Relation of $\det(M(x)\delta(x-y))$ to $\det(M(x))$?

Question

The two functional determinants $\det(M(x)\delta(x-y))$ to $\det(M(x))$ often appear in path integration (aka functional integration). From the form in which they appear in these integrals it is apparent that: $$ \det(M(x)\delta(x-y))\propto \det(M(x))$$ is it possible to show this? And does it only hold when the determinants are evaluated on certain special spaces or does it hold more generally?

Answer 1

Consider a standard matrix $M_{ij}$ the determinant of this matrix is given by: $$\det(M_{ij})=\prod_k \lambda_K \tag{1}$$ where $\lambda_k$ are the eigenvalues of $M_{ij}$ defined by: $$M_{ij}v_j^{(k)}=\lambda_k v^{(k)}_i$$ for the continuous 'matrix' of the form $M(x,y)$ the analogous to the above equation is: $$\int dy M(x,y) f_k(y)=\lambda_kf_k(x)$$ ($k$ here and in the above labels eigenvectors). Thus for $M(x,y)=M(x) \delta(x-y)$ we get: $$\int dy M(x) \delta(x-y)f_k(y)=\lambda_kf_k(x)$$ $$M(x)f_k(x)=\lambda_kf_k(x)\tag{2}$$ (2) is thus the eigenvalue equation solved when finding the determinate of $M(x,y)$. It is also that solved when finding $\det(M(x))$ which I now realize is a slight abuse of notation. Thus we have that: $$\det(M(x)\delta(x-y))= \det(M(x))$$ which holds generally. I am even tempted to write that: $$\det(M(x)\delta(x-y))\equiv \det(M(x))$$ since $M(x)$ is not really a 'matrix' acting on its own.