Jordan Form of a 3x3 Matrix with an eigenvalue of multiplicity 3…












0












$begingroup$


Let



$$A=
begin{bmatrix}
2&2&3\
1&3&3\
-1&-2&-2
end{bmatrix}
.
$$



Find the Jordan Form, $J$, of this matrix, and an invertible matrix $Q$ such that $A = QJQ^{-1}$. I have already found the Jordan Form of this matrix, that is,



$$J = begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}.$$



The part that I am confused about is finding the matrix $Q$. I know that the columns of $Q$ will consist of the eigenvectors, and generalized eigenvectors of $A - lambda I$. The characteristic polynomial of $A$ is



$$p_A(lambda) = lambda^3 - 3lambda^2 + 3lambda + 1 = (lambda - 1)^3.$$



I have found the eigenvector associated with $lambda = 1$ to be



$$v = (-5, 1, 1).$$



However, $(A - I)^2 = 0$, so I am confused on how to find the generalized eigenvectors. Thanks in advance!










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$endgroup$








  • 2




    $begingroup$
    This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
    $endgroup$
    – Dietrich Burde
    Jan 7 at 20:32


















0












$begingroup$


Let



$$A=
begin{bmatrix}
2&2&3\
1&3&3\
-1&-2&-2
end{bmatrix}
.
$$



Find the Jordan Form, $J$, of this matrix, and an invertible matrix $Q$ such that $A = QJQ^{-1}$. I have already found the Jordan Form of this matrix, that is,



$$J = begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}.$$



The part that I am confused about is finding the matrix $Q$. I know that the columns of $Q$ will consist of the eigenvectors, and generalized eigenvectors of $A - lambda I$. The characteristic polynomial of $A$ is



$$p_A(lambda) = lambda^3 - 3lambda^2 + 3lambda + 1 = (lambda - 1)^3.$$



I have found the eigenvector associated with $lambda = 1$ to be



$$v = (-5, 1, 1).$$



However, $(A - I)^2 = 0$, so I am confused on how to find the generalized eigenvectors. Thanks in advance!










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
    $endgroup$
    – Dietrich Burde
    Jan 7 at 20:32
















0












0








0


1



$begingroup$


Let



$$A=
begin{bmatrix}
2&2&3\
1&3&3\
-1&-2&-2
end{bmatrix}
.
$$



Find the Jordan Form, $J$, of this matrix, and an invertible matrix $Q$ such that $A = QJQ^{-1}$. I have already found the Jordan Form of this matrix, that is,



$$J = begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}.$$



The part that I am confused about is finding the matrix $Q$. I know that the columns of $Q$ will consist of the eigenvectors, and generalized eigenvectors of $A - lambda I$. The characteristic polynomial of $A$ is



$$p_A(lambda) = lambda^3 - 3lambda^2 + 3lambda + 1 = (lambda - 1)^3.$$



I have found the eigenvector associated with $lambda = 1$ to be



$$v = (-5, 1, 1).$$



However, $(A - I)^2 = 0$, so I am confused on how to find the generalized eigenvectors. Thanks in advance!










share|cite|improve this question









$endgroup$




Let



$$A=
begin{bmatrix}
2&2&3\
1&3&3\
-1&-2&-2
end{bmatrix}
.
$$



Find the Jordan Form, $J$, of this matrix, and an invertible matrix $Q$ such that $A = QJQ^{-1}$. I have already found the Jordan Form of this matrix, that is,



$$J = begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}.$$



The part that I am confused about is finding the matrix $Q$. I know that the columns of $Q$ will consist of the eigenvectors, and generalized eigenvectors of $A - lambda I$. The characteristic polynomial of $A$ is



$$p_A(lambda) = lambda^3 - 3lambda^2 + 3lambda + 1 = (lambda - 1)^3.$$



I have found the eigenvector associated with $lambda = 1$ to be



$$v = (-5, 1, 1).$$



However, $(A - I)^2 = 0$, so I am confused on how to find the generalized eigenvectors. Thanks in advance!







linear-algebra jordan-normal-form






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asked Jan 7 at 20:18









Taylor McMillanTaylor McMillan

695




695








  • 2




    $begingroup$
    This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
    $endgroup$
    – Dietrich Burde
    Jan 7 at 20:32
















  • 2




    $begingroup$
    This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
    $endgroup$
    – Dietrich Burde
    Jan 7 at 20:32










2




2




$begingroup$
This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
$endgroup$
– Dietrich Burde
Jan 7 at 20:32






$begingroup$
This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
$endgroup$
– Dietrich Burde
Jan 7 at 20:32












1 Answer
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2












$begingroup$

Hint:



You should proceed backwards:




  • take any vector $u_3$ in $ker(A-I)^2smallsetminusker(A-I)$, i.e. any vector in $mathbf R^3$ which does not satisfy the equation $;x+2y+3z=0$, e.g. $u_3=(1,0,0)$.

  • set $u_2=(A-I)u_3$. This vector is an eigenvector.

  • complete $u_2$ with a linearly independent vector $u_1$, so as to obtain a basis of the eigenspace.






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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

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    2












    $begingroup$

    Hint:



    You should proceed backwards:




    • take any vector $u_3$ in $ker(A-I)^2smallsetminusker(A-I)$, i.e. any vector in $mathbf R^3$ which does not satisfy the equation $;x+2y+3z=0$, e.g. $u_3=(1,0,0)$.

    • set $u_2=(A-I)u_3$. This vector is an eigenvector.

    • complete $u_2$ with a linearly independent vector $u_1$, so as to obtain a basis of the eigenspace.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint:



      You should proceed backwards:




      • take any vector $u_3$ in $ker(A-I)^2smallsetminusker(A-I)$, i.e. any vector in $mathbf R^3$ which does not satisfy the equation $;x+2y+3z=0$, e.g. $u_3=(1,0,0)$.

      • set $u_2=(A-I)u_3$. This vector is an eigenvector.

      • complete $u_2$ with a linearly independent vector $u_1$, so as to obtain a basis of the eigenspace.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint:



        You should proceed backwards:




        • take any vector $u_3$ in $ker(A-I)^2smallsetminusker(A-I)$, i.e. any vector in $mathbf R^3$ which does not satisfy the equation $;x+2y+3z=0$, e.g. $u_3=(1,0,0)$.

        • set $u_2=(A-I)u_3$. This vector is an eigenvector.

        • complete $u_2$ with a linearly independent vector $u_1$, so as to obtain a basis of the eigenspace.






        share|cite|improve this answer









        $endgroup$



        Hint:



        You should proceed backwards:




        • take any vector $u_3$ in $ker(A-I)^2smallsetminusker(A-I)$, i.e. any vector in $mathbf R^3$ which does not satisfy the equation $;x+2y+3z=0$, e.g. $u_3=(1,0,0)$.

        • set $u_2=(A-I)u_3$. This vector is an eigenvector.

        • complete $u_2$ with a linearly independent vector $u_1$, so as to obtain a basis of the eigenspace.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 21:13









        BernardBernard

        121k740116




        121k740116






























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