If $A$ is a linear transformation with minimal polynomial $p(x)$, and if $q(x)$ is a polynomial such that $q(A)=0$, then show that $q$ is divisible by $p$.
so I am trying to show that $f(x)$ divides $g(x)$ for all polynomials $g(x)$ satisfying that $g(A)=0$ where $f(x)$ is the minimal polynomial of a square matrix $A$.  I know from my professor that $f(x)$ is the minimal polynomial of $A$ so $f(A)=0$ so then $g(A)=0$. Therefore, $f(x)$ divides $g(x)$, i.e. $g(x)=f(x)h(x)$ where $h(x)$ is some other polynomial.  My prof told us to solve this question we need to let $b(x)=gcd((f(x),g(x))$ and by using Bezout's Identity, we need to show that $b(A)=0$. I am just having a little trouble trying to prove this. I know that this identity says:     For two polynomials $f_1(x)$ and $f_2(x)$, there exist polynomials   $g_1(x)$ and $g_2(x)$ such that $g_1(x)f_1(x) + g_2(x)f_2(x) = &gt; gcd(f_1(x), f_2(x))$.   And the uniqueness of minimal polynomial theorem is:     For a square matrix $A$, if $f_1(A) = f_2(A) = 0$ for two polynomials   $f_1(x)$ and $f_2(x)$, then $g(A) = 0$ for $g(x) = gcd(f_1(x), &gt; f_2(x))$. So the minimal polynomial of $A$ is unique up to a scalar.   So since $f(x)$ is already the minimal polynomial, I know that the $deg(b(x))=deg(f(x))$. I know how to prove the Bezout identity using the Euclidean algorithm, I'm just not exactly sure how I would show all of this using the notation he gave us of $b(x)=gcd((f(x),g(x))$. Any help at all would be appreciated.
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