[proofplan] We decompose $\mathbb{R}^p$ into the target spectral subspace of $A$ and its orthogonal complement. For each unit eigenvector of $\widehat A$ in the perturbed cluster, the component outside the target subspace is controlled by the spectral gap of $A$ and the perturbation $E$. This gives a bound on $(I-P)\widehat P$, which is the sine of the largest canonical angle between the two subspaces. The operator-norm projector bound follows from an operator-level residual estimate for separated spectral subspaces, rather than from summing one-vector estimates. The Frobenius bound follows by summing the one-vector sine estimates and then using the equal-rank projection identity. [/proofplan]

[step:Decompose the space into the target subspace and its orthogonal complement] Let $I \in \mathbb{R}^{p\times p}$ denote the identity matrix, and let \begin{align*} U := \operatorname{Range}(P), \qquad U^\perp := \operatorname{Range}(I-P). \end{align*} Since $A$ is symmetric and $P$ is a spectral projector of $A$, both $U$ and $U^\perp$ are invariant under $A$, and $A$ is block diagonal with respect to the [orthogonal decomposition](/theorems/436) \begin{align*} \mathbb{R}^p = U \oplus U^\perp. \end{align*} Let $\Lambda \subset \mathbb{R}$ denote the eigenvalue cluster of $A$ onto which $P$ projects. Let $J \subset \mathbb{R}$ denote a closed interval containing $\Lambda$ such that \begin{align*} \operatorname{dist}\bigl(J,\sigma(A)\setminus\Lambda\bigr) \geq \Delta. \end{align*} For a symmetric matrix $B \in \mathbb{R}^{m\times m}$, let $\sigma(B) \subset \mathbb{R}$ denote the set of its eigenvalues, counted without multiplicity. Thus $\sigma(A|_U)=\Lambda$ and $\sigma(A|_{U^\perp})=\sigma(A)\setminus\Lambda$ as sets. For every eigenvalue $\lambda$ of $A|_{U}$ and every eigenvalue $\nu$ of $A|_{U^\perp}$, the cluster separation assumption gives \begin{align*} |\lambda-\nu| \geq \Delta. \end{align*} Let \begin{align*} \widehat U := \operatorname{Range}(\widehat P). \end{align*} By hypothesis, $\dim \widehat U = \operatorname{rank}(\widehat P)=r=\operatorname{rank}(P)=\dim U$. [/step]

[step:Control the component of a perturbed eigenvector outside the target subspace]Let $\widehat u \in \widehat U$ be a unit eigenvector of $\widehat A$ with eigenvalue $\widehat\lambda$ belonging to the perturbed cluster. Define \begin{align*} x := P\widehat u \in U, \qquad z := (I-P)\widehat u \in U^\perp. \end{align*} Then $\widehat u=x+z$ and \begin{align*} (A+E)\widehat u = \widehat\lambda \widehat u. \end{align*} Applying $I-P$ to this equation and using $(I-P)A P=0$ gives \begin{align*} (A|_{U^\perp}-\widehat\lambda I)z = -(I-P)E\widehat u. \end{align*} List the eigenvalues of $A$ and $\widehat A$ in nondecreasing order, repeated with algebraic multiplicity, as $\lambda_1 \leq \cdots \leq \lambda_p$ and $\widehat\lambda_1 \leq \cdots \leq \widehat\lambda_p$. The finite-dimensional Weyl eigenvalue perturbation bound for symmetric matrices gives \begin{align*} |\widehat\lambda_k-\lambda_k| \leq \|E\|_{\mathrm{op}} \end{align*} for every $k\in\{1,\dots,p\}$. The perturbed cluster is chosen using the same ordered index set as the eigenvalues of $A$ lying in $\Lambda$; hence every eigenvalue $\widehat\lambda$ in the perturbed cluster lies within $\|E\|_{\mathrm{op}}$ of an eigenvalue in $\Lambda\subset J$. Every eigenvalue of $A|_{U^\perp}$ lies at distance at least $\Delta$ from $J$. Hence \begin{align*} \operatorname{dist}\bigl(\widehat\lambda,\sigma(A|_{U^\perp})\bigr) \geq \Delta-\|E\|_{\mathrm{op}} > \frac{\Delta}{2}. \end{align*} Therefore $A|_{U^\perp}-\widehat\lambda I$ is invertible on $U^\perp$. Since $A|_{U^\perp}$ is symmetric, the spectral theorem gives the inverse norm formula \begin{align*} \|(A|_{U^\perp}-\widehat\lambda I)^{-1}\|_{\mathrm{op}} = \frac{1}{\operatorname{dist}\bigl(\widehat\lambda,\sigma(A|_{U^\perp})\bigr)} \leq \frac{1}{\Delta-\|E\|_{\mathrm{op}}} \leq \frac{2}{\Delta}. \end{align*} Taking the Euclidean norm in $\mathbb{R}^p$ and then using the operator norm bounds for $(A|_{U^\perp}-\widehat\lambda I)^{-1}$, $I-P$, and $E$, we obtain \begin{align*} \|(I-P)\widehat u\| = \|z\| \leq \|(A|_{U^\perp}-\widehat\lambda I)^{-1}\|_{\mathrm{op}}\,\|(I-P)E\widehat u\|. \end{align*} Since $I-P$ is an [orthogonal projection](/theorems/437), $\|I-P\|_{\mathrm{op}}=1$, and since $\|\widehat u\|=1$, this gives \begin{align*} \|(I-P)\widehat u\| \leq \frac{2\|E\|_{\mathrm{op}}}{\Delta}. \end{align*}[/step]

[guided]The goal is to measure how much a perturbed eigenvector $\widehat u$ leaves the original target subspace $U$. We split it orthogonally as \begin{align*} \widehat u = x+z, \qquad x:=P\widehat u \in U, \qquad z:=(I-P)\widehat u \in U^\perp. \end{align*} The quantity to bound is exactly $\|z\|=\|(I-P)\widehat u\|$. Since $\widehat u$ is an eigenvector of $\widehat A=A+E$ with eigenvalue $\widehat\lambda$, we have \begin{align*} (A+E)\widehat u=\widehat\lambda \widehat u. \end{align*} Apply $I-P$ to both sides. Because $P$ is a spectral projector of $A$, the subspaces $U$ and $U^\perp$ are invariant under $A$, so the off-diagonal block $(I-P)AP$ is zero. Hence \begin{align*} (I-P)A(x+z)+(I-P)E\widehat u=\widehat\lambda z \end{align*} reduces to \begin{align*} A|_{U^\perp}z+(I-P)E\widehat u=\widehat\lambda z. \end{align*} Rearranging gives the resolvent equation \begin{align*} (A|_{U^\perp}-\widehat\lambda I)z=-(I-P)E\widehat u. \end{align*} The spectral gap now enters. Eigenvalues of $A|_U$ lie in the formally defined cluster $\Lambda$, while eigenvalues of $A|_{U^\perp}$ lie in $\sigma(A)\setminus\Lambda$, and these two sets are separated by at least $\Delta$. The finite-dimensional Weyl eigenvalue perturbation bound for symmetric matrices says that perturbing $A$ by the symmetric matrix $E$ moves each eigenvalue by at most $\|E\|_{\mathrm{op}}$ after ordering the eigenvalues increasingly. Since the perturbed cluster is chosen to correspond to $\Lambda$, this implies that the perturbed eigenvalue $\widehat\lambda$ is at distance at most $\|E\|_{\mathrm{op}}$ from $\Lambda$. Therefore \begin{align*} \operatorname{dist}\bigl(\widehat\lambda,\sigma(A|_{U^\perp})\bigr) \geq \Delta-\|E\|_{\mathrm{op}} > \frac{\Delta}{2}. \end{align*} Since $A|_{U^\perp}$ is symmetric, the spectral theorem diagonalizes $A|_{U^\perp}$ by an [orthonormal