Suppose $\phi:V\rightarrow W$ is a Lie algebra isomorphism between matrix Lie algebras $V$ and $W$. Since $V$ and $W$ are real vector spaces of matrices, we can consider the traces of their elements.
My question is, do we have $\operatorname{tr} \phi(X) = \lambda\cdot\operatorname{tr} X$ for some $\lambda\ne 0$? If not, is there a good counterexample?
I am thinking we can choose a basis $\{E_{1}, \ldots, E_{n}\}$ for $V$, take $F_{j} := \phi(E_{j})$ for all $j$, and then consider structure constants
$$ [E_{i}, E_{j}] = c_{ijk}E_{k} $$
where we sum over repeated indices (Einstein summation notation).
Since $\phi$ is a Lie algebra isomorphism, we have
$$ [F_{i}, F_{j}] = c_{ijk}F_{k}. $$
By taking the traces, and noting that the trace of commutators is zero, we find
$$ c_{ijk}e_{k} = 0 \qquad\text{ and }\qquad c_{ijk}f_{k} = 0 $$
where $e_{k} := \operatorname{tr} E_{k}$ and $f_{k} := \operatorname{tr} F_{k}$.
Now my idea is to note that the $c_{ijk}$'s are determined, and our question is simply what is the null space / kernel of the $n^{2}\times n$ matrix $C = (c_{ij, k})$ ($ij$ determine row; $k$ determines column), but it's not clear how to proceed from here, or if this approach is any good to begin with. Unfortunately, I don't know of many non-trivial matrix Lie algebra examples at the moment.
Best Answer
If $V$ and $W$ are abelian Lie algebras, then any linear map $V\to W$ is a Lie algebra homomorphism. So, for instance, if $V$ is some vector space of commuting matrices (say, all the diagonal matrices, or all matrices that are polynomials in some given matrix), you are asking whether every invertible linear map $V\to V$ preserves the trace up to scaling. This is obviously false as long as $V$ contains two linearly independent elements with nonzero trace, since an invertible linear map can (for instance) scale those elements by different factors.