Definition of characteristic polynomial The question is very simple but it's giving me a hard time. I've been given the following definition for the characteristic polynomial of a linear transformation $$p_c(x)=\det(xI-T)$$
But what does it mean to take the determinant of a linear transformation? I understand that you can find the characteristic polynomial by computing $\det(xI-A)$ where $A=[T]_B$ for $B$ an ordered basis of the space but that's taking the determinant of a matrix, which I understand.
 A: For the purposes of computing the determinant it suffices to represent the matrix according to any basis you like since it can be shown the result is independent of the chosen basis. However, this does not explain what the determinant of a linear transformation "really" is. 
For a linear transformation $T:V\to V$, where $V$ is a vector space in which it makes sense to speak of the determinant of $n$ vectors (e.g., by means of taking the determinant of the corresponding matrix, and specifically if $V=\mathbb R^n$ then this is related to volume) you can construct the multi-linear function $F:V\times V\cdots \times V \to \mathbb R$ by $F(v_1,\ldots ,v_n)={\rm det}(Tv_1,\ldots, Tv_n)$. It can then be shown that no matter which $n$ vectors you choose you will get that $F(v_1,\ldots, v_n)=\alpha \cdot {\rm det}(v_1,\ldots, v_n)$. This $\alpha $ is the determinant of the transformation T. 
In particular, if you choose $v_1,\ldots, v_n$ to be the standard basis in $\mathbb R^n$, then the determinant of $T$ is the determinant of the vectors $Tv_1,\ldots, Tv_n$. Thinking of determinants as expressing volume we see that the determinant of a transformation is the factor by which the transformation distorts the volume of standard cube. By the above, its also the factor by which it distorts any other parallelpipe determined by any $n$ vectors. This is geometrically what the determinant of a transformation is.   
A: It seems to me the problem is that you think the determinant is a property of a matrix, and a matrix alone, but it's commonly taken that the determinant is a property of a linear operator, and as such, the computation of the determinant using a matrix is just one way to figure out the value of this property.
You can define the determinant of a linear operator without reference to some matrix representation.  A common way to do this is with exterior algebra and wedge products:  the wedge product (denoted $\wedge$) produces elements of the exterior algebra.  If $k$ vectors are wedged together, then the object is an element of the grade-$k$ exterior algebra on $V$, denoted $\bigwedge^k V$.
Now, define the action of a linear operator on an element of $\bigwedge^2 V$ like so:  if $B = a \wedge b$, then
$$T(B) = T(a \wedge b) \equiv T(a) \wedge T(b)$$
and do so recursively for any $k > 2$ to generalize this to $\bigwedge^k V$.
The vector space $\bigwedge^n V$, where $n = \dim V$, is "one-dimensional", in the sense that all the elements are scalar multiples of each other.  Hence, if $I \in \bigwedge^n V$, then it must be true that
$$T(I) = \alpha I$$
for some scalar $\alpha$.  $\alpha$ obeys all the usual properties of the determinant of the matrix representation for $T$, so we can say it is the determinant.
A: Given an ordered basis $\beta= (b_1,...,b_n)$ you can find a corresponding matrix $T_\beta$, and we define the characteristic polynomial as $x \mapsto \det (xI-T_\beta)$.
To see that it is well defined, suppose $\beta'= (b_1',...,b_n')$ is another basis, then for some matrix $A$, we have $T_{\beta'} = A^{-1} T_\beta A$, and then
\begin{eqnarray}
\det (xI-T_{\beta'}) &=& \det (xI-A^{-1}T_{\beta}A) \\
&=& \det (xA^{-1}A-A^{-1}T_{\beta}A) \\
&=& {1 \over \det A} \det (xI-T_\beta) \det A \\
&=&\det (xI-T_\beta)
\end{eqnarray}
Hence the characteristic polynomial is independent of the choice of basis.
Another way of viewing it is to notice that if we choose a basis $\beta_J$ so 
that $T_{\beta_J}$ is the Jordan normal form, then
$\det(xI - T_{\beta_J}) = (x-\lambda_1)\cdots (x -\lambda_n)$, where $\lambda_k$ are the eigenvalues of $T$. This is, perhaps, a more intrinsic viewpoint?
A: Determinants of linear maps in real vector spaces can be understood geometrically. For example, if $u, v, w$ are three linearly independent vectors belonging to a three-dimensional vector space $\mathcal{V}$ associated with Euclidean space, then the determinant of the linear map $S$ 
$$
\det(S) = \frac{Su.(Sv \times Sw)}{u. (v \times w)}
$$
which is a measure of the ratios of the volumes formed by the triad $u,v,w$ after and before the linear map $S$ acts on them. 
Note that this definition is basis independent. 
A:  
Conceptually, taking $\det(xI−T)$ means that, no matter what basis B you use to obtain the matrix $A=[T]_B$, the resulting characteristic polynomial will be the same. So $\det(xI−T)$ is defined with respect to a basis, perhaps, but is ultimately basis independent. This is done a lot when mathematical entities are defined; e.g., if $H$is a normal subgroup of $G$, we can define coset multiplication in the factor group using coset representatives, then show that the product coset is the same no matter what representatives we choose. Representation theory is BIG.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: Every linear transformation $\mathbb{R}^n \to \mathbb{R}^n$ can be represented by a square matrix.  So the determinant can be calculated using this matrix in place of the linear transformation.  Its just substitution really.
A: Every linear operator (i.e. from a vector space into itself) on a finite-dimensional vector space can be realized as a square matrix. Say you have a linear transformation:
$$
   f: \mathbb{R}^n \to \mathbb{R}^n
$$
Then, given an ordered basis $B = ( b_1, \ldots, b_n)$ define $T_f^B$ to be
$$
T_f^B = 
\begin{bmatrix}
    f(b_1) &f(b_2)& \ldots& f(b_n)
\end{bmatrix}
$$
where $f(b_i)$ is the column vector of $b_i$ transformed by $f$. You may now use the matrix $T_f^B$ as the 'linear transformation', since in the context of this basis, they are indistinguishable.
