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


2 Answers 2


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 $G$.

  • $\begingroup$ 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. $\endgroup$
    – BenB
    Commented Jan 19, 2019 at 21:01
  • $\begingroup$ @BenB I think so. Perhaps such manipulations were the key. I hope this will help. $\endgroup$ Commented Jan 19, 2019 at 21:03

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.


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