I'm trying to prove that if an invertible n-by-n matrix $A$ has characteristic polynomial $$\chi_A(t)=(-1)^nt^n+a_{n-1}t^{n-1}+\ldots+a_2t^2+a_1t+a_0$$ with $a_0\not=0$then $$A^{-1}=\frac{-1}{a_0}((-1)^nA^{n-1}+a_{n-1}A^{n-2}+\ldots+a_2A+a_1I$$
So using Cayley-Hamilton theorem we have: $$\chi_A(A)=(-1)^nA^n+a_{n-1}A^{n-1}+\ldots+a_2A^2+a_1A+a_0I=0 \\ \chi_A(A)=(-1)^nA^n+a_{n-1}A^{n-1}+\ldots+a_2A^2+a_1AI+a_0AA^{-1}=0 \\ \chi_A(A)=A[(-1)^nA^{n-1}+a_{n-1}A^{n-2}+\ldots+a_2A+a_1I+a_0A^{-1}]=0 $$ And now I would love to be able to say something like this:
"since $A\not=0$, it must be the case that $(-1)^nA^{n-1}+a_{n-1}A^{n-2}+\ldots+a_2A+a_1I+a_0A^{-1}=0$" and from there it would be easy to get the desired result, the only problem is the zero-product property doesn't hold for matrices so I don't think I can conclude it this way.
My question: is there a reason for which in this particular case the zero-product propery would apply? Or what any other useful fact would help me to solve it?
Thanks
You're right to start with the identity $$\chi_A(A)=(-1)^nA^n+a_{n-1}A^{n-1}+\ldots+a_2A^2+a_1A+a_0I=0$$ but instead of factoring out an $A$, you can subtract $a_0I$ from both sides and multiply by $A^{-1}$ to obtain $$ -a_0A^{-1}=(-1)^nA^{n-1}+a_{n-1}A^{n-2}+\ldots+a_2A+a_1I$$
Then multiply both sides by $-\frac{1}{a_0}$ to conclude.