Monotonicity of $\log \det R(d_i, d_j)$

Solution 1:

The result follows from some properties of the positive semidefinite (Loewner) ordering ($X\geq Y$ and $X>Y$ mean that $X-Y$ are positive semidefinite and positive definite, respectively).

Let $A=H_iv_iv_i^*H_i^*$, $B=H_jv_j v_j^*H_j^*$, C=$\sigma^2 I$, $x=d_i^{-\alpha}$, $y=d_j^{\alpha}$. We shall only need that $A\geq 0$, $B\geq 0$, $C>0$ in the following.

The function simplifies to $\log\det(I+ xA(yB+C)^{-1})$. As argued by Vedran, we may get rid of the log. The required determinant is equal to that of $I+x(yB+C)^{-1/2}A(yB+C)^{-1/2}$ (using the identity $\det(I+XY)=\det(I+YX)$).

If $0<x_1<x_2$, $0 \leq x_1 (yB+C)^{-1/2}A(yB+C)^{-1/2} \leq x_2 (yB+C)^{-1/2}A(yB+C)^{-1/2}$, hence all eigenvalues of the LHS are smaller than those of the RHS and the required determinant is smaller.

We may also replace that determinant with the one of $I+A^{1/2}(yB+C)^{-1}A^{1/2}$.

If $0<y_1<y_2$ then $0<y_1B+C<y_2B+C$ and $0<(y_2B+C)^{-1}<(y_1B+C)^{-1}$, hence also $0<A^{1/2}(y_2B+C)^{-1}A^{1/2}<A^{1/2}(y_1B+C)^{-1}A^{1/2}$ and we conclude as above.

Solution 2:

Obviously, $\log$ is of no interest here, as it is a monotonous function. So, we observe

$$f(d_i, d_j) = \det \left( {\rm I} + (d_{i}^{-\alpha} H_i v_i v_i^* H_i^*)(d_j^{-\alpha} H_j v_j v_j^* H_j^* + \sigma^2 {\rm I})^{-1} \right).$$

Since $H_i$, $H_j$, $v_i$, and $v_j$ are constant matrices/vectors, we can simplify this a bit by introducing $u_i := H_i v_i$ and $u_j := H_j v_j$:

$$f(d_i, d_j) = \det \left( {\rm I} + (d_{i}^{-\alpha} u_i u_i^*)(d_j^{-\alpha} u_j u_j^* + \sigma^2 {\rm I})^{-1} \right).$$

Using Sylvester's determinant theorem or, more precisely, its consequence (taken from here)

$$\det(X + AB) = \det(X) \det( I_n + BX^{-1}A ),$$

for $A = {\rm I}$, $B = d_{i}^{-\alpha} u_i u_i^*$, and $X = d_j^{-\alpha} u_j u_j^* + \sigma^2 {\rm I}$, we see that

\begin{align*} f(d_i, d_j) &= \det \left( {\rm I} + (d_{i}^{-\alpha} u_i u_i^*)(d_j^{-\alpha} u_j u_j^* + \sigma^2 {\rm I})^{-1} \right) \\ &= \frac{\det \left( (d_j^{-\alpha} u_j u_j^* + \sigma^2 {\rm I}) + (d_{i}^{-\alpha} u_i u_i^*) \right)}{\det (d_j^{-\alpha} u_j u_j^* + \sigma^2 {\rm I})}. \end{align*}

Using the fact that $u_i u_i^*$ is positive semidefinite, and that $d_i^{-\alpha}$ decreases when $d_i$ increases, it should be fairly easy to show that the numerator also decreases when $d_i$ increases.

Right now I don't have a good idea for $d_j$, so I'll just leave this half answered. However, it does seem like the statement holds.