Proof. We prove the results in three steps.

Step 1  \(a, \lambda _N\) and \(\phi \) are unchanged under orthogonal/unitary transformations of matrix \(C\). We do two transformations:

• Rotate \(v\) ti be the first basis vector \((1, 0^{N-1})\).
• Rotate in the orthogonal complement of \((1, 0^{N-1})\), so that the \((N-1)\times (N-1)\) bottom random corner of G\(\beta \)E becomes diagonal, while the law of the first row and column of G\(\beta \)E preserved.

Step 2  Now we brought \(C\) into \[ \hat {C} = \begin {pmatrix} a \delta _N + \mathcal {N}(0, \frac {2}{\beta }) & \xi _2 & \xi _3 & \dots & \xi _N \\ \bar {\xi }_2 & \mu _2 & & & \\ \bar {\xi }_3 & & \mu _3 & & \\ \vdots & & & \ddots & \\ \bar {\xi }_N & & & & \mu _N \end {pmatrix} \] and \(\mu _2,\cdots ,\mu _N\) be the eigenvalues of \(\sqrt {\frac {2}{\beta }} G\beta E\) and \(\xi _2,\cdots ,\xi _N \overset {i.i.d}{\sim } \sqrt {\frac {2}{\beta }}\) corresponding Normal distribution. We find an eigenvector \((x_1,\cdots ,x_N)\) of \(\tilde {C}\) with eigenvalues \(\lambda \) \[ \begin {cases} x_1\left (a\sqrt {N}+{\mathcal{N}} \left (0,\frac {2}{\beta }\right )\right )+\sum _{i=1}^N\xi _ix_i=\lambda x_1\\ x_1\bar {\xi }_2+\mu _2x_2=\lambda x_2\Rightarrow x_2=\frac {x_1\bar {\xi }_2}{\lambda -\mu _2}\\ \qquad \qquad \qquad \vdots \\ x_1\bar {\xi }_N+\mu _Nx_N=\lambda x_N\Rightarrow x_N=\frac {x_1\bar {\xi }_N}{\lambda -\mu _N} \end {cases} \Longrightarrow \begin {cases} x_2 = \frac {\bar {\xi }_2}{\lambda -\mu _2}\\ \qquad \vdots \\ x_N = \frac {\bar {\xi }_N}{\lambda -\mu _N} \end {cases} \] Plug the \(2 \sim N\) equations into the first line and we get \begin{equation} x_1 \left [a\sqrt {N}+{\mathcal{N}} \left (0,\frac {2}{\beta }\right ) - \lambda + \sum _{i=1}^N\frac {\xi _i\bar {\xi }_i}{\lambda -\mu _i} \right ] = 0. \tag {$\ast $} \end{equation} By interpolating of eigenvalues, \(\widetilde{C}\) has 1 eigenvalue larger than \(\mu _N\). Only this eigenvalue has chance to become larger than \(2\sqrt {N}\) as \(\frac {\mu _N}{\sqrt {N}} \to 2\). Denote \(y = \frac {\lambda _N}{\sqrt {N}}\) and investigate (\(\ast \)) for \(y>2\). \[ a + \frac {{\mathcal{N}} \left (0,\frac {2}{\beta }\right )}{\sqrt {N}} - y + \frac {1}{N} \sum _{i=2}^N \frac {\xi _i\bar {\xi }_i}{y - \frac {\mu _i}{\sqrt {N}}} = 0. \tag {$\ast \ast $} \] By LLN, we have \[ \frac {1}{N} \sum _{i=2}^N \frac {\xi _i\bar {\xi }_i}{y - \frac {\mu _i}{\sqrt {N}}} \approx \frac {1}{N} \sum _{i=2}^N \frac {1}{y - \frac {\mu _i}{\sqrt {N}}} \to G(y) = \frac {1}{2}(y- \sqrt {y^2-4}). \] Then we have \[ a - y + \frac {1}{2}(y- \sqrt {y^2-4}) = 0 \Longrightarrow y = a+ \frac {1}{a}, \] and this proves Theorem 8.17.

Step 3  For Theorem 8.19, we notice that the eigenvalues are \[ (1,0,\cdots ,0) \text { and } \left (1, \frac {\bar {\xi }_2}{\lambda -\mu _2},\cdots ,\frac {\bar {\xi }_N}{\lambda -\mu _N} \right ), \] and we have \[ \cos ^2 \phi = \frac {1}{1+\sum _{i=2}^N\frac {\xi _i\bar {\xi }_i}{(\lambda -\mu _i)^2}} \to \frac {1}{1+\frac {1}{2}\frac {y}{\sqrt {y^2-4}}-\frac {1}{2}} = \frac {2\sqrt {y^2-4}}{y+\sqrt {y^2-4}}. \] As \(y = a+\frac {1}{a}\), we have \[ \sin ^2 \phi \to \frac {1}{a^2}. \] □