Is a characteristic polynomial we consider in Linear Algebra a polynomial or a polynomial function? In Linear Algebra, we consider characteristic polynomials.
Is a characteristic polynomial we consider in Linear Algebra a polynomial or a polynomial function?  
I think it is a polynomial function.  
I am reading "Introduction to Linear Algebra" (in Japanese) by Kazuo Matsuzaka.  
In this book, the characteristic polynomial of a linear map $F$ is defined by $\det(A - \lambda I)$, where $A$ is a matrix which represents $F$.  
And in this book, the author defines a determinant only for a matrix whose elements belong to a some field $K$.  
If $\det(A - \lambda I)$ is a polynomial, then the elements of $A - \lambda I$ are polynomials too. But the author didn't define a determinant for a matrix whose elements are polynomials.  
 A: The characteristic polynomial of $T$ (either a matrix or a linear transformation, depending on your preference) is a polynomial, not a function. What we really care about are its coefficients. For instance, the leading coefficient is always $1$ (so that is boring) but the degree of the polynomial is the dimension of the ambient vector space. The next coefficient is (up to a sign) the trace of $T$. The free coefficient is the determinant. The other coefficients also have meaning directly expressed in $T$. All of this will be lost if you considered the polynomial merely as a function since over certain fields this process destroys the coefficients. 
A: The characteristic polynomial is actually ... a polynomial!
Here are more details about the definition of the determinant and of the characteristic polynomial in general case. In the case of Linear Algebra, $M$ would be an $n$-dimensional vector space over $R$ (a field).
For every free unital module $M$ of finite rank $n$ over a commutative
unital ring $R$ and for every endomorphism $a$ of $M$, the
determinant of $a$ is defined by the identity
$$
  ax_1\wedge\dotsb\wedge ax_n
  = (\det a)(x_1\wedge\dotsb\wedge x_n)\qquad
  (x_1,\dotsc,x_n\in M).
$$
If $S$ is a unital $R$-algebra, then there is a natural homomorphism
$$
  \operatorname{End}_{R}(M)\otimes_RS\to\operatorname{End}_{S}(M\otimes_RS).
$$
Since $M$ is assumed to be free of finite rank, it can be shown that
this homomorphism is an isomorphism:
$$
  \operatorname{End}_{R}(M)\otimes_RS\cong\operatorname{End}_{S}(M\otimes_RS).
$$
The characteristic polynomial of $a\in\operatorname{End}_R(M)$ is
$\chi_a\in R[X]$ defined by
$$
  \chi_a =\det(a - X),
$$
where $a - X = (a\operatorname{id}_M)\otimes 1 -\operatorname{id}_M\otimes X\in\operatorname{End}_{R}(M)[X] =\operatorname{End}_{R}(M)\otimes_RR[X]$
is viewed as an element of $\operatorname{End}_{R[X]}(M[X])$, where $M[X] = M\otimes_RR[X]$.
A: Nice question! In many cases, that distinction is irrelevant, but in some cases it matters. And, when it matters, you are not right: it is a polynomial, not a polynomial function. For instance, polynomials have degrees, whereas polynomial functions don't (for instance, over $\mathbb F_2$ the polynomial function $x\mapsto x^2+x$ is the null function, but the polynomial $x^2+x$ still has degree $2$, whereas the null polynomial still has degree $0$). And the degree of the characteristic polynomial of a $n\times n$ matrix is $n$.
