Consider the eigenvalue equation $AX=\lambda X$ where A is a $n\times n$ square matrix and $\lambda$ is the eigenvalue corresponding to the eigenvector $\lambda$. For two square matrices, A and $\det(AB)=\det A.\det B$. How to take the determinant of both sides of the equation $(A-\lambda I)X=0$ where $A-\lambda I$ is a $n\times n$ matrix and $X$ is a column vector.
My question is how to show that for nontrivial $X$, $\det(A-\lambda I)=0$? This is a matrix equation and not an algebraic equation of the form $ax=0$.