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I know that the roots of a Lie Algebra are functionals such that if $\alpha$ is a root and $h \in \mathfrak h$ is an element of the Cartan subalgebra, then $\alpha(h)$ is an eigenvalue.

I'm looking for an explanation as to why a root evaluated on a Cartan element is equal to the Killing form of that Cartan element with the root's associated dual basis element in the Cartan subalgebra - in other words for any $h \in \mathfrak h$, $$\alpha (h)=K (h, \text{basis of cartan dualed with alpha})$$

I tested this with $\mathfrak{sl}_2$ (only basis element diag $(2,0,-2)$ and $h$=diag $(4,0,-4)$) the Killing form of these two is $16$, when $\alpha(h)=4$.

Thanks

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If I've understood your question correctly, the answer lies in how we define the dual basis for $\mathfrak h$.

We can define a map $\mathfrak h \to \mathfrak h^*$ given by $$h^*(x) = K(h, x)$$for $h \in \mathfrak h$. This map is linear, and we can prove it gives an isomorphism $\mathfrak h \to \mathfrak h^*$.

Now, the set of roots $\Phi$ is a subset of $\mathfrak h^*$, and hence, under the above isomorphism, for each $\alpha \in \Phi$, $\exists h_\alpha \in \mathfrak h$ such that $$\alpha(x) = K(h_\alpha, x)$$

We must choose our basis with respect to the Killing form. Note that for $\mathfrak {sl}_2$, in order for the basis you've described to be the correct basis, the Killing form will actually be $$K(x,y) = \frac14 \mathrm{Tr}(ad_x\circ ad_y)$$In this case, you'll find that the numbers do add up.

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