norm of each block above is $O(N^{A})$ with probability $1-O(N^{-2})$. The treatment of each block is similar; as an illustration, we will show that $${\mathbb{P}}\left( \left\| \left(\frac{1}{N} {\mathbf}{Y}_{N,1} {\mathbf}{Y}_{N,2} - z^2 {\mathbf}{I} \right)^{-1} \frac{1}{\sqrt{N}} {\mathbf}{Y}_{N,1} \right\| \geq N^{A} \right) = O(N^{-2}).$$

Indeed, since $${\mathbb{P}}\left( \left\| \frac{1}{\sqrt{N}} {\mathbf}{Y}_{N,1} \right\| \geq N^{100} \right) \leq N^{-200} {\mathbb{E}}\left\| \frac{1}{\sqrt{N}} {\mathbf}{Y}_{N,1} \right\|^2_2 = O(N^{-2}),$$ it suffices to show that there exists $A>200$ such that $$\label{eq:show2prodbnd}
    {\mathbb{P}}\left( \left\| \left( \frac{1}{N} {\mathbf}{Y}_{N,1} {\mathbf}{Y}_{N,2} - z^2 {\mathbf}{I} \right)^{-1} \right\| \geq N^{A} \right) = O(N^{-2}).$$

By Theorem \[thm:NgLSV\], there exists $A'>0$ such that the event $$\Omega_N := \left\{ \left\| \left( \frac{1}{\sqrt{N}} {\mathbf}{Y}_{N,2} \right)^{-1} \right\| \leq N^{A'} \right\}$$ holds with probability $1-O(N^{-2})$. On this event, we observe that $$\left( \frac{1}{N} {\mathbf}{Y}_{N,1} {\mathbf}{Y}_{N,2} - z^2 {\mathbf}{I} \right)^{-1} = \left( \frac{1}{\sqrt{N}} {\mathbf}{Y}_{N,2} \right)^{-1} \left( \frac{1}{\sqrt{N}} {\mathbf}{Y}_{N,1} - z^2 \left(\frac{1}{\sqrt{N}} {\mathbf}{Y}_{N,2} \right)^{-1} \right)^{-1}.$$ Therefore, we conclude that $$\begin{aligned}
    {\mathbb{P}}&\left( \left\| \left( \frac{1}{N} {\mathbf}{Y}_{N,1} {\mathbf}{Y}_{N,2} - z^2 {\mathbf}{I} \right)^{-1} \right\| \geq N^{A} \right) \\
    &\leq {\mathbb{P}}\left( \left\| \left( \frac{1}{N} {\mathbf}{Y}_{N,1} {\mathbf}{Y}_{N,2} - z^2 {\mathbf}{I} \right)^{-1} \right\| \geq N^{A} \bigg| \Omega_N \right) {\mathbb{P}}(\Omega_N) + {\mathbb{P}}(\Omega_N^C) \\
    &\leq O(N^{-2}) + {\mathbb{P}}\left( \left\| \left( \frac{1}{\sqrt{N}} {\mathbf}{Y}_{N,1} - z^2 \left( \frac{1}{\sqrt{N}} {\mathbf}{Y}_{N,2}\right)^{-1} \right)^{-1} \right\| \geq N^{A/2} \bigg| \Omega_N 