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wu :: forums    general    complex analysis (Moderators: towr, Eigenray, william wu, SMQ, Grimbal, Icarus, ThudnBlunder)    Cauchy's Integral Formula and Cayley-Hamilton thm « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Cauchy's Integral Formula and Cayley-Hamilton thm  (Read 6189 times) immanuel78 Newbie     Gender: Posts: 23 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 wu::riddles Moderator Uberpuzzler Boldly going where even angels fear to tread.     Gender: Posts: 4863 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! IP Logged "Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous immanuel78 Newbie     Gender: Posts: 23 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 Uberpuzzler Pie = pi * e     Gender: Posts: 1261 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 "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 pex Uberpuzzler     Gender: Posts: 880 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 IP Logged pex Uberpuzzler     Gender: Posts: 880 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. IP Logged Sameer Uberpuzzler Pie = pi * e     Gender: Posts: 1261 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     [edit] Corrected the matrix [/edit] IP Logged "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 Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth => complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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