the $\log$ function has two poles, one at infinity and one at zero. The largest singular value can easily be bounded by a polynomial in $N$. The main difficulty is controlling the least singular value.

In order to study $\nu_{{\mathbf}M}$ it is useful to note that it is also the empirical spectral measure for the Hermitization of ${\mathbf}M$. The *Hermitization* of ${\mathbf}M$ is defined to be $${\mathbf}H:=\begin{bmatrix} 
                0 &  {\mathbf}M\\
                            {\mathbf}M^*&  0
            \end{bmatrix}.$$ For an $N \times N$ matrix, the Stieltjes transform of $\nu_{{\mathbf}{M} - z {\mathbf}{I}}$ is also the trace of the Hermitized resolvent. That is, for $\eta \in \mathbb{C}^{+} := \{ w \in \mathbb{C} : {\operatorname{Im}}(w) > 0\}$, we have $$\int \frac{ 1  }{x-\eta}\nu_{{\mathbf}{M} - z {\mathbf}{I}}(dx) = \frac{1}{2N} \operatorname{tr}({\mathbf}R( {\mathbf}q)),$$ where $${\mathbf}{R}({\mathbf}q) := ( {\mathbf}{H} -   {\mathbf}{q} \otimes {\mathbf}{I}_N)^{-1}, \quad {\mathbf}{q} := \begin{bmatrix}  \eta  & z  \\ \bar{z}  & \eta  \end{bmatrix}.$$ Here ${\mathbf}{q} \otimes {\mathbf}{I}_N$ denotes the Kronecker product of the matrix ${\mathbf}{q}$ and the identity matrix ${\mathbf}{I}_N$.

Typically, in order to estimate the measures $\nu_{{\mathbf}{M} -z {\mathbf}{I}}$, one shows that the Stieltjes transform approximately satisfies a fixed point equation. Then one can show that this Stieltjes transform is close to the Stieltjes transform that exactly solves 