Why Is the Trace of a Linear Operator Well-Defined?

[strugglinginmat]

I understand the definition of trace and linear operator individually but I don't seem to understand as to what does it mean by trace of a linear operator on a finite dimensional linear space.
What I have found out is that trace of a linear operator on a finite dimensional linear space is the trace of any matrix which represents the operator relative to an ordered basis of the space. I am confused as why is this definition well defined.
If T:V->V is the linear operator defined on V by T(A)=BA for all A in V and B is a fixed matrix. How do I represent T relative to standard ordered basis for V where V is the linear space of all 2X2 real matrices.

 

[CompuChip]

As for the first part: you can prove it as follows: Let L be a linear operator and A its matrix representation w.r.t. some chosen basis.
- Check that the representation w.r.t. any other basis can be written as $$D A D^{-1}$$, where D is an invertible ("change of basis") matrix.
- Now check the cyclic property for the trace (e.g. by writing out in components) $$\mathrm{Tr}(A B C) = \mathrm{Tr}(B C A) = \mathrm{Tr}(C A B)$$
- Combine them, you'll see that the trace is the same in any basis. So it's well-defined.

Now in general, you can write out the matrix of a linear transformation by finding out how it acts on the basis vectors. For example, suppose A mirrors the plane in the origin. Take the standard basis i = (1, 0), j = (0, 1). Now Ai = (-1, 0) and Aj = (0, -1). Putting these as columns of A gives $$A =\left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right)$$. Now check for yourself, that this does indeed produce the correct result for any vector (hint: write it out in components w.r.t. to the basis {i, j}).

Hope that gets you started. I left the details out on purpose, if you get stuck anywhere just ask :)