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A problem about nxn Matrixby universedrill
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#1
Dec508, 08:03 PM

P: 3

1. The problem statement, all variables and given/known data
1) Prove that: nxn real matrix A is a root of f(X)= a[n].X^n+...+a[0].I, where a[n],...,a[0] are coefficients of the polynomial P(t)= det [At.I] 2) Let 5x5 real matrix A be satisfied: A^2008 = 0. Prove that: A^5=0. 2. The attempt at a solution I tried to solve problem 2 with an general idea: nxn matrix A: A^m=0 (m>n). Prove: A^n=0. Let P(t)=det [At.I]. So, deg P(t)=n, t is a real number. Let t is a root of P(t), we get: det[AtI]=0 > the equation: (AtI)X=0 has a root X which is different from 0 > AX = tIX=tX > A(AX)=A(tX) >A^2.X=t(AX)=t(tX)=t^2.X >..... > A^m.X=t^m.X Because X differ from 0 and A^m =0, we find out t^m =0 > t=0 Thus, P(t)= t^n. Now, the important thing is proving problem 1. I remember that the problem 1 seem to be a theorem? Can you help me prove that, or find meterials saying that? Thanks 


#2
Dec508, 09:11 PM

HW Helper
P: 2,616

1) In other words, they are asking for a proof of the CayleyHamilton theorem? That I believe should be rather difficult, since proof of this theorem was omitted when I took my intermediate linear algebra course this semester.
2) Use result 1 to prove it. Instead of evaluating det(AtI), what is det(A^2008 tI) ? P.S. Use of square brackets [ ] can be confusing. Use the normal parantheses instead. 


#3
Dec508, 09:26 PM

P: 3

Thanks, but
And, is there any solution where theorem 1 isn't used for problem 2? 


#4
Dec508, 09:58 PM

HW Helper
P: 2,616

A problem about nxn Matrix
Well, it appears that the problem has been set up in such a way such that you can use the result of theorem 1 (even if you do not know how to prove it) to do 2). And I don't know which part of what I wrote you do not understand. What don't you understand about finding det(A^2008 tI) ?



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