Does Cayley Hamilton Theorem apply for non-diagonalizable matrices as well?
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Cayley Hamilton Theorem says that a matrix $A$ satisfies its characteristic equation. My professor proved this for diagonalizable matrices. What happens if $A$ is not diagonalizable? Does the C-H Theorem still hold? Can you give a proof why it holds or why it does not?
Thanks!
linear-algebra matrices characteristic-functions cayley-hamilton
asked Nov 2 '18 at 2:21
Nagabhushan S NNagabhushan S N
260214
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add a comment |
$begingroup$
Cayley Hamilton Theorem says that a matrix $A$ satisfies its characteristic equation. My professor proved this for diagonalizable matrices. What happens if $A$ is not diagonalizable? Does the C-H Theorem still hold? Can you give a proof why it holds or why it does not?
Thanks!
linear-algebra matrices characteristic-functions cayley-hamilton
asked Nov 2 '18 at 2:21
Nagabhushan S NNagabhushan S N
260214
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It does indeed hold in general
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– RhythmInk
Nov 2 '18 at 2:22
1
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Yes, it still holds. One way to see this is to use the fact that diagonalizable matrices are dense in all matrices.
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– Qiaochu Yuan
Nov 2 '18 at 2:22
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@QiaochuYuan, can you please elaborate?
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– Nagabhushan S N
Nov 2 '18 at 2:43
add a comment |
$begingroup$
Cayley Hamilton Theorem says that a matrix $A$ satisfies its characteristic equation. My professor proved this for diagonalizable matrices. What happens if $A$ is not diagonalizable? Does the C-H Theorem still hold? Can you give a proof why it holds or why it does not?
Thanks!
linear-algebra matrices characteristic-functions cayley-hamilton
asked Nov 2 '18 at 2:21
Nagabhushan S NNagabhushan S N
260214
$endgroup$
Cayley Hamilton Theorem says that a matrix $A$ satisfies its characteristic equation. My professor proved this for diagonalizable matrices. What happens if $A$ is not diagonalizable? Does the C-H Theorem still hold? Can you give a proof why it holds or why it does not?
Thanks!
linear-algebra matrices characteristic-functions cayley-hamilton
linear-algebra matrices characteristic-functions cayley-hamilton
asked Nov 2 '18 at 2:21
Nagabhushan S NNagabhushan S N
260214
asked Nov 2 '18 at 2:21
Nagabhushan S NNagabhushan S N
260214
asked Nov 2 '18 at 2:21
Nagabhushan S NNagabhushan S N
260214
asked Nov 2 '18 at 2:21
Nagabhushan S NNagabhushan S N
260214
asked Nov 2 '18 at 2:21
Nagabhushan S NNagabhushan S N
260214
260214
$begingroup$
It does indeed hold in general
$endgroup$
– RhythmInk
Nov 2 '18 at 2:22
1
$begingroup$
Yes, it still holds. One way to see this is to use the fact that diagonalizable matrices are dense in all matrices.
$endgroup$
– Qiaochu Yuan
Nov 2 '18 at 2:22
$begingroup$
@QiaochuYuan, can you please elaborate?
$endgroup$
– Nagabhushan S N
Nov 2 '18 at 2:43
add a comment |
$begingroup$
It does indeed hold in general
$endgroup$
– RhythmInk
Nov 2 '18 at 2:22
1
$begingroup$
Yes, it still holds. One way to see this is to use the fact that diagonalizable matrices are dense in all matrices.
$endgroup$
– Qiaochu Yuan
Nov 2 '18 at 2:22
$begingroup$
@QiaochuYuan, can you please elaborate?
$endgroup$
– Nagabhushan S N
Nov 2 '18 at 2:43
$begingroup$
It does indeed hold in general
$endgroup$
– RhythmInk
Nov 2 '18 at 2:22
$begingroup$
It does indeed hold in general
$endgroup$
– RhythmInk
Nov 2 '18 at 2:22
1
1
$begingroup$
Yes, it still holds. One way to see this is to use the fact that diagonalizable matrices are dense in all matrices.
$endgroup$
– Qiaochu Yuan
Nov 2 '18 at 2:22
$begingroup$
Yes, it still holds. One way to see this is to use the fact that diagonalizable matrices are dense in all matrices.
$endgroup$
– Qiaochu Yuan
Nov 2 '18 at 2:22
$begingroup$
@QiaochuYuan, can you please elaborate?
$endgroup$
– Nagabhushan S N
Nov 2 '18 at 2:43
$begingroup$
@QiaochuYuan, can you please elaborate?
$endgroup$
– Nagabhushan S N
Nov 2 '18 at 2:43
add a comment |
1 Answer
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It holds in general. Also the proof can be found in https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
answered Nov 24 '18 at 17:08
Mostafa AyazMostafa Ayaz
15.3k3939
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add a comment |
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$begingroup$
It holds in general. Also the proof can be found in https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
answered Nov 24 '18 at 17:08
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
It holds in general. Also the proof can be found in https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
answered Nov 24 '18 at 17:08
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
It holds in general. Also the proof can be found in https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
answered Nov 24 '18 at 17:08
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
It holds in general. Also the proof can be found in https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
answered Nov 24 '18 at 17:08
Mostafa AyazMostafa Ayaz
15.3k3939
answered Nov 24 '18 at 17:08
Mostafa AyazMostafa Ayaz
15.3k3939
answered Nov 24 '18 at 17:08
Mostafa AyazMostafa Ayaz
15.3k3939
answered Nov 24 '18 at 17:08
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
add a comment |
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$begingroup$
It does indeed hold in general
$endgroup$
– RhythmInk
Nov 2 '18 at 2:22
1
$begingroup$
Yes, it still holds. One way to see this is to use the fact that diagonalizable matrices are dense in all matrices.
$endgroup$
– Qiaochu Yuan
Nov 2 '18 at 2:22
$begingroup$
@QiaochuYuan, can you please elaborate?
$endgroup$
– Nagabhushan S N
Nov 2 '18 at 2:43