# Surjective lie algebra homomorphism and diagonalisation of ad map

I was wondering if anyone could check my solution to the following exercise (from Erdmann and Wildon's Introduction to Lie Algebras):

True or false: Let $$\phi: L_1 \to L_2$$ be a surjective Lie algebra homomorphism. Let $$h \in L_1$$. If ad $$h$$ is diagonalisable, then ad $$\phi(h)$$ is diagonalisable. What is different if $$\phi$$ is an isomorphism?

My solution

I claim the statement is true if $$\phi$$ is an isomorphism. Proof:

Let $$\lambda_1, \cdots, \lambda_n$$ be the n (not necessarily distinct) eigenvalues of ad $$h$$, and let $$x_1, \cdots, x_n$$ be associated linearly independant eigenvectors (which exist because ad $$h$$ is diagonalisable). Then $$(x_1, \cdots, x_n)$$ is a basis for $$L_1$$. So $$(\phi(x_1), \cdots, \phi(x_n))$$ is a basis for $$L_2$$. They are eigenvectors because we have ad $$\phi(h)(\phi(x_i)) = [\phi(h), \phi(x_i)] = \phi([h, x_i]) = \phi(\lambda_ix_i) = \lambda_i\phi(x_i)$$. So
ad $$\phi(h)$$ has a basis of eigenvectors, so is diagonalisable.

My guess is the statement would be false if $$\phi$$ is only surjective, because then there is no guarantee that the image of a set of eigenvectors of ad $$h$$ is linearly independant. However I can't come up with a counterexample; any ideas?

I'd say your proof shows in general that a Lie algebra homomorphism $$\phi$$ maps eigenvectors of $$\mathrm{ad}(h)$$ to eigenvectors of $$\mathrm{ad}(\phi(h))$$; besides, any surjective linear map $$\phi$$ between vector spaces maps any basis to a generating system. So if $$x_1, ..., x_n$$ is a basis of $$L_1$$ consisting of $$\mathrm{ad}(h)$$-eigenvectors, then $$\phi(x_1), ..., \phi(x_n)$$ are a generating system of $$L_2$$ consisting of $$\mathrm{ad}(\phi(h))$$-eigenvectors. True, they might not be linearly independent any more, but by basic linear algebra they contain some basis of $$L_2$$, and that basis still consists of $$\mathrm{ad}(\phi(h))$$-eigenvectors.