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I'm trying to show:

Let $A\in \mathcal{M}_n(\mathbb{F})$ a matrix with characteristic polynomial $p_A(t)$ and minimal poliynomial $q_A(t)$. Let $d_{n-1}(t)$ 'greatest common divisor' of the slots of $\operatorname{adj}(tI_n-A)$.

a) Show that $d_{n-1}$ divided to $p_A(t)$. If $q^*(t)=\frac{p_A(t)}{d_{n-1}(t)}$ show that $q^*(t)=0$ and $q_A(t)$ divided $q^*(t)=$.

b) If $q^*(t)=s(t)q_A(t)$ in $\mathbb{F}(t)$, show that $s(t)\equiv 1$.

I was trying to use the equivalency: $$B^{-1}=\frac{1}{\det B} \operatorname{adj} B$$ $$\det B (B)^{-1}=\operatorname{adj}B$$ with $B=tI-A$, but I have two problems: if $t$ is a eigenvalues of $A$, the matrix $tI-A$ is not invertible and I dont know if $(tI-A)^{-1}$ have a polynomials slots.

Thanks for your help.

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What do you mean by "slot"? –  Bill Dubuque Apr 11 '12 at 17:08
    
@BillDubuque, I mean... "entry". Excuse my English. –  Hiperion Apr 11 '12 at 18:00
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