Understanding Gillman's proof of the Chernoff bound for expander graphs My question is about the proof of Claim 1 in this paper: Gillman (1993).
At the end of the proof, the author says:

The matrix product $U^\top\sqrt{D^{-1}}(P+(\mathrm{e}^x-1)B(0)-\mu I)\sqrt{D}U$, which is equal to $(D'-\mu I)(I+(D'-\mu I)^{-1}(\mathrm{e}^x-1)D'U^\top D_A U)$, is singular. Therefore,
\begin{align*}
1&\leq \lVert (D'-\mu I)^{-1}(\mathrm{e}^x-1)D'U^\top D_A U \rVert_2 \\
&\leq \frac{1}{\mu-\lambda_2}(\mathrm{e}^x -1).
\end{align*}
(The first inequality uses the continuity of the function
  $\lambda_2(y)$.) 

I understand why the two expressions at the beginning are equal and I understand the second inequality, but I do not understand the first inequality and also why the matrix product is singular.
Let me provide the definitions so you can avoid reading the whole paper. There is a weighted undirected graph $G=(V, E, w)$ where $w_{ij}=0$ if $\{i,j\}\notin E$. Denote $w_i:=\sum_j w_{ij}$. Let $P$ denote the transition matrix, so $P_{ij}:=\frac{w_{ij}}{w_i}$. Denote by $\lambda_2$ the second largest eigenvalue of $P$ and by $\epsilon:=1-\lambda_2$ the spectral gap. Next, let $M$ be the weighted adjacency matrix $M_{ij}:=w_{ij}$. Let $A$ be a set of vertices and $\chi_A$ be an indicator function. Some more definitions are:
\begin{align*}
&E_r:=\operatorname{diag}(\mathrm{e}^{r\chi_A}) & &P(r):=PE_r \\
&D:=\operatorname{diag}(\frac{1}{w_i}) & &S:=\sqrt{D}M\sqrt{D} \\
&S_r:=\sqrt{DE_r}M\sqrt{DE_r} & & B(r):=\frac{1}{\mathrm{e}-1}(P(r+1)-P(r))
\end{align*}
Moreover, since $S$ is unitarily diagonalizable, there exist a unitary matrix $U$ and a diagonal matrix $D'$ such that $D'=U^\top SU$. Furthermore, there exists a diagonal matrix $D_A$ such that $B(0)=PD_A$.
Define $\lambda(r)$ to be the largest eigenvalue of $P(r)$ and $\lambda_2(r)$ to be its second largest eigenvalue. As before, $\epsilon_r := \lambda(r)-\lambda_2(r)$ is the spectral gap.
In Claim 1 the author lets $0\leq x\leq r$ be some number. He also defines $\mu<\lambda(x)$ to be any other eigenvalue of $P(x)$. At the end of the proof, we are only interested in $\mu>\lambda_2$.
Some other facts are:
\begin{align*}
&\lVert D' \rVert_2 = \lVert D_A \rVert_2 = 1 & &D'=U^\top\sqrt{D^{-1}}P\sqrt{D}U \\
&P(0)=P & &\lambda(0)=1 & &\lambda_2(0)=\lambda_2 \\
&P=\sqrt{D}S\sqrt{D^{-1}} & &P(r)=\sqrt{DE_r^{-1}}S_r\sqrt{E_rD^{-1}}
\end{align*}
I hope I didn't miss anything relevant. Thank you for your help.
 A: First, we'll explain why $U^\top\sqrt{D^{-1}}(P+(\mathrm{e}^x-1)B(0)-\mu I)\sqrt{D}U$ is sungular. It is enough to show that $P+(\mathrm{e}^x-1)B(0)$ has eigenvalue $\mu$. We know that $\mu$ is an eigenvalue of $P(x)$, so we'll show why $P+(\mathrm{e}^x-1)B(0) = P(x)$.
Note that $E_r:=\operatorname{diag}(\mathrm{e}^{r\chi_A})=(\mathrm{e}^r-1)\operatorname{diag}(\chi_A)+I$ because $\chi_A$ is binary. Therefore,
\begin{align*}
B(r)&:=\frac{1}{\mathrm{e}-1}(P(r+1)-P(r)) = \frac{1}{\mathrm{e}-1}P\cdot(E_{r+1}-E_r) \\
&= \frac{1}{\mathrm{e}-1}P\cdot((\mathrm{e}^{r+1}-1)\operatorname{diag}(\chi_A) - (\mathrm{e}^r-1)\operatorname{diag}(\chi_A)) \\
&= \frac{1}{\mathrm{e}-1}P\cdot(\mathrm{e}^{r+1}-\mathrm{e}^r)\operatorname{diag}(\chi_A) = \mathrm{e}^rP\operatorname{diag}(\chi_A)
\end{align*}
And specifically $B(0)=P\operatorname{diag}(\chi_A)$. Now,
\begin{align*}
P+(\mathrm{e}^x-1)B(0) &= P+(\mathrm{e}^x-1)P\operatorname{diag}(\chi_A) \\
&= P((\mathrm{e}^x-1)\operatorname{diag}(\chi_A)+I) \\
&= PE_x = P(x)
\end{align*}
Therefore $\mu$ is an eigenvalue of $P+(\mathrm{e}^x-1)B(0)$ and hence $U^\top\sqrt{D^{-1}}(P+(\mathrm{e}^x-1)B(0)-\mu I)\sqrt{D}U$ is sungular.
Second, denote $C:=(D'-\mu I)^{-1}(\mathrm{e}^x-1)D'U^\top D_A U$. We'll explain why $1\leq \lVert C \rVert_2$. We know that 
\begin{align*}
(D'-\mu I)(I+C) &= (D'-\mu I)(I+(D'-\mu I)^{-1}(\mathrm{e}^x-1)D'U^\top D_A U) \\
&= U^\top\sqrt{D^{-1}}(P+(\mathrm{e}^x-1)B(0)-\mu I)\sqrt{D}U 
\end{align*}
which we shown is singular, so $(D'-\mu I)(I+C)$ is singular. But $D'-\mu I$ is invertible, hence $I+C$ is singular. Now, there exists a vector $v\neq 0$ such that $(I+C)v=0$. So $Cv=-v$, and then $\lVert Cv \rVert = \lVert v \rVert$. Therefore $\frac{\lVert Cv \rVert}{\lVert v \rVert} = 1$ for some vector $v\neq 0$.
Finally, by definition
\begin{equation*}
\lVert C \rVert := \sup_{w\neq 0}\frac{\lVert Cw \rVert}{\lVert w \rVert}\geq \frac{\lVert Cv \rVert}{\lVert v \rVert} = 1
\end{equation*}
