# Equality concerning the norm of rows of a resolvent matrix.

This problem showed up on UCLA's basic exam for Fall 2018:

Let $$X$$ be an $$n \times n$$ symmetric (real) matrix and $$z \in \mathbb{C}$$ with $$\text{Im } z > 0$$. Define $$G = (X - zI)^{-1}.$$ Show that $$\sum_{1 \leq j \leq n} |G_{ij}|^2 = \frac{\text{Im } G_{ii}}{\text{Im }z}.$$

I worked on this for a little bit where I applied the real spectral theorem to $$X$$ which in turn gives you that $$G = Q^T D Q$$ where $$Q$$ is real orthogonal ($$Q^T Q = Q Q^T = I$$) and $$D$$ is diagonal satisfying $$D_{ii} = (\lambda_i - z)^{-1}$$. Where $$\lambda_1, \ldots, \lambda_n$$ are the eigenvalues for $$X$$. Couldn't really see any immediate way out from there. Would be interested in seeing peoples' solutions to this problem and any connections to the study of matrix resolvents: https://en.wikipedia.org/wiki/Resolvent_formalism

By considering $$\tilde{X}=X-\text{Re}(z)I$$ in place of $$X$$, we may assume that $$z=\mu i$$ for some $$\mu>0$$. Now observe that $$(X-\mu iI)(X+\mu iI) = X^2+\mu^2 I>0.$$ This gives $$G=(X-\mu iI)^{-1}=(X^2+\mu^2 I)^{-1}(X+\mu iI).$$ Let $$e_i$$ be the vector whose $$i$$-th coordinate is $$1$$ and other coordinates are all $$0$$'s. We can observe that $$\begin{eqnarray} G_{ii}=e_i'Ge_i&=&e_i'(X^2+\mu^2 I)^{-1}(X+\mu iI)e_i\\&=&e_i'(X^2+\mu^2 I)^{-1}Xe_i+i\mu \cdot e_i'(X^2+\mu^2 I)^{-1}e_i \end{eqnarray}$$ and hence $$\text{Im}(G_{ii})=\mu\cdot e_i'(X^2+\mu^2 I)^{-1}e_i.$$ This gives $$\frac{\text{Im}(G_{ii})}{\text{Im}( z)}=\frac{\mu\cdot e_i'(X^2+\mu^2 I)^{-1}e_i}{\mu}= e_i'(X^2+\mu^2 I)^{-1}e_i.$$ We can also see that $$GG^*=(X-\mu iI)^{-1}(X+\mu iI)^{-1}=(X^2+\mu^2 I)^{-1}$$ and $$\sum_{j=1}^n |G_{ij}|^2$$ can be represented as $$\sum_{j=1}^n |G_{ij}|^2 =|G^*e_i|^2=e_i'GG^*e_i.$$ Thus it follows $$\sum_{j=1}^n |G_{ij}|^2 =e_i'GG^*e_i=e_i'(X^2+\mu^2 I)^{-1}e_i=\frac{\text{Im}(G_{ii})}{\text{Im}( z)}.$$ This proves the desired result.
Note: $$e_i'$$ means the transpose of $$e_i$$ and $$G^*$$ means conjugate transpose of $$G$$.
• Awesome. So it seems to be the key insight is that the diagonal entries of $G G^{*}$ give us the values for the sums we are concerned with, and by considering what $G G^{*}$ looks like in terms of $X$ with some clever manipulation we get what we want. – BenB Jan 19 at 21:01
We first note that $$(GG^*)_{ii} = \sum_{j = 1}^n |G_{ij}|^2$$. Thus our statement reduces to showing that $$(GG^*)_{ii} = \frac{\text{Im } G_{ii}}{\text{Im }z} \iff \text{Im }[(zG^*G)_{ii}] = \text{Im }G_{ii}.$$ Well $$G = (X - zI)^{-1} \implies G(X - zI) = I \implies GXG^* - zGG^* = G^* \implies GXG^* - G^* = zGG^*$$ Thus $$\text{Im }[(zGG^*)_{ii}] = \text{Im }[(GXG^* - G^*)_{ii}] = \text{Im }[(GXG^*)_{ii}] + \text{Im }[(-G^*)_{ii}] = \text{Im }[(GXG^*)_{ii}] + \text{Im }[G_{ii}].$$ Well we then notice that since $$GXG^*$$ is self-adjoint that $$(GXG^*)_{ii} = e_i^*GXG^*e_i = \langle GXG^*e_i, e_i\rangle \in \mathbb{R}$$ where $$e_i$$ be the $$i$$th standard basis vector, i.e. the vector containing all zeros except for a one in the $$i$$th component. Thus $$\text{Im }[(zGG^*)_{ii}] = \text{Im }[(GXG^*)_{ii}] + \text{Im }[G_{ii}] = \text{Im }[G_{ii}]$$ as desired.