Show that the determinant of jordan normal form wtih spatial weight matrix by its eigenvalues I try to figure out the proof of determinant of matrix by eigen decomposition, $$det(I_n-\lambda W)=det(QQ^{-1}(I_n-\lambda W))$$ $$=det(Q(I_n-\lambda W)Q^{-1})$$ $$=(1-\lambda \upsilon_1)(1-\lambda \upsilon_2)...(1-\lambda \upsilon_n)$$ where $\lambda$ is spatial autoregressive coefficient, $W$ is normalized weight matrix, $\upsilon$ is eigenvalues of $W$, $Q$ is the matrix that trinagularizes $W$. 1.i can't figure out how changes taking place of the $Q^{-1}$ matrix in front of $(I_n-\lambda W)$ in the first equation to behind of $(I_n-\lambda W)$ in the second equation? is there any proporties for that? 2.why do need $W$ is triangularized? can it be proof without triangularized? I would really appreciate some help.

So we know (given) that there is a $Q$ such that $QWQ^{-1}$ is triangular. The first step is to introduce $Q$

$$\det(I - \lambda W) = \det( Q^{-1}Q(I-\lambda W))$$ Second let's use the property $\det(AB) = \det(BA)$ where $A = Q^{-1}$ and $B = Q(I-\lambda W)$ hence

$$\det(I - \lambda W) = \det(Q(I-\lambda W) Q^{-1})$$ Third, expand the above $$\det(I - \lambda W) = \det(QQ^{-1}-\lambda Q W Q^{-1} )$$ Notice that $QQ^{-1} = I$ again so $$\det(I - \lambda W) = \det(I-\lambda Q W Q^{-1} )$$ Fifth denote our triangular matrix $QWQ^{-1} = T$ hence $$\det(I - \lambda W) = \det(I-\lambda T )$$ Notice that $I - \lambda T$ is also **triangular** with diagonal elements $1 - \lambda v_1$, where $v_k$ appear on the diagonal entries of $T$. Using the fact that **the determinant of a triangular matrix is nothing other than the product of it's diagonal entries** , then $$\det(I - \lambda W) =(1 - \lambda v_1)\ldots(1-\lambda v_n)$$