lambda)^2(1+lambda)$ and found the roots $lambda_1 = 2$ with algebraic multiplicity $mu_1 = 2$ and $lambda_2 = -1$ with algebraic multiplicity $mu_2 = -1$, respectively. Then, for $lambda_1$, I found that a basis for the kernel of $A - 2 I$ is the vector $left(begin{array}{c} 1 \ 0 \ 1end{array}right).$ Clearly, this has subspace has dimension $gamma_{11} = 1$ which is less than $mu_1 = 2$, so I have to continue and calculate the kernel of $(A - 2I)^2$. A basis for this space is given by $left(begin{array}{c} 1 \ 0 \ 1end{array}right), left(begin{array}{c} 1 \ 1 \ 0end{array}right).$ Since now the geometric multiplicity equals the algebraic multiplicity, I am finished with calculating kernels. Now I have to pick some vector $w_{12}$ in the kernel of $(A-2I)^2$ which is not in $(A-2 I)$. An obvious choice is $w_{12} = left(begin{array}{c} 1 \ 1 \ 0end{array}right)$. Then: $$w_{11} = (A - 2I) = left(begin{array}{c}1 \ -3 \ 4end{array}right).$$
Now turning to $lambda_2$, a basis for the kernel is $left(begin{array}{c} 1 \ -3 \ 4end{array}right)$.



But then I get stuck because I have two times the exact same vector in my basis which of course is not enough to span a 3 dimensional space. I cannot see what I did wrong or where my mistake comes from. What do I do in such a situation?






linear-algebra jordan-normal-form







share|cite|improve this question









$endgroup$












  • $begingroup$
    You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
    $endgroup$
    – amd
    Dec 13 '18 at 0:49














0












0








0





$begingroup$


I have 3 dimensional matrix $$A =
left(begin{array}{c} 2 & 1 & 0 \ -1 & 0 & 1 \ 1 & 3 & 1end{array}right)$$
and want to find a Jordan Form for it and a basis for the Jordan Form. My procedure: I calculated the characteristic polynomial $chi_A(lambda) = -(2-lambda)^2(1+lambda)$ and found the roots $lambda_1 = 2$ with algebraic multiplicity $mu_1 = 2$ and $lambda_2 = -1$ with algebraic multiplicity $mu_2 = -1$, respectively. Then, for $lambda_1$, I found that a basis for the kernel of $A - 2 I$ is the vector $left(begin{array}{c} 1 \ 0 \ 1end{array}right).$ Clearly, this has subspace has dimension $gamma_{11} = 1$ which is less than $mu_1 = 2$, so I have to continue and calculate the kernel of $(A - 2I)^2$. A basis for this space is given by $left(begin{array}{c} 1 \ 0 \ 1end{array}right), left(begin{array}{c} 1 \ 1 \ 0end{array}right).$ Since now the geometric multiplicity equals the algebraic multiplicity, I am finished with calculating kernels. Now I have to pick some vector $w_{12}$ in the kernel of $(A-2I)^2$ which is not in $(A-2 I)$. An obvious choice is $w_{12} = left(begin{array}{c} 1 \ 1 \ 0end{array}right)$. Then: $$w_{11} = (A - 2I) = left(begin{array}{c}1 \ -3 \ 4end{array}right).$$
Now turning to $lambda_2$, a basis for the kernel is $left(begin{array}{c} 1 \ -3 \ 4end{array}right)$.



But then I get stuck because I have two times the exact same vector in my basis which of course is not enough to span a 3 dimensional space. I cannot see what I did wrong or where my mistake comes from. What do I do in such a situation?






linear-algebra jordan-normal-form







share|cite|improve this question









$endgroup$




I have 3 dimensional matrix $$A =
left(begin{array}{c} 2 & 1 & 0 \ -1 & 0 & 1 \ 1 & 3 & 1end{array}right)$$
and want to find a Jordan Form for it and a basis for the Jordan Form. My procedure: I calculated the characteristic polynomial $chi_A(lambda) = -(2-lambda)^2(1+lambda)$ and found the roots $lambda_1 = 2$ with algebraic multiplicity $mu_1 = 2$ and $lambda_2 = -1$ with algebraic multiplicity $mu_2 = -1$, respectively. Then, for $lambda_1$, I found that a basis for the kernel of $A - 2 I$ is the vector $left(begin{array}{c} 1 \ 0 \ 1end{array}right).$ Clearly, this has subspace has dimension $gamma_{11} = 1$ which is less than $mu_1 = 2$, so I have to continue and calculate the kernel of $(A - 2I)^2$. A basis for this space is given by $left(begin{array}{c} 1 \ 0 \ 1end{array}right), left(begin{array}{c} 1 \ 1 \ 0end{array}right).$ Since now the geometric multiplicity equals the algebraic multiplicity, I am finished with calculating kernels. Now I have to pick some vector $w_{12}$ in the kernel of $(A-2I)^2$ which is not in $(A-2 I)$. An obvious choice is $w_{12} = left(begin{array}{c} 1 \ 1 \ 0end{array}right)$. Then: $$w_{11} = (A - 2I) = left(begin{array}{c}1 \ -3 \ 4end{array}right).$$
Now turning to $lambda_2$, a basis for the kernel is $left(begin{array}{c} 1 \ -3 \ 4end{array}right)$.



