I have a matrix $A$ as follows: $$ A=\begin{pmatrix} 0 & \boldsymbol{W} \\ \boldsymbol{W}^{\dagger} & \boldsymbol{H} \end{pmatrix} $$ where $H$ and $W$ are a random Hermitian $N\times N$ matrix and an $N$-component vector of independently distributed complex variables, respectively. The matrix elements have zero mean and variances $$ \langle H_{kl}H^{*}_{mn}\rangle_H=\frac{\lambda^2}{N}\delta_{km}\delta_{ln},\ \langle W_kW^{*}_l\rangle_W=\frac{g\lambda^2}{N}\delta_{kl}. $$ The definition of resolvent is $$ G(z)=\frac{1}{z-A} $$ after a ensemble average. My question is: Why the diagonal element $G_{11}(z)=\left(1\ 0\ 0\ \dots\right)G(z)\left(1\ 0\ 0\ \dots\right)^T$ is $$ G_{11}(z)=\frac{1}{z-ig\lambda} $$ as written in the textbook.

## My solution

A generic way to calculate the GF is using projection operators method. Denoting the projector $$ P=\left(1\ 0\ 0\ \dots\right)^T\left(1\ 0\ 0\ \dots\right),Q=I_{N+1,N+1}-P, $$ that satisfy $P^2=P,Q^2=Q,QP=PQ=0$. Rewritting $A$ as $A=H_0+V$ with $$ H_0=\begin{pmatrix} 0 & 0 \\ 0 & H \end{pmatrix}, $$ which satisfies $QH_0P=PH_0Q=0,QVQ=PVP=0$. After some algebras, the projected GF $PG(z)P$, which is a $1\times1$ matrix with element given by $G_{11}(z)$, reads as $$ PG(z)P=\frac{P}{z-PH_0P-PR(z)P}. $$ Here the matrix $R(z)$ is $$ R(z)=V+V\frac{Q}{z-H_0}V+V\frac{Q}{z-H_0}V\frac{Q}{z-H_0}V+\cdots. $$ Using the relation $QVQ=PVP=0$, the projected GF has this form $$ PG(z)P=\frac{P}{z-PV\frac{Q}{z-H_0}VP}. $$ Now calculating the denominator $$ PV\frac{Q}{z-H_0}VP=\sum_{i,j,m}W_iW^*_j\frac{c^m_i{c^m_j}^*}{z-E_m}=\sum_{i,j,m}W_iW^*_j\frac{1}{z-E_m}\frac{1}{N}, $$ where $E_m$ is the $m$-th eigenvalue of $H$ and $c^m_i$ the $i$-th component of the normalized eigenvector associated with the jth eigenvalue of H. Taking a ensemble average, this gives $$ G_{11}(z)=\frac{1}{z-g\lambda^2\left(z-\sqrt{z^2-4\lambda^2}\right)}. $$ This answer is obvious not consistent with $1/\left(z-ig\lambda\right)$ in the limit $z=0$ for $PR(z)P$. Where did I go wrong with the calculation?