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Topic: Cauchy's Integral Formula and CayleyHamilton thm (Read 5903 times) 

immanuel78
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Cauchy's Integral Formula and CayleyHamilton thm
« on: Aug 27^{th}, 2006, 10:16pm » 
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There is another problem that I can't solve. Use Cauchy's Integral Formula to prove CayleyHamilton Theorem. CayleyHamilton Theorem : Let A be an nxn matrix over C and let f(z)=det(zA) be a characteristic polynomial of A. Then f(A)=O

« Last Edit: Aug 28^{th}, 2006, 6:06pm by immanuel78 » 
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Icarus
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Re: Cauchy's Integral Formula and CayleyHamilton
« Reply #1 on: Aug 28^{th}, 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 CayleyHamilton
« Reply #2 on: Aug 29^{th}, 2006, 8:45pm » 
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If Icarus think as follows, the proof seems to be false. Since f(z)=det(zIA), f(A)=det(AIA)=det(O)=0 Because f(z)=det(zIA) : 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 29^{th}, 2006, 10:25pm by immanuel78 » 
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Sameer
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Re: Cauchy's Integral Formula and CayleyHamilton
« Reply #3 on: Aug 29^{th}, 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 30^{th}, 2006, 10:35am by Sameer » 
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pex
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Re: Cauchy's Integral Formula and CayleyHamilton
« Reply #4 on: Aug 30^{th}, 2006, 12:20am » 
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on Aug 29^{th}, 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... 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 CayleyHamilton
« Reply #5 on: Aug 30^{th}, 2006, 12:35am » 
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on Aug 29^{th}, 2006, 8:45pm, immanuel78 wrote:If Icarus think as follows, the proof seems to be false. Since f(z)=det(zIA), f(A)=det(AIA)=det(O)=0 Because f(z)=det(zIA) : 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) = (z1)(z4)  (2)*(3) = z^{2}  5z  2. The CayleyHamilton Theorem now states that A^{2}  5A  2I = O, which is, indeed, true.


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Sameer
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Re: Cauchy's Integral Formula and CayleyHamilton
« Reply #6 on: Aug 30^{th}, 2006, 10:35am » 
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on Aug 30^{th}, 2006, 12:20am, pex wrote: Funny identity matrix you've got there... 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 [edit] Corrected the matrix [/edit]


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"Obvious" is the most dangerous word in mathematics. Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity



