Let $A$ be a square matrix. Its eigenvalues satisfy $det(\lambda I-A)=0$. But is this the most basic definition of eigenvalues?

NO!

The basic definition is: $\lambda$ is an eigenvalue of $A$ if and only if there is a vector $v$ such that

$Av=\lambda v$

And it is easy to show $Av=\lambda v$ is equivalent to $det(\lambda I-A)=0$.