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 
 A: 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 $G$.
A: Here's an alternative solution:
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.
