Proof of Jordan-Chevally-Decomposition












1












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Let A be a square matrix over $mathbb{C}$, prove there are matrices $D$ and $N$ such that $A = D + N$ such that $D$ is diagonalizable, $N$ is Nilpotent and $DN = ND$.




I can see that any diagonalizable matrix has to have a power such that $D^t=I$ where I is the identity and any Nilpotent matrix has to satisfy $N^l=0$. I'm not sure how to go about proving that all these conditions hold for any square matrix A.






linear-algebra matrices diagonalization matrix-decomposition nilpotence







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$endgroup$












  • $begingroup$
    Look at en.m.wikipedia.org/wiki/Jordan–Chevalley_decomposition
    $endgroup$
    – Mindlack
    Jan 10 at 17:19










  • $begingroup$
    It is a special case of Primary decomposition theorem.
    $endgroup$
    – Thomas Shelby
    Jan 10 at 17:20










  • $begingroup$
    @Mindlack The country Jordan is very interesting but that's not why I'm here haha, seriously though thanks for the heads up i'll take a peak
    $endgroup$
    – L G
    Jan 10 at 17:32


















1












$begingroup$



Let A be a square matrix over $mathbb{C}$, prove there are matrices $D$ and $N$ such that $A = D + N$ such that $D$ is diagonalizable, $N$ is Nilpotent and $DN = ND$.




I can see that any diagonalizable matrix has to have a power such that $D^t=I$ where I is the identity and any Nilpotent matrix has to satisfy $N^l=0$. I'm not sure how to go about proving that all these conditions hold for any square matrix A.






linear-algebra matrices diagonalization matrix-decomposition nilpotence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Look at en.m.wikipedia.org/wiki/Jordan–Chevalley_decomposition
    $endgroup$
    – Mindlack
    Jan 10 at 17:19










  • $begingroup$
    It is a special case of Primary decomposition theorem.
    $endgroup$
    – Thomas Shelby
    Jan 10 at 17:20










  • $begingroup$
    @Mindlack The country Jordan is very interesting but that's not why I'm here haha, seriously though thanks for the heads up i'll take a peak
    $endgroup$
    – L G
    Jan 10 at 17:32
















1












1








1





$begingroup$



Let A be a square matrix over $mathbb{C}$, prove there are matrices $D$ and $N$ such that $A = D + N$ such that $D$ is diagonalizable, $N$ is Nilpotent and $DN = ND$.




I can see that any diagonalizable matrix has to have a power such that $D^t=I$ where I is the identity and any Nilpotent matrix has to satisfy $N^l=0$. I'm not sure how to go about proving that all these conditions hold for any square matrix A.






linear-algebra matrices diagonalization matrix-decomposition nilpotence







share|cite|improve this question











$endgroup$





Let A be a square matrix over $mathbb{C}$, prove there are matrices $D$ and $N$ such that $A = D + N$ such that $D$ is diagonalizable, $N$ is Nilpotent and $DN = ND$.




I can see that any diagonalizable matrix has to have a power such that $D^t=I$ where I is the identity and any Nilpotent matrix has to satisfy $N^l=0$. I'm not sure how to go about proving that all these conditions hold for any square matrix A.






linear-algebra matrices diagonalization matrix-decomposition nilpotence




linear-algebra matrices diagonalization matrix-decomposition nilpotence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 18:46









Viktor Glombik

8211527




8211527










asked Jan 10 at 17:17









L GL G

248




248












  • $begingroup$
    Look at en.m.wikipedia.org/wiki/Jordan–Chevalley_decomposition
    $endgroup$
    – Mindlack
    Jan 10 at 17:19










  • $begingroup$
    It is a special case of Primary decomposition theorem.
    $endgroup$
    – Thomas Shelby
    Jan 10 at 17:20










