I want to know whether $GL_n(\mathbb R)$ is a unimodular Lie group, that is, whether it has a Haar measure that is both left and right invariant.
What I've tried so far is seeing $GL_n(\mathbb R)$ as an open submanifold of $\mathbb R^{n^2}$ and I've tried to prove that for any $A\in GL_n(\mathbb R)$ and any open $U\subseteq GL_n(\mathbb R)$ we have $\mu(A\cdot U)=\mu(U)$, where $\mu$ is the Lebesgue measure of $\mathbb R^{n^2}$, but I've had no success.
I don't know whether the same strategy works for $GL_n(\mathbb C)$.
Yes, both of them are unimodular. A (both left and right) Haar measure on $GL_n(\mathbb{R})$ is $\frac{\mathrm dx}{|\det x|^n}$, where $\mathrm dx$ is the Lebesgue measure on the set of $n\times n$ matrices. A similar argument works for $GL_n(\mathbb{C})$; this time, a Haar measure will be $\frac{\mathrm dx}{|\det x|^{2n}}$.