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immanuel78
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 Cauchy's Integral Formula and Cayley-Hamilton thm   « on: Aug 27th, 2006, 10:16pm » Quote Modify

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
Icarus
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 Re: Cauchy's Integral Formula and Cayley-Hamilton   « Reply #1 on: Aug 28th, 2006, 3:36pm » Quote Modify

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 » Quote Modify

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 » Quote Modify

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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Proof is an idol before which the mathematician tortures himself.
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pex
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 Re: Cauchy's Integral Formula and Cayley-Hamilton   « Reply #4 on: Aug 30th, 2006, 12:20am » Quote Modify

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...

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 » Quote Modify

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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Sameer
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 Re: Cauchy's Integral Formula and Cayley-Hamilton   « Reply #6 on: Aug 30th, 2006, 10:35am » Quote Modify

on Aug 30th, 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

 Corrected the matrix [/edit]
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"Obvious" is the most dangerous word in mathematics.
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Proof is an idol before which the mathematician tortures himself.
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