Diagonalizable matrix

A matrix is diagonalizable exactly when its eigenvectors span the whole vector space.

A diagonalization is a factorization $A=PD{P}^{-1}$ in which $D$ is diagonal and the columns of $P$ form an eigenvector basis.

Let $A$ be invertible over an algebraically closed field. If $A^m$ is diagonalizable for some $m>0$, then $A$ is diagonalizable. Indeed, a nontrivial Jordan block $A{|}_E=\lambda I+N$ has $\lambda \ne 0$ andThe first nonzero nilpotent term prevents this power from being diagonalizable.

Suppose real matrices $A,B$ satisfy $AP=PB$ for an invertible complex matrix $P=X+iY$. Then $AX=XB$ and $AY=YB$. The real polynomial $\det (X+tY)$ is not identically zero because its value at $t=i$ is nonzero, so some real $t$ makes $X+tY$ invertible. This real matrix intertwines $A$ and $B$.

Eigenvectors belonging to distinct eigenvalues are linearly independent. Consequently, an $n$-dimensional operator with $n$ distinct eigenvalues is diagonalizable.

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

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.

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.