[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*}
[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]
[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]
