Show that the determinant of jordan normal form wtih spatial weight matrix by its eigenvalues

Question

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.

Answer 1

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 $$\det(I - \lambda W) =(1 - \lambda v_1)\ldots(1-\lambda v_n)$$

Answer 2

1. Just think of a theorem saying that $$\det(AB)=\det(A)\det(B)$$so we have $$\det(AB)=\det(BA)$$2. sure it can be proved in such way. Probably in your problem the matrix is particularly triangular but it doesn't require to remain so. To show that the key idea is that $$\text{the determinant of a matrix is multiplication of all its eigenvalues}$$can you finish this one?