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   Cauchy's Integral Formula and Cayley-Hamilton thm
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   Author  Topic: Cauchy's Integral Formula and Cayley-Hamilton thm  (Read 5903 times)
immanuel78
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Cauchy's Integral Formula and Cayley-Hamilton thm  
« on: Aug 27th, 2006, 10:16pm »
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There is another problem that I can't solve.
 
Use Cauchy's Integral Formula to prove Cayley-Hamilton Theorem.
 
Cayley-Hamilton Theorem :
Let A be an nxn matrix over C and let f(z)=det(z-A) be a characteristic polynomial of A.
Then f(A)=O
 
 
« Last Edit: Aug 28th, 2006, 6:06pm by immanuel78 » IP Logged
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Re: Cauchy's Integral Formula and Cayley-Hamilton  
« Reply #1 on: Aug 28th, 2006, 3:36pm »
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Better make that f(z) = det(zI - A), or else the result is trivial!
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immanuel78
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Re: Cauchy's Integral Formula and Cayley-Hamilton  
« Reply #2 on: Aug 29th, 2006, 8:45pm »
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If Icarus think as follows, the proof seems to be false.
 
Since f(z)=det(zI-A), f(A)=det(AI-A)=det(O)=0
 
Because f(z)=det(zI-A) : C -> C is defined but f(A) is defined on the set of square matrices.
In other words, f(z) and f(A) are actually different functions.
« Last Edit: Aug 29th, 2006, 10:25pm by immanuel78 » IP Logged
Sameer
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Re: Cauchy's Integral Formula and Cayley-Hamilton  
« Reply #3 on: Aug 29th, 2006, 9:48pm »
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zI is right.. z in this case is a complex number and zI is a complex matrix defined over CxC. A is a subset of CxC which is required to define the function..  
 
zI - A in this case will be
 
z - a11 - a12 .... - a1n
- a21 z - a22 .... - a2n
..
 
- an1 - an2 .... z - ann
 
Determinant of this will be f(z). Cayley Hamilton theorem simply says that A wil satisfy its own characterictic equation...
(Above a1..n 1..n can be real or complex numbers)
« Last Edit: Aug 30th, 2006, 10:35am by Sameer » IP Logged

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Re: Cauchy's Integral Formula and Cayley-Hamilton  
« Reply #4 on: Aug 30th, 2006, 12:20am »
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on Aug 29th, 2006, 9:48pm, Sameer wrote:
zI - A in this case will be
 
z - a11   z - a12 .... z - a1n
z - a21   z - a22 .... z - a2n
..
 
z - an1   z - an2 .... z - ann

 
Funny identity matrix you've got there... Wink
 
Code:
zI - A =
 
z - a11   - a12   ...   - a1n
 - a21   z - a22  ...   - a2n
  ...
 - an1    - an2   ...  z - ann
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pex
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Re: Cauchy's Integral Formula and Cayley-Hamilton  
« Reply #5 on: Aug 30th, 2006, 12:35am »
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on Aug 29th, 2006, 8:45pm, immanuel78 wrote:
If Icarus think as follows, the proof seems to be false.
 
Since f(z)=det(zI-A), f(A)=det(AI-A)=det(O)=0
 
Because  f(z)=det(zI-A) : C -> C is defined  but f(A) is defined on the set of square matrices.
In other words, f(z) and f(A) are actually different functions.

 
The definition is rather dirty. f(A) is not supposed to be det(AI - A) (which would make the theorem trivial). Instead, it is what you get when you find the characteristic polynomial f(z) and then substitute A for z.
 
Small example: let A =
1 2
3 4.
Then f(z) = det(zI - A) = (z-1)(z-4) - (-2)*(-3) = z2 - 5z - 2.
The Cayley-Hamilton Theorem now states that A2 - 5A - 2I = O, which is, indeed, true.
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Re: Cauchy's Integral Formula and Cayley-Hamilton  
« Reply #6 on: Aug 30th, 2006, 10:35am »
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on Aug 30th, 2006, 12:20am, pex wrote:

 
Funny identity matrix you've got there... Wink
 
Code:
zI - A =
 
z - a11 - a12 ... - a1n
 - a21 z - a22 ... - a2n
 ...
 - an1 - an2 ... z - ann

 
 
Yes, sorry I wrote this late at night and then when I woke up I realised I did this wrong and came here to correct this mistake before it was too late.. i see i was too late  Wink
 
[edit] Corrected the matrix [/edit]
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