But then I get stuck because I have two times the exact same vector in my basis which of course is not enough to span a 3 dimensional space. I cannot see what I did wrong or where my mistake comes from. What do I do in such a situation?






linear-algebra jordan-normal-form




linear-algebra jordan-normal-form






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asked Dec 12 '18 at 23:05









Syd AmerikanerSyd Amerikaner

138213




138213












  • $begingroup$
    You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
    $endgroup$
    – amd
    Dec 13 '18 at 0:49


















  • $begingroup$
    You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
    $endgroup$
    – amd
    Dec 13 '18 at 0:49
















$begingroup$
You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
$endgroup$
– amd
Dec 13 '18 at 0:49




$begingroup$
You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
$endgroup$
– amd
Dec 13 '18 at 0:49










2 Answers
2






active

oldest

votes


















1












$begingroup$

$$ (A-2I)^2 =
left(
begin{array}{ccc}
-1&-2&1 \
3&6&-3 \
-4&-8&4 \
end{array}
right)
$$
of rank one, with row echelon form
$$ (A-2I)^2 Longrightarrow
left(
begin{array}{ccc}
1&2&-1 \
0&0&0 \
0&0&0 \
end{array}
right)
$$



Your vector $w_{12}$ is not in the kernel of $ (A-2I)^2 ; ; ;$ your basis for that kernel is wrong.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
    $endgroup$
    – Will Jagy
    Dec 15 '18 at 18:49



















0












$begingroup$

Where do you get $pmatrix{1\1\0}$?



What do you have for $(A-2I)^2$?



I have $(A-2I)^2 = pmatrix{-1&-2&1\3&6&-3\-4&-8&4}$



And in addition to $pmatrix{1\0\1}$, I see $pmatrix{0\1\2}$ as a candidate for the second eigenvector.



$Apmatrix{0\1\2} = pmatrix{1\2\5} = 2pmatrix{0\1\2}+pmatrix{1\0\1}$ which is exactly what we were hoping for.



$Apmatrix{1&1&0\-3&0&1\4&2&1} = pmatrix{1&1&0\-3&0&1\4&2&1}pmatrix{-1&0&0\0&2&1\0&0&2}$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $$ (A-2I)^2 =
    left(
    begin{array}{ccc}
    -1&-2&1 \
    3&6&-3 \
    -4&-8&4 \
    end{array}
    right)
    $$
    of rank one, with row echelon form
    $$ (A-2I)^2 Longrightarrow
    left(
    begin{array}{ccc}
    1&2&-1 \
    0&0&0 \
    0&0&0 \
    end{array}
    right)
    $$



    Your vector $w_{12}$ is not in the kernel of $ (A-2I)^2 ; ; ;$ your basis for that kernel is wrong.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
      $endgroup$
      – Will Jagy
      Dec 15 '18 at 18:49
















    1












    $begingroup$

    $$ (A-2I)^2 =
    left(
    begin{array}{ccc}
    -1&-2&1 \
    3&6&-3 \
    -4&-8&4 \
    end{array}
    right)
    $$
    of rank one, with row echelon form
    $$ (A-2I)^2 Longrightarrow
    left(
    begin{array}{ccc}
    1&2&-1 \
    0&0&0 \
    0&0&0 \
    end{array}
    right)
    $$



    Your vector $w_{12}$ is not in the kernel of $ (A-2I)^2 ; ; ;$ your basis for that kernel is wrong.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
      $endgroup$
      – Will Jagy
      Dec 15 '18 at 18:49














    1












    1








    1





    $begingroup$

    $$ (A-2I)^2 =
    left(
    begin{array}{ccc}
    -1&-2&1 \
    3&6&-3 \
    -4&-8&4 \
    end{array}
    right)
    $$
    of rank one, with row echelon form
    $$ (A-2I)^2 Longrightarrow
    left(
    begin{array}{ccc}
    1&2&-1 \
    0&0&0 \
    0&0&0 \
    end{array}
    right)
    $$



