An operator that commutes with another operator $T$ with distinct characteristic values is a polynomial in $T$

Note: we will prove that the commutant of $T$ is equal to $K[T]$. This property is actually equivalent to the fact that the characteristic and minimal polynomials are equal. That's also equivalent to similarity to a companion matrix.

Let $\lambda_j$ be the $n$ pairwise distinct eigenvalues of $T$. Such an operator must be diagonalizable, and its eigenspaces are all one-dimensional. Since $U$ commute with $T$, it leaves the eigenspaces of $T$ invariant. Since they are one-dimensional, this implies that every eigenvector of $T$ is an eigenvector for $U$. So $T$ and $U$ are simultaneously diagonalizable.

In a basis of simultaneous diagonalization, $T=\mbox{diag}(\lambda_1,\ldots,\lambda_n)$ and $U=\mbox{diag}(\mu_1,\ldots,\mu_n)$.

Now we will do Lagrange interpolation. Since the $\lambda_j$ are pairwise distinct, we can consider the degree $n$ polynomials $$ L_j(x)=\prod_{i\neq j}\frac{x-\lambda_i}{\lambda_j-\lambda_i}\qquad L_j(\lambda_i)=\delta_{ij}. $$ Now $$p(x)=\sum_{j=1}^n\mu_jL_j(x)\qquad\mbox{satisfies }\quad p(\lambda_j)=\mu_j \;\forall j. $$ Therefore $p(T)=\mbox{diag}(p(\lambda_1),\ldots,p(\lambda_n))=U$ belongs to $K[T]$. The converse is clear, so the commutant of $T$ is $$ \{T\}'=\{U\in L(V)\,;\,UT=TU\}=K[T] $$ the subagebra of all polynomials in $T$.

Note: for a less explicit argument, just compare the dimensions of $K[T]$ and the subspace of all diagonal operators in this basis. The latter is clearly $n$. The former is $n$ by minimal polynomial consideration and Euclidean division. And clearly $K[T]$ is contained in the commutant, so they must be equal. Oh...but that's TTS argument...

It seems to me that you've found necessary and sufficient conditions on $U$. What do you think the dimension of the space of such $U$ is? You won't yet need to use that the eigenvalues are distinct.

Now, consider the space of polynomials in $T$. Elements of this space always commute with $T$. What's the dimension?