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        general >> complex analysis >> Cauchy's Integral Formula and Cayley-Hamilton thm
        
(Message started by: immanuel78 on Aug 27th, 2006, 10:16pm)
Title: Cauchy's Integral Formula and Cayley-Hamilton thm
Post by immanuel78 on Aug 27th, 2006, 10:16pm
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
Title: Re: Cauchy's Integral Formula and Cayley-Hamilton
Post by Icarus on Aug 28th, 2006, 3:36pm
Better make that f(z) = det(zI - A), or else the result is trivial!
Title: Re: Cauchy's Integral Formula and Cayley-Hamilton
Post by immanuel78 on Aug 29th, 2006, 8:45pm
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.
Title: Re: Cauchy's Integral Formula and Cayley-Hamilton
Post by Sameer on Aug 29th, 2006, 9:48pm
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)
Title: Re: Cauchy's Integral Formula and Cayley-Hamilton
Post by pex on Aug 30th, 2006, 12:20am

on 08/29/06 at 21:48:04, 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
Title: Re: Cauchy's Integral Formula and Cayley-Hamilton
Post by pex on Aug 30th, 2006, 12:35am

on 08/29/06 at 20:45:20, 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.
Title: Re: Cauchy's Integral Formula and Cayley-Hamilton
Post by Sameer on Aug 30th, 2006, 10:35am

on 08/30/06 at 00:20:11, 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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