Showing that linear transformations $1, T, T^2, T^3 ,\dots $ do not span the set of linear transformations of $ \mathbb C^n$ into $ \mathbb C^n$ Suppose $T : \mathbb C^n \rightarrow \mathbb C^n$, $n \geq 2$ is a linear transformation. Show that the linear transformations $1,T, T^2, \dots$ do NOT span $L(\mathbb  C^n, \mathbb C^n)$, the set of linear transformations of $ \mathbb C^n$ into $ \mathbb C^n$. 
So I think one way to prove this is to show an upper bound on the dimension of the space spanned by the powers of $T$? However, I'm not sure how to do this.
1 is the identity map, i.e., $1(x) = x $.
 A: The dimension of the space of linear operators is $n^2$, whereas the space consisting of polynomials of $T$ is of dimension at most $n$, by the Cayley-Hamilton Theorem. Thus, since $n<n^2$ for $n\geq 2$, powers of $T$ don't span the space of linear operators.
Alternatively, note that all polynomials of $T$ commute with each other. If these generated all linear operators, then all linear operators would commute with each other, which is false if $n\geq 2$.
A: I'll talk about matrices, which is equivalent, and over any field $k$.
Suppose that the powers of a matrix $T\in M_n(k)$ span $M_n(k)$ as a vector space. It follows at once that $T$ commutes with every element of $M_n(k)$, so that it is in the center of $M_n(k)$.
Now everyone should know that the center of $M_n(k)$ is one-dimensional and generated as a vector space by the identity matrix $I$. It follows that our $T$ is in fact a scalar multiple of the $I$ and, of course, so are its powers, so that the subspace generated by them is one-dimesional. In view of out hypothesis, this implies that $n=1$. This proves what you want.
N.B. The powers of a matrix span a subspace of dimension exactly equal to the degree of the minimal poynomial of the matrix. 
A: Alternatively, every one-dimensional eigenspace of $T$ is an invariant subspace of every polynomial in $T$. So, whenever a linear operator maps an eigenvector $v$ of $T$ to something that is linearly independent of $v$ (such a linear map exists because $n\ge2$), it cannot be written as a polynomial in $T$.
A: If the powers of a matrix span $M_n(k)$, then it is follows easily that we have $AB=BA$ for all $A,B\in M_n(k)$. Since there are elements in $M_n(k)$ which do not commute as soon as $n\geq2$, what you want follows.
