Spectral decomposition

A spectral decomposition expresses a linear operator through its eigenvalues and projections onto its eigenspaces.

For a diagonalizable matrix with distinct eigenvalues $\lambda ,\mu ,\dots$, this polynomial is the projection operator onto the eigenspace for $\lambda$ along the direct sum of the other eigenspaces. It is one on its own eigenspace and zero on the others. Consequently $P_{\lambda }{P}_{\mu }=0$ for different eigenvalues, $\sum P_{\lambda }=I$ and $A=\sum \lambda P_{\lambda }$. These projections need not be orthogonal projections; that stronger property holds for a normal matrix. This is the polynomial interpolation form of the spectral decomposition.

An eigenvalue is dominant when its modulus is greater than or equal to that of every other eigenvalue. It is uniquely dominant when the inequality is strict.

A spectral gap is a positive separation between designated parts of the spectrum of a linear operator. For the power method, the relevant gap separates the modulus of the dominant eigenvalue from the other eigenvalue moduli.

An energy gap is a positive separation between a ground-state sector and its lowest excitations, with the sector, momentum and thermodynamic limit specified. A local positive curvature of an auxiliary field is not alone a proof of a dynamical gap: the kinetic terms and conjugate variables determine the excitation poles.

Let $B$ be a finite-dimensional Hermitian matrix and $a<b$ real. The spectral theorem gives $v={\sum }_j{c}_j{x}_j$ in an orthonormal eigenbasis, soIf the spectrum avoids the closed interval, every factor product is positive. Consequently the quadratic form is positive for $v\ne 0$ and zero for $v=0$. More precisely it is bounded below by ${\min }_j[({\lambda }_j-a)({\lambda }_j-b)]\Vert v{\Vert }^2$. This is a finite-dimensional instance of testing a spectral gap with a polynomial in the operator.

Suppose $H\ge 0$ has ground energy zero and positive spectral gap at least $g$, and $0\le V\le I$. Let $\nu$ be the minimum of $V$ restricted to the ground space. Splitting a normalized vector into ground and excited components bounds its energy below by $\delta \nu +(g-\delta )y^2-2\delta y$, with $y$ the excited norm. Completing the square yields the displayed bound for $0<\delta <g$. The error is controlled by the gap and cannot be treated as a uniform $O({\delta }^2)$ when $g$ closes.

The Davis-Kahan theorem bounds the change of an invariant eigenspace of a symmetric matrix in terms of a matrix perturbation and an appropriate spectral gap. The rank-one eigenprojector perturbation bound below depends only on the unperturbed gap. Yu, Wang and Samworth's formulation gives population-gap bounds in greater generality.

If a symmetric matrix $H$ has largest eigenvalue ${\lambda }_1$ separated from ${\lambda }_2$ by $\gamma >0$, and $P=v{v}^{\top }$ is its leading rank-one orthogonal projection matrix, then every unit vector $u$, with $Q=u{u}^{\top }$, satisfies the displayed inequality. Expanding $u$ in an orthonormal eigenbasis gives ${\lambda }_1-u^{\top }Hu\ge \gamma (1-|u^{\top }v{|}^2)$, while $\Vert P-Q{\Vert }_F^2=2(1-|u^{\top }v{|}^2)$. The analogous statement for a smallest eigenvalue follows by replacing $H$ by $-H$.

Let $\hat{P}$ project onto a leading unit eigenvector of $H+W$, with $H$ as in the Davis-Kahan curvature lemma. Its maximizing Rayleigh quotient gives $\mathrm{tr}(H(P-\hat{P}))\le \mathrm{tr}(W(\hat{P}-P))$. Duality of the operator norm and trace norm, together with the rank-two bound $\Vert \hat{P}-P{\Vert }_{\ast }\le \sqrt{2}\Vert \hat{P}-P{\Vert }_F$, proves the estimate. No gap condition on the perturbed matrix is needed.