[proofplan] We prove the bound deterministically on a fixed sample outcome satisfying the operator-norm perturbation assumption. The population covariance $\Sigma = I_d+\lambda vv^\top$ acts by eigenvalue $1+\lambda$ on $\operatorname{span}\{v\}$ and by eigenvalue $1$ on $v^\perp$. Decomposing the empirical leading eigenvector into its component along $v$ and its orthogonal component, the projected empirical eigenvalue equation controls the orthogonal component by the perturbation size. A lower bound on the empirical leading eigenvalue supplies the eigengap denominator. [/proofplan]

[step:Pass to deterministic notation on the good sample event] Fix $\omega \in \Omega$ satisfying \begin{align*} \|\widehat{\Sigma}(\omega)-\Sigma\|_{\mathrm{op}} \leq \frac{\lambda}{4}. \end{align*} For this proof write \begin{align*} A := \widehat{\Sigma}(\omega) \in \mathbb{R}^{d \times d}, \qquad E := A-\Sigma \in \mathbb{R}^{d \times d}, \qquad \delta := \|E\|_{\mathrm{op}}. \end{align*} Then $A=\Sigma+E$, $A$ is real symmetric, and $\delta \leq \lambda/4$. Let $\widehat\mu \in \mathbb{R}$ denote the largest eigenvalue of $A$, so that \begin{align*} A\widehat v = \widehat\mu \widehat v, \qquad |\widehat v|=1. \end{align*} [/step]

[step:Decompose the empirical eigenvector into signal and orthogonal components] Define the [orthogonal projection](/theorems/437) $P_{v^\perp}: \mathbb{R}^d \to \mathbb{R}^d$ by \begin{align*} P_{v^\perp}(x) = x-(x\cdot v)v \end{align*} for each $x \in \mathbb{R}^d$. Let \begin{align*} a := \widehat v \cdot v \in \mathbb{R}, \qquad w := P_{v^\perp}\widehat v = \widehat v-a v \in v^\perp. \end{align*} Then $\widehat v = a v+w$ and, since $|v|=1$ and $w\cdot v=0$, \begin{align*} 1 = |\widehat v|^2 = a^2+|w|^2. \end{align*} Therefore \begin{align*} \sin\angle(\widehat v,v) = \sqrt{1-(\widehat v\cdot v)^2} = \sqrt{1-a^2} = |w|. \end{align*} [/step]

[step:Lower-bound the empirical leading eigenvalue by testing on the population spike]Since $\widehat\mu$ is the largest eigenvalue of the real symmetric matrix $A$, its Rayleigh quotient dominates the Rayleigh quotient at the unit vector $v$: \begin{align*} \widehat\mu \geq v^\top A v. \end{align*} Using $A=\Sigma+E$ and $\Sigma v=(1+\lambda)v$, we obtain \begin{align*} v^\top A v = v^\top \Sigma v + v^\top E v = 1+\lambda+v^\top E v. \end{align*} By the definition of the operator norm for the real symmetric matrix $E$ and the fact that $|v|=1$, \begin{align*} |v^\top E v| \leq \|E\|_{\mathrm{op}}=\delta. \end{align*} Thus \begin{align*} \widehat\mu \geq 1+\lambda-\delta, \qquad \widehat\mu-1 \geq \lambda-\delta. \end{align*}[/step]

[guided]The empirical top eigenvalue must remain close to the population top eigenvalue because we can test the empirical quadratic form on the true spike direction $v$. Since $A$ is real symmetric and $\widehat\mu$ is its largest eigenvalue, the variational characterization of the largest eigenvalue gives \begin{align*} \widehat\mu \geq v^\top A v, \end{align*} because $v$ is a unit vector. Now substitute the perturbation decomposition $A=\Sigma+E$: \begin{align*} v^\top A v = v^\top \Sigma v + v^\top E v. \end{align*} The population covariance satisfies \begin{align*} \Sigma v = (I_d+\lambda vv^\top)v = v+\lambda v(v^\top v) = (1+\lambda)v, \end{align*} because $|v|=1$. Hence \begin{align*} v^\top \Sigma v = 1+\lambda. \end{align*} The perturbation term is controlled by the operator norm. Since $E$ is real symmetric and $|v|=1$, \begin{align*} |v^\top E v| \leq \|E\|_{\mathrm{op}}|v|^2=\delta. \end{align*} Combining these estimates gives \begin{align*} \widehat\mu \geq v^\top A v = 1+\lambda+v^\top E v \geq 1+\lambda-\delta. \end{align*} Subtracting $1$ from both sides yields \begin{align*} \widehat\mu-1 \geq \lambda-\delta. \end{align*} This lower bound is the empirical version of the population eigengap: it says that the empirical leading eigenvalue is still separated from the flat eigenvalue level $1$ by at least $\lambda-\delta$.[/guided]

[step:Project the eigenvalue equation onto the orthogonal complement of the spike] Apply $P_{v^\perp}$ to the eigenvalue equation $A\widehat v=\widehat\mu\widehat v$. Since $A=\Sigma+E$, this gives \begin{align*} P_{v^\perp}\Sigma\widehat v + P_{v^\perp}E\widehat v = \widehat\mu P_{v^\perp}\widehat v. \end{align*} Using $\widehat v=a v+w$, $w\in v^\perp$, and \begin{align*} \Sigma(a v+w)=(1+\lambda)a v+w, \end{align*} we obtain \begin{align*} P_{v^\perp}\Sigma\widehat v = w. \end{align*} Therefore \begin{align*} w + P_{v^\perp}E\widehat v = \widehat\mu w, \end{align*} and hence \begin{align*} (\widehat\mu-1)w = P_{v^\perp}E\widehat v. \end{align*} Taking Euclidean norms and using $\|P_{v^\perp}\|_{\mathrm{op}}=1$ and $|\widehat v|=1$ gives \begin{align*} (\widehat\mu-1)|w| = |P_{v^\perp}E\widehat v| \leq \|P_{v^\perp}\|_{\mathrm{op}}\|E\|_{\mathrm{op}}|\widehat v| = \delta. \end{align*} Since $\widehat\mu-1 \geq \lambda-\delta > 0$, it follows that \begin{align*} |w| \leq \frac{\delta}{\lambda-\delta}. \end{align*} [/step]

[step:Use the perturbation assumption to obtain the stated sine bound] The assumption $\delta \leq \lambda/4$ implies \begin{align*} \lambda-\delta \geq \frac{3\lambda}{4}. \end{align*} Therefore \begin{align*} \sin\angle(\widehat v,v) = |w| \leq \frac{\delta}{\lambda-\delta} \leq \frac{4\delta}{3\lambda} \leq \frac{2\delta}{\lambda}. \end{align*} Recalling that $\delta=\|\widehat{\Sigma}(\omega)-\Sigma\|_{\mathrm{op}}$, we obtain \begin{align*} \sin\angle(\widehat v,v) \leq \frac{2\|\widehat{\Sigma}(\omega)-\Sigma\|_{\mathrm{op}}}{\lambda}. \end{align*} This is the desired bound. [/step]