So if the minimal polynomial of some linear transformation is say $\ \mu (x) = x^2+x+2$, then if we put in a matrix $\ A$ in for $\ X$ instead, we would write this is as $\ \mu(A) = A^2 + A + 2I$ with the identity matrix multiplying our scalar. Obviously we can put any linear transformation in for the variable x using the definition that $\ f^2(v) = f(f(v))$, but in this case what would we put next to our scalar? I have a feeling it would be the identity transformation.
A specific example: Say we have $\ f(B) = AB $ for some fixed matrix A. What would be $\ \mu(f)$ in my example? A solution to a problem I am working on (past exam) would suggest that $\ \mu(f(B)) = A^2B + AB + 2B$. I am confused as to why my 2 was transformed to a 2B.
We are using an evaluation morphism at $a$, $\phi_a$, which goes from $K[x]$ to a commutative unital algebra $A$ in such a way that
What we have here is evaluation from $(K[x],+,\cdot)$ to $A:=(K[f],+,\circ)$, where $f$ is a linear endomorphism of the vector space $V$ and the product $\circ$ is the composition. For this algebra we get $1_A=\text{id}$, i.e., the identity map is the identity element (because the product is composition).
Then, after evaluation of the polynomial $p$ at the linear endomorphism $f$, you get a new linear endomorphism $p(f)$, in which the independent term of $p$ generates a multiple of the identity map.