If $A$ and $B$ are Hermitian non-commuting matrices then it is known that they do not have a common eigenbasis. However, I believe that it is still true that they may share some eigenvectors. Assume $\vec{x}$ is a normalised common eigenvector of $A,B$ with eigenvalues $\lambda_A, \lambda_B$ respectively. If this is true then, $$ \vec{x}^\top A^2 \vec{x}= \left(\vec{x}^\top A \, \vec{x}\right)^2= \lambda_A^2 \,, \quad \vec{x}^\top B^2 \vec{x}= \left(\vec{x}^\top B \, \vec{x}\right)^2= \lambda_B^2 \,. $$ That in turn implies that the variances (e.g. $ Var (A) = \vec{x}^\top A^2 \vec{x}- \left(\vec{x}^\top A\, \vec{x}\right)^2$) of matrices $A$ and $B$ are both zero at the same time.
Now, assuming that what I have mentioned so far is correct and noting that observables in quantum mechanics are represented by Hermitian operators acting on a Hilbert space (let's just focus on the finite dimensional case), how doesn't this contradict the usual intuition from quantum mechanics where it is thought impossible to have states that assign precise values to non-commuting observables?
Is there something wrong in my understanding?
If you have $A$ and $B$ self adjoint operators on an Hilbert space $H$, and $[A,B]$ is a multiple of the identity in the domain $Dom(AB)\cap Dom(BA)$, let's say that $[A,B]_{|Dom(AB)\cap Dom(BA)}= \lambda Id$, then $\forall \psi \in Dom(AB)\cap Dom(BA) $, it holds that $$||A\psi||||B\psi||\geq \frac{|\lambda|}2$$ and $$ Var(A,\psi)Var(B,\psi) \geq \frac{|\lambda|^2}4.$$
For the observables position and momentum we have indeed that $\pi([q,p])=i\hbar \ Id$ (Heisenberg commutation relation), so we obtain the well known result that you cite.
In general as in your case, this theorem can be not applicable, or simply $\lambda = 0$ so there is no lower bound on the product of the variances (or there is a lower bound but this theorem will not tell you).
I hope this will help you. I recommend also to ask this question on https://physics.stackexchange.com/ where IMHO it belongs.