    Your vector $w_{12}$ is not in the kernel of $ (A-2I)^2 ; ; ;$ your basis for that kernel is wrong.






    share|cite|improve this answer









    $endgroup$



    $$ (A-2I)^2 =
    left(
    begin{array}{ccc}
    -1&-2&1 \
    3&6&-3 \
    -4&-8&4 \
    end{array}
    right)
    $$
    of rank one, with row echelon form
    $$ (A-2I)^2 Longrightarrow
    left(
    begin{array}{ccc}
    1&2&-1 \
    0&0&0 \
    0&0&0 \
    end{array}
    right)
    $$



    Your vector $w_{12}$ is not in the kernel of $ (A-2I)^2 ; ; ;$ your basis for that kernel is wrong.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 12 '18 at 23:22









    Will JagyWill Jagy

    104k5102201




    104k5102201












    • $begingroup$
      @Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
      $endgroup$
      – Will Jagy
      Dec 15 '18 at 18:49


















    • $begingroup$
      @Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
      $endgroup$
      – Will Jagy
      Dec 15 '18 at 18:49
















    $begingroup$
    @Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
    $endgroup$
    – Will Jagy
    Dec 15 '18 at 18:49




    $begingroup$
    @Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
    $endgroup$
    – Will Jagy
    Dec 15 '18 at 18:49











    0












    $begingroup$

    Where do you get $pmatrix{1\1\0}$?



    What do you have for $(A-2I)^2$?



    I have $(A-2I)^2 = pmatrix{-1&-2&1\3&6&-3\-4&-8&4}$



    And in addition to $pmatrix{1\0\1}$, I see $pmatrix{0\1\2}$ as a candidate for the second eigenvector.



    $Apmatrix{0\1\2} = pmatrix{1\2\5} = 2pmatrix{0\1\2}+pmatrix{1\0\1}$ which is exactly what we were hoping for.



    $Apmatrix{1&1&0\-3&0&1\4&2&1} = pmatrix{1&1&0\-3&0&1\4&2&1}pmatrix{-1&0&0\0&2&1\0&0&2}$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Where do you get $pmatrix{1\1\0}$?



      What do you have for $(A-2I)^2$?



      I have $(A-2I)^2 = pmatrix{-1&-2&1\3&6&-3\-4&-8&4}$



      And in addition to $pmatrix{1\0\1}$, I see $pmatrix{0\1\2}$ as a candidate for the second eigenvector.



      $Apmatrix{0\1\2} = pmatrix{1\2\5} = 2pmatrix{0\1\2}+pmatrix{1\0\1}$ which is exactly what we were hoping for.



      $Apmatrix{1&1&0\-3&0&1\4&2&1} = pmatrix{1&1&0\-3&0&1\4&2&1}pmatrix{-1&0&0\0&2&1\0&0&2}$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Where do you get $pmatrix{1\1\0}$?



        What do you have for $(A-2I)^2$?



        I have $(A-2I)^2 = pmatrix{-1&-2&1\3&6&-3\-4&-8&4}$



        And in addition to $pmatrix{1\0\1}$, I see $pmatrix{0\1\2}$ as a candidate for the second eigenvector.



        $Apmatrix{0\1\2} = pmatrix{1\2\5} = 2pmatrix{0\1\2}+pmatrix{1\0\1}$ which is exactly what we were hoping for.



        $Apmatrix{1&1&0\-3&0&1\4&2&1} = pmatrix{1&1&0\-3&0&1\4&2&1}pmatrix{-1&0&0\0&2&1\0&0&2}$






        share|cite|improve this answer









        $endgroup$



        Where do you get $pmatrix{1\1\0}$?



        What do you have for $(A-2I)^2$?



        I have $(A-2I)^2 = pmatrix{-1&-2&1\3&6&-3\-4&-8&4}$



        And in addition to $pmatrix{1\0\1}$, I see $pmatrix{0\1\2}$ as a candidate for the second eigenvector.



        $Apmatrix{0\1\2} = pmatrix{1\2\5} = 2pmatrix{0\1\2}+pmatrix{1\0\1}$ which is exactly what we were hoping for.



        $Apmatrix{1&1&0\-3&0&1\4&2&1} = pmatrix{1&1&0\-3&0&1\4&2&1}pmatrix{-1&0&0\0&2&1\0&0&2}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 0:38









        Doug MDoug M

        45.4k31954




        45.4k31954






























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