basis](/page/Orthonormal%20Basis) of $U^\perp$. Hence the operator norm of the inverse of $A|_{U^\perp}-\widehat\lambda I$ is the reciprocal of this spectral distance: \begin{align*} \|(A|_{U^\perp}-\widehat\lambda I)^{-1}\|_{\mathrm{op}} \leq \frac{1}{\Delta-\|E\|_{\mathrm{op}}} \leq \frac{2}{\Delta}. \end{align*} Applying this inverse to the resolvent equation gives \begin{align*} z=-(A|_{U^\perp}-\widehat\lambda I)^{-1}(I-P)E\widehat u. \end{align*} Taking Euclidean norms and using $\|I-P\|_{\mathrm{op}}=1$ and $\|\widehat u\|=1$, we obtain \begin{align*} \|(I-P)\widehat u\| = \|z\| \leq \|(A|_{U^\perp}-\widehat\lambda I)^{-1}\|_{\mathrm{op}}\,\|(I-P)E\widehat u\|. \end{align*} Because $I-P$ is an orthogonal projection, $\|I-P\|_{\mathrm{op}}=1$. Combining this with $\|\widehat u\|=1$ and the inverse bound gives \begin{align*} \|(I-P)\widehat u\| \leq \frac{2\|E\|_{\mathrm{op}}}{\Delta}. \end{align*} This is the sine-angle estimate for one perturbed eigenvector.[/guided]

[step:Apply the separated invariant subspace residual estimate]Define the [linear map](/page/Linear%20Map) $T: \widehat U\to U^\perp$ by $T\widehat u=(I-P)\widehat u$. Define the restriction $S:\widehat U\to\widehat U$ by $S\widehat u=\widehat A\widehat u$; this is well-defined because $\widehat U$ is a spectral subspace of the symmetric matrix $\widehat A$. The spectrum of $S$ is exactly the perturbed cluster, and the preceding step gives \begin{align*} \operatorname{dist}\bigl(\sigma(S),\sigma(A|_{U^\perp})\bigr) \geq \Delta-\|E\|_{\mathrm{op}}. \end{align*} For every $\widehat u\in\widehat U$, applying $I-P$ to $\widehat A\widehat u=S\widehat u$ gives the Sylvester equation \begin{align*} A|_{U^\perp}T\widehat u-TS\widehat u=-(I-P)E\widehat u. \end{align*} The finite-dimensional separated Sylvester estimate for [self-adjoint operators](/page/Self-Adjoint%20Operators) says that if self-adjoint operators have spectra separated by $\delta>0$, then the solution $X$ of $BX-XC=Y$ satisfies $\|X\|_{\mathrm{op}}\leq \delta^{-1}\|Y\|_{\mathrm{op}}$. Applying this estimate with $B=A|_{U^\perp}$, $C=S$, $X=T$, $Y=-(I-P)E|_{\widehat U}$, and $\delta=\Delta-\|E\|_{\mathrm{op}}$, we obtain \begin{align*} \|T\|_{\mathrm{op}} \leq \frac{\|(I-P)E|_{\widehat U}\|_{\mathrm{op}}}{\Delta-\|E\|_{\mathrm{op}}} \leq \frac{\|E\|_{\mathrm{op}}}{\Delta-\|E\|_{\mathrm{op}}} \leq \frac{2\|E\|_{\mathrm{op}}}{\Delta}. \end{align*} Since $T=(I-P)\widehat P$ as a map from $\mathbb R^p$ to $U^\perp$ after restricting the domain to $\widehat U=\operatorname{Range}(\widehat P)$, this gives \begin{align*} \|(I-P)\widehat P\|_{\mathrm{op}} \leq \frac{2\|E\|_{\mathrm{op}}}{\Delta}. \end{align*} For orthogonal projections $P$ and $\widehat P$ of the same finite rank, the canonical-angle identity for two subspaces gives \begin{align*} \|\widehat P-P\|_{\mathrm{op}}=\|(I-P)\widehat P\|_{\mathrm{op}}. \end{align*} Therefore \begin{align*} \|\widehat P-P\|_{\mathrm{op}} \leq \frac{2\|E\|_{\mathrm{op}}}{\Delta}. \end{align*}[/step]