  • $begingroup$
    @Mindlack The country Jordan is very interesting but that's not why I'm here haha, seriously though thanks for the heads up i'll take a peak
    $endgroup$
    – L G
    Jan 10 at 17:32




















  • $begingroup$
    Look at en.m.wikipedia.org/wiki/Jordan–Chevalley_decomposition
    $endgroup$
    – Mindlack
    Jan 10 at 17:19










  • $begingroup$
    It is a special case of Primary decomposition theorem.
    $endgroup$
    – Thomas Shelby
    Jan 10 at 17:20










  • $begingroup$
    @Mindlack The country Jordan is very interesting but that's not why I'm here haha, seriously though thanks for the heads up i'll take a peak
    $endgroup$
    – L G
    Jan 10 at 17:32


















$begingroup$
Look at en.m.wikipedia.org/wiki/Jordan–Chevalley_decomposition
$endgroup$
– Mindlack
Jan 10 at 17:19




$begingroup$
Look at en.m.wikipedia.org/wiki/Jordan–Chevalley_decomposition
$endgroup$
– Mindlack
Jan 10 at 17:19












$begingroup$
It is a special case of Primary decomposition theorem.
$endgroup$
– Thomas Shelby
Jan 10 at 17:20




$begingroup$
It is a special case of Primary decomposition theorem.
$endgroup$
– Thomas Shelby
Jan 10 at 17:20












$begingroup$
@Mindlack The country Jordan is very interesting but that's not why I'm here haha, seriously though thanks for the heads up i'll take a peak
$endgroup$
– L G
Jan 10 at 17:32






$begingroup$
@Mindlack The country Jordan is very interesting but that's not why I'm here haha, seriously though thanks for the heads up i'll take a peak
$endgroup$
– L G
Jan 10 at 17:32












2 Answers
2






active

oldest

votes


















2












$begingroup$

"Intuition":
We construct $D$ and $N$ in the following fashion: We want to show, that for any invertible matrix $S in text{GL}(n,K)$ we have
$$
A = S^{-1} tilde{D} S + S^{-1} ( J(A) - tilde{D}) S,
$$
where $J(A)$ is the Jordan decomposition of $A$.



This is true because from Jordan decomposition we that the exists an invertible matrix $S in text{GL}(n,K)$, so that
begin{align*}
A = S^{-1} J(A) S
= S^{-1} (tilde{D} + J(A) - tilde{D}) S
= S^{-1} tilde{D} S + S^{-1} ( J(A) - tilde{D}) S.
end{align*}



Existence:
Let $tilde{D}$ be the diagonal matrix whose entries are the eigenvalues of $A$.
Because every diagonalisable matrix is similar to a diagonal matrix, we know, that $S^{-1} tilde{D} S$ is diagonalisable und let $D := S^{-1} tilde{D} S$.



Now $J(A) - tilde{D}$ is a upper triangular matrix, whose diagonal only contains zeros and therefore nilpotent.
Also, $N := S^{-1}( J(A) - tilde{D}) S$ is nilpotent and we have $A = D + N$.



Commutativity:



Because $tilde{D}$ is a diagonal matrix we have
begin{align*}
DN
& = S^{-1} tilde{D} S S^{-1} ( J(A) - tilde{D}) S
= S^{-1} tilde{D} ( J(A) - tilde{D}) S
= S^{-1} ( J(A) - tilde{D}) tilde{D} S \
& = S^{-1} ( J(A) - tilde{D}) S S^{-1} tilde{D} S
= ND
end{align*}






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    A good thing to remember is that complex numbers form an algebraically closed field. So A is similar to a triangular matrix. So I think you can narrow your proof by only proving the result for a triangular matrix and concluding by using similarity.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      "Intuition":
      We construct $D$ and $N$ in the following fashion: We want to show, that for any invertible matrix $S in text{GL}(n,K)$ we have
      $$
      A = S^{-1} tilde{D} S + S^{-1} ( J(A) - tilde{D}) S,
      $$
      where $J(A)$ is the Jordan decomposition of $A$.



