Prove existence of Diagonalizable Matrix Suppose R, T $\in L(F^3)$ each have 2, 6, 7 as eigenvalues. Prove that there exists an invertible operator S $\in L(F^3)$ such that $R=S^{-1}TS$.

What I got so far is that since R and T have three eigenvalues each then R and T must be diagonalizable? I'm not really sure how to approach this question help?
Since both $R$ and $T$ have 3 distinct eigenvalues, they are diagonalizable i.e. there exists invertible matrices $P, Q$ such that $R = P D P^{-1}$ and $T = Q D Q^{-1}$, where $D = \textrm{diag}(2,3,5)$. Substituting one equation into the other, we get that
$$
R = P (Q^{-1} T Q)P^{-1} = (PQ^{-1}) T (PQ^{-1})^{-1}.
$$
Thus, $S = PQ^{-1}$ is the invertible matrix you're after.
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Since R and T both have 3 distinct eigenvalues, R and T are both diagonalizable. Hence, there exists invertible operators A and B such that $R = ADA^{-1}$ and $T = BDB^{-1}.$ $$B^{-1}TB = B^{-1}BDB^{-1}B$$ $$AB^{-1}TBA^{-1} = ADA^{-1}$$ $$(AB^{-1})T(BA^{-1}) = R $$ $$S^{-1}TS = R$$