Mixing
Proof of Jordan-Chevally-Decomposition
$begingroup$
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
edited Jan 10 at 18:46
Viktor Glombik
8211527
asked Jan 10 at 17:17
L GL G
248
$endgroup$
add a comment |
$begingroup$
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
edited Jan 10 at 18:46
Viktor Glombik
8211527
asked Jan 10 at 17:17
L GL G
248
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 10 at 18:46
Viktor Glombik
8211527
asked Jan 10 at 17:17
L GL G
248
$endgroup$
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
edited Jan 10 at 18:46
Viktor Glombik
8211527
asked Jan 10 at 17:17
L GL G
248
edited Jan 10 at 18:46
Viktor Glombik
8211527
asked Jan 10 at 17:17
L GL G
248
edited Jan 10 at 18:46
Viktor Glombik
8211527
edited Jan 10 at 18:46
Viktor Glombik
8211527
edited Jan 10 at 18:46
Viktor Glombik
8211527
8211527
asked Jan 10 at 17:17
L GL G
248
asked Jan 10 at 17:17
L GL G
248
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
"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*}
answered Jan 10 at 17:44
Viktor GlombikViktor Glombik
8211527
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 10 at 17:33
NassoumoNassoumo
812
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
"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*}
answered Jan 10 at 17:44
Viktor GlombikViktor Glombik
8211527
$endgroup$
add a comment |
$begingroup$
"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*}
answered Jan 10 at 17:44
Viktor GlombikViktor Glombik
8211527
$endgroup$
add a comment |
$begingroup$
"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*}
answered Jan 10 at 17:44
Viktor GlombikViktor Glombik
8211527
$endgroup$
"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*}
answered Jan 10 at 17:44
Viktor GlombikViktor Glombik
8211527
edited Jan 10 at 17:50
answered Jan 10 at 17:44
Viktor GlombikViktor Glombik
8211527
answered Jan 10 at 17:44
Viktor GlombikViktor Glombik
8211527
answered Jan 10 at 17:44
Viktor GlombikViktor Glombik
8211527
8211527
add a comment |
add a comment |
$begingroup$
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.
answered Jan 10 at 17:33
NassoumoNassoumo
812
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 10 at 17:33
NassoumoNassoumo
812
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 10 at 17:33
NassoumoNassoumo
812
$endgroup$
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.
answered Jan 10 at 17:33
NassoumoNassoumo
812
answered Jan 10 at 17:33
NassoumoNassoumo
812
answered Jan 10 at 17:33
NassoumoNassoumo
812
answered Jan 10 at 17:33
NassoumoNassoumo
812
812
add a comment |
add a comment |
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$begingroup$
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