      This is true because from Jordan decomposition we that the exists an invertible matrix $S in text{GL}(n,K)$, so that
      begin{align*}
      A = S^{-1} J(A) S
      = S^{-1} (tilde{D} + J(A) - tilde{D}) S
      = S^{-1} tilde{D} S + S^{-1} ( J(A) - tilde{D}) S.
      end{align*}



      Existence:
      Let $tilde{D}$ be the diagonal matrix whose entries are the eigenvalues of $A$.
      Because every diagonalisable matrix is similar to a diagonal matrix, we know, that $S^{-1} tilde{D} S$ is diagonalisable und let $D := S^{-1} tilde{D} S$.



      Now $J(A) - tilde{D}$ is a upper triangular matrix, whose diagonal only contains zeros and therefore nilpotent.
      Also, $N := S^{-1}( J(A) - tilde{D}) S$ is nilpotent and we have $A = D + N$.



      Commutativity:



      Because $tilde{D}$ is a diagonal matrix we have
      begin{align*}
      DN
      & = S^{-1} tilde{D} S S^{-1} ( J(A) - tilde{D}) S
      = S^{-1} tilde{D} ( J(A) - tilde{D}) S
      = S^{-1} ( J(A) - tilde{D}) tilde{D} S \
      & = S^{-1} ( J(A) - tilde{D}) S S^{-1} tilde{D} S
      = ND
      end{align*}






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        "Intuition":
        We construct $D$ and $N$ in the following fashion: We want to show, that for any invertible matrix $S in text{GL}(n,K)$ we have
        $$
        A = S^{-1} tilde{D} S + S^{-1} ( J(A) - tilde{D}) S,
        $$
        where $J(A)$ is the Jordan decomposition of $A$.



        This is true because from Jordan decomposition we that the exists an invertible matrix $S in text{GL}(n,K)$, so that
        begin{align*}
        A = S^{-1} J(A) S
        = S^{-1} (tilde{D} + J(A) - tilde{D}) S
        = S^{-1} tilde{D} S + S^{-1} ( J(A) - tilde{D}) S.
        end{align*}



        Existence:
        Let $tilde{D}$ be the diagonal matrix whose entries are the eigenvalues of $A$.
        Because every diagonalisable matrix is similar to a diagonal matrix, we know, that $S^{-1} tilde{D} S$ is diagonalisable und let $D := S^{-1} tilde{D} S$.



        Now $J(A) - tilde{D}$ is a upper triangular matrix, whose diagonal only contains zeros and therefore nilpotent.
        Also, $N := S^{-1}( J(A) - tilde{D}) S$ is nilpotent and we have $A = D + N$.



        Commutativity:



        Because $tilde{D}$ is a diagonal matrix we have
        begin{align*}
        DN
        & = S^{-1} tilde{D} S S^{-1} ( J(A) - tilde{D}) S
        = S^{-1} tilde{D} ( J(A) - tilde{D}) S
        = S^{-1} ( J(A) - tilde{D}) tilde{D} S \
        & = S^{-1} ( J(A) - tilde{D}) S S^{-1} tilde{D} S
        = ND
        end{align*}






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          "Intuition":
          We construct $D$ and $N$ in the following fashion: We want to show, that for any invertible matrix $S in text{GL}(n,K)$ we have
          $$
          A = S^{-1} tilde{D} S + S^{-1} ( J(A) - tilde{D}) S,
          $$
          where $J(A)$ is the Jordan decomposition of $A$.



          This is true because from Jordan decomposition we that the exists an invertible matrix $S in text{GL}(n,K)$, so that
          begin{align*}
          A = S^{-1} J(A) S
          = S^{-1} (tilde{D} + J(A) - tilde{D}) S
          = S^{-1} tilde{D} S + S^{-1} ( J(A) - tilde{D}) S.
          end{align*}



          Existence:
          Let $tilde{D}$ be the diagonal matrix whose entries are the eigenvalues of $A$.
          Because every diagonalisable matrix is similar to a diagonal matrix, we know, that $S^{-1} tilde{D} S$ is diagonalisable und let $D := S^{-1} tilde{D} S$.



