The Cartan Generators of $SU(3)$ in the three dimensional rep have eigenvalues $(1,-1,0)$ and $\frac{1}{\sqrt{3}} (1,1,-2)$. Therefore we have the weights:
$$ (1,\frac{1}{\sqrt{3}}) \quad (-1,\frac{1}{\sqrt{3}}) \quad (1,\frac{-2}{\sqrt{3}}) $$
In the Dynkin basis the weights of the 3 dimensional rep are
$$ [1,0] \quad [-1,1] \quad [0,1] $$
How are these connected to the eigenvalues of the Cartan generators quoted above? I thought that I get them by multiplying the weights in the Dynkin basis with the corresponding metric tensor (=the inverse of the Cartan matrix) of $SU(3)$:
$$ G=\frac{1}{3} \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} $$
For example,
$$ G [1,0] = \frac{1}{3} \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix}1 \\ 0 \end{pmatrix} = \begin{pmatrix}\frac{2}{3} \\ \frac{1}{3}\end{pmatrix} $$
Unfortunately this yields the wrong weights. What is wrong and how can I compute correctly the eigenvalues of the Cartan generators from the weights in the Dynkin basis?
The thing is, by multiplying the weights in den Dynkin basis with the metric tensor, we get the weights in the simple root basis. If we want to know the weights in terms of the Cartan generator $H_1,H_2,...$ eigenvalues, we need to use the simple roots in this basis.
For $SU(3)$ the simple roots are $\alpha_1= (\frac{1}{2}, \frac{\sqrt{3}}{2})=\frac{1}{2} H_1 + \frac{\sqrt{3}}{2} H_2 $ (the last equality should be understood symbolically) and $\alpha_2= (\frac{1}{2}, \frac{-\sqrt{3}}{2})$.
Then we can rewrite the weight I computed in the OP $ \begin{pmatrix}\frac{2}{3} \\ \frac{1}{3}\end{pmatrix} = \frac{2}{3} \alpha_1 + \frac{1}{3} \alpha_2$. Using the explicit form of the simple roots then yields
$$\frac{2}{3} \alpha_1 + \frac{1}{3} \alpha_2 = \frac{2}{3} \begin{pmatrix}\frac{1}{2} \\ \frac{\sqrt{3}}{2}\end{pmatrix} + \frac{1}{3} \begin{pmatrix}\frac{1}{2} \\ \frac{-\sqrt{3}}{2}\end{pmatrix} =\begin{pmatrix}\frac{1}{2} \\ \frac{1}{2\sqrt{3}}\end{pmatrix}$$
Now, recalling that our generators are in fact $ \frac{1}{2} $ times the Gell-Mann matrices this is the correct weight in terms of eigenvalues of the Cartan generators.