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 $.

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 $n\geq 2$.