          Now $J(A) - tilde{D}$ is a upper triangular matrix, whose diagonal only contains zeros and therefore nilpotent.
          Also, $N := S^{-1}( J(A) - tilde{D}) S$ is nilpotent and we have $A = D + N$.



          Commutativity:



          Because $tilde{D}$ is a diagonal matrix we have
          begin{align*}
          DN
          & = S^{-1} tilde{D} S S^{-1} ( J(A) - tilde{D}) S
          = S^{-1} tilde{D} ( J(A) - tilde{D}) S
          = S^{-1} ( J(A) - tilde{D}) tilde{D} S \
          & = S^{-1} ( J(A) - tilde{D}) S S^{-1} tilde{D} S
          = ND
          end{align*}






          share|cite|improve this answer











          $endgroup$



          "Intuition":
          We construct $D$ and $N$ in the following fashion: We want to show, that for any invertible matrix $S in text{GL}(n,K)$ we have
          $$
          A = S^{-1} tilde{D} S + S^{-1} ( J(A) - tilde{D}) S,
          $$
          where $J(A)$ is the Jordan decomposition of $A$.



          This is true because from Jordan decomposition we that the exists an invertible matrix $S in text{GL}(n,K)$, so that
          begin{align*}
          A = S^{-1} J(A) S
          = S^{-1} (tilde{D} + J(A) - tilde{D}) S
          = S^{-1} tilde{D} S + S^{-1} ( J(A) - tilde{D}) S.
          end{align*}



          Existence:
          Let $tilde{D}$ be the diagonal matrix whose entries are the eigenvalues of $A$.
          Because every diagonalisable matrix is similar to a diagonal matrix, we know, that $S^{-1} tilde{D} S$ is diagonalisable und let $D := S^{-1} tilde{D} S$.



          Now $J(A) - tilde{D}$ is a upper triangular matrix, whose diagonal only contains zeros and therefore nilpotent.
          Also, $N := S^{-1}( J(A) - tilde{D}) S$ is nilpotent and we have $A = D + N$.



          Commutativity:



          Because $tilde{D}$ is a diagonal matrix we have
          begin{align*}
          DN
          & = S^{-1} tilde{D} S S^{-1} ( J(A) - tilde{D}) S
          = S^{-1} tilde{D} ( J(A) - tilde{D}) S
          = S^{-1} ( J(A) - tilde{D}) tilde{D} S \
          & = S^{-1} ( J(A) - tilde{D}) S S^{-1} tilde{D} S
          = ND
          end{align*}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 10 at 17:50

























          answered Jan 10 at 17:44









          Viktor GlombikViktor Glombik

          8211527




          8211527























              1












              $begingroup$

              A good thing to remember is that complex numbers form an algebraically closed field. So A is similar to a triangular matrix. So I think you can narrow your proof by only proving the result for a triangular matrix and concluding by using similarity.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                A good thing to remember is that complex numbers form an algebraically closed field. So A is similar to a triangular matrix. So I think you can narrow your proof by only proving the result for a triangular matrix and concluding by using similarity.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  A good thing to remember is that complex numbers form an algebraically closed field. So A is similar to a triangular matrix. So I think you can narrow your proof by only proving the result for a triangular matrix and concluding by using similarity.






                  share|cite|improve this answer









                  $endgroup$



                  A good thing to remember is that complex numbers form an algebraically closed field. So A is similar to a triangular matrix. So I think you can narrow your proof by only proving the result for a triangular matrix and concluding by using similarity.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 at 17:33









                  NassoumoNassoumo

                  812




                  812






























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