[guided]The one-eigenvector estimate from the previous step cannot simply be extended to an arbitrary linear combination of eigenvectors by summing, because that would lose a dimension-dependent factor. The correct object is the whole map $T:\widehat U\to U^\perp$ defined by $T\widehat u=(I-P)\widehat u$. This map records the component of every vector in the perturbed subspace that lies outside the original target subspace. Define $S:\widehat U\to\widehat U$ by $S\widehat u=\widehat A\widehat u$. Since $\widehat U$ is a spectral subspace of the symmetric matrix $\widehat A$, the map $S$ is self-adjoint on $\widehat U$, and its spectrum is the perturbed cluster. The Weyl step already proved that every point of this perturbed cluster is at distance at least $\Delta-\|E\|_{\mathrm{op}}$ from $\sigma(A|_{U^\perp})$. Now take an arbitrary $\widehat u\in\widehat U$. Because $\widehat A\widehat u=S\widehat u$ and $\widehat A=A+E$, applying $I-P$ gives \begin{align*} (I-P)A\widehat u+(I-P)E\widehat u=(I-P)S\widehat u. \end{align*} The identity $(I-P)AP=0$ and the definition of $T$ turn this into \begin{align*} A|_{U^\perp}T\widehat u+ (I-P)E\widehat u=TS\widehat u. \end{align*} Equivalently, \begin{align*} A|_{U^\perp}T\widehat u-TS\widehat u=-(I-P)E\widehat u. \end{align*} This is a Sylvester equation for the operator $T$, not a separate scalar resolvent equation for each eigenvector. That distinction is what prevents a spurious factor of $\sqrt r$. We apply the finite-dimensional separated Sylvester estimate for self-adjoint operators. Its hypotheses are satisfied because $A|_{U^\perp}$ and $S$ are self-adjoint, and their spectra are separated by at least $\delta:=\Delta-\|E\|_{\mathrm{op}}>0$. Therefore \begin{align*} \|T\|_{\mathrm{op}} \leq \frac{\|(I-P)E|_{\widehat U}\|_{\mathrm{op}}}{\delta}. \end{align*} Since $I-P$ is an orthogonal projection, $\|I-P\|_{\mathrm{op}}=1$, so $\|(I-P)E|_{\widehat U}\|_{\mathrm{op}}\leq\|E\|_{\mathrm{op}}$. Using $\|E\|_{\mathrm{op}}<\Delta/2$, we get \begin{align*} \|T\|_{\mathrm{op}} \leq \frac{\|E\|_{\mathrm{op}}}{\Delta-\|E\|_{\mathrm{op}}} \leq \frac{2\|E\|_{\mathrm{op}}}{\Delta}. \end{align*} Finally, $T$ is exactly $(I-P)\widehat P$ on the range of $\widehat P$ and is zero on the orthogonal complement of that range. Hence \begin{align*} \|(I-P)\widehat P\|_{\mathrm{op}} \leq \frac{2\|E\|_{\mathrm{op}}}{\Delta}. \end{align*}[/guided]

[step:Sum the sine-angle bounds to obtain the Frobenius estimate] Using the orthonormal basis $\widehat u_1,\dots,\widehat u_r$ of $\widehat U$, we compute \begin{align*} \|(I-P)\widehat P\|_F^2 = \sum_{j=1}^r \|(I-P)\widehat u_j\|^2. \end{align*} By the estimate from the second step, \begin{align*} \|(I-P)\widehat P\|_F^2 \leq \sum_{j=1}^r \left(\frac{2\|E\|_{\mathrm{op}}}{\Delta}\right)^2 = r\left(\frac{2\|E\|_{\mathrm{op}}}{\Delta}\right)^2. \end{align*} Taking square roots gives \begin{align*} \|(I-P)\widehat P\|_F \leq \frac{2\sqrt r\,\|E\|_{\mathrm{op}}}{\Delta}. \end{align*} For orthogonal projections of equal rank, the canonical-angle identity gives \begin{align*} \|\widehat P-P\|_F^2 = 2\|(I-P)\widehat P\|_F^2. \end{align*} Therefore \begin{align*} \|\widehat P-P\|_F \leq \frac{2\sqrt{2r}\,\|E\|_{\mathrm{op}}}{\Delta}. \end{align*} This is the Frobenius estimate in the corrected statement, and the proof is complete. [/step]