f(t)=t^n+a_n-1t^(n-1)+...+a1t+a0 there's a square matrix of order n, A: [tex]\bordermatrix{ & & & & \cr 0 & 0 & ... & 0& -a_0 \cr 1 &0 & ... & 0 & -a_1 \cr ... & ... & ... & ... & ... \cr 0 & 0 & ... & 1 & -a_{n-1}\cr}[/tex] show that f(t) is the minimal polynomial of A. i know that f(t) is m.p when f(A)=0, or perhaps all that i should prove here, is that f(t) divides the charectraistic polynomial of A?
No, it is not sufficient to prove that f(t) divides the characteristic polynomial. What the minimal polynomial must divide the characteristic polynomial, it is not the only polynomial to do so. For example, if the characteristic polynomial is (t-1)(t-2), both t-1 and t-2 divide the characteristic polynomial but neither is the minimal polynomial ((t-1)(t-2) itself is). To prove a given polynomial is the minimal polynomial you must show that it satisfies the definition of minimal polynomial: that it is the monic polynomial of lowest degree satisfied by A. (which is not the same as "f(t) is m.p. when f(A)= 0"- again, there are many polynomials satisfying that.)
then, in order to prove that, i should calculate f(A). i know that m.p is the polynomial which is highest degree has a coefficient of 1. so what i need to do, is to find the charectaristic plynomial of A, and then divide the options for the minimal polynomial from c.p. the answer should be f(t), right?
Am I alone in not seeing exactly what that matrix looks like? It ought to have the a_i in the last column, but what about the singleton 1 that appears in the second row. Should there not be more 1's than that? All on the diagonal immediately below the main diagonal, perhaps? I have no idea what 'divide the options' means. f(t) is the characteristic poly by the way: if it is the minimal it divides the char poly, but the char poly is a degree n monic poly, so they must be the same.
i mean you have for example the c.p like this: (t-k)^n then the option to m.p are (t-k)^n, (t-k)^(n-1), etc.