How can we prove that a nxn real matrix A is a root of a given polynomial?

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Homework Statement


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 [A-t.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 [A-t.I]. So, deg P(t)=n, t is a real number.
Let t is a root of P(t), we get:
det[A-tI]=0 -> the equation: (A-tI)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
 
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1) In other words, they are asking for a proof of the Cayley-Hamilton 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(A-tI), what is det(A^2008 -tI) ?

P.S. Use of square brackets [ ] can be confusing. Use the normal parantheses instead.
 
Thanks, but
Defennder said:
2) Instead of evaluating det(A-tI), what is det(A^2008 -tI) ?
I don't understand clearly what you mean.
And, is there any solution where theorem 1 isn't used for problem 2?
 
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) ?
 
There are a few different methods to prove (1). Have you studied adjugate matrices yet? If so, can you reason that \text{adj} (A-tI_n) exists? If so, what can you say about (A-tI_n) \cdot \text{adj} (A-tI_n)?
 
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