How does minimal polynomial divide characteristic polynomial of matrix? Given a matrix $A_{n\times n}$ with entries coming from a field, say $\mathbb{C}$, the characteristic polynomial is defined as 
$$\det(A - \lambda I)=a_n\lambda^n+\cdots+a_1\lambda + a_0\in\mathbb{C}[\lambda].$$
On the other hand, the minimal polynomial of $A$ is defined as "the polynomial $p$ such that $p(A)=0$ with the least degree".
There is a theorem saying that the minimal polynomial divides characteristic polynomial. My confusion is that these two polynomials don't seem to be the "same kind" of polynomials. Namely, in characteristic polynomial we have $a_0$ because characteristic polynomial is in $\mathbb{C}[\lambda]$ which is a polynomial ring. 
But in minimal polynomial we must have $a_0 I$. It doesn't seem to be in some sort of polynomial ring. Without the identity matrix, we would end up with a matrix plus a number which does not make sense to me? 
Hope this question makes sense.
 A: It comes down to this: if we're going to talk about applying polynomials to matrices, we need a way of applying any polynomial to a matrix.  Your textbook should, at some point, have a statement that equivalent to the following:

Let $p(t) = \sum_{k=0}^d a_k t^k = t^d + \cdots + a_2t^2 + a_1t + a_0$ be a polynomial, where we define $t^0 = 1$ (for convenience).  Then for a square matrix $A$, the matrix $p(A)$ is defined to be
  $$
p(A) = \sum_{k=0}^d a_kA^k
$$
  where we define $\color{red}{A^0 = I}$ (for convenience).  That is,
  $$
p(A) = a_dA^d + \cdots + a_2 A^2 + a_1 A + \color{red}{a_0 I} 
$$

Note that it is only with a definition like this that we can sensibly interpret the Cayley-Hamilton theorem, a result that you have almost certainly seen and used by this point.
A: No one is talking here about "polynomials with matrix coefficients". It is just that given a polynomial $p(t)=a_nt^n+\cdots+a_1t+a_0$, one evaluates that polynomial on a matrix $A$ (actually, on any element of a ring with unity) by doing
$$
p(A):=a_nA^n+\cdots+a_1A+a_0I. 
$$
The identification $c\longleftrightarrow cI$ is so natural that most of the time it is not worth mentioning. 
