Here' s an alternative way to find characteristic polynomials of operators.  It depends on there being some vector v whose images v, Tv, T^2 v, etc generate the whole vector space, so it doesn't always work.  However, when the dimension of the vector space is large, it's much more practical than the determinant method for finding characteristic polynomials.

Anyhow, let's consider the operator T:R^4 -> R^4 with standard matrix

MatrixForm[A={{1,2,3,1},{-1,3,2,0},{0,1,3,2},{4,-1,2,1}}]

I take the first standard basis vector for my "v" and then compute its image under T, T^2, etc.

v0 = {1,0,0,0}

v1 = A.v0

v2 = A.v1

v3 = A.v2

Now I stop for a second to see if I' ve gone far enough

Solve[c3 v3 + c2 v2 + c1 v1 + c0 v0 == 0,{c0,c1,c2,c3}]

The vectors are still independent.  So I need to keep going (note that this is *good* news).  Since the vector space at hand is only four dimensional, I know that the next vector will be the last one and that v, Tv, T^2 v, T^3 v will give me a basis for R^4.

v4 = A.v3

Solve[v4 +c3 v3 + c2 v2 + c1 v1 + c0 v0 == 0,{c0,c1,c2,c3}]

So there' s my dependence relation.  The characteristic polynomial of my operator should accordingly be:  x^4 - 8 x^3 + 14 x^2 - 6 x - 3

Just in case, I recompute it the old way.

Det[x IdentityMatrix[4]-A]

Tada