Unitary and differential identities for finite matrix dynamics

If $Hv=\lambda v$ with $v\mathrm{\ne }0$, then self-adjointness gives $\lambda ⟨v,v⟩=⟨Hv,v⟩=⟨v,Hv⟩=\bar{\lambda }⟨v,v⟩$ by [F1]–[F3], so every eigenvalue $\lambda$ is real. Since $H$ is self-adjoint it is normal, and [F4] supplies an orthonormal eigenbasis. In dimension zero the assertions below hold for the unique endomorphism.

Write the spectral resolution $H={\sum }_{\lambda }\lambda P_{\lambda }$ over distinct eigenvalues. The projections satisfy $P_{\lambda }^{\ast }=P_{\lambda }$, $P_{\lambda }{P}_{\mu }=0$ for $\lambda \mathrm{\ne }\mu$, and ${\sum }_{\lambda }{P}_{\lambda }=I_V$. Definition Finite Hermitian matrix dynamics is the functional-calculus sum in [F5]. By [F11], at $z=0$ the exponential series has first term $1$ and every later term $0$, so every partial sum is $1$ and $e^0=1$.

The adjoint of this sum is $U_H(t{)}^{\ast }={\sum }_{\lambda }{e}^{it\lambda }{P}_{\lambda }$. Thus $U_H(t{)}^{\ast }{U}_H(t)={\sum }_{\lambda }\mathrm{\mid }{e}^{-it\lambda }{\mathrm{\mid }}^2{P}_{\lambda }={\sum }_{\lambda }{P}_{\lambda }=I_V$, since $t\lambda$ is real and [F7] gives $\mathrm{\mid }{e}^{-it\lambda }\mathrm{\mid }=e^0=1$ by step 2.1. At $t=0$ every coefficient is $1$ by step 2.1, so $U_H(0)=I_V$.

The product $U_H(t)U_H(s)$ has coefficient $e^{-it\lambda }{e}^{-is\lambda }=e^{-i(t+s)\lambda }$ on each $P_{\lambda }$ by [F6]; all cross terms vanish by orthogonality. Hence $U_H(t+s)=U_H(t)U_H(s)$.

For each real eigenvalue $\lambda$, the difference quotient of $t\mapsto e^{-it\lambda }$ at $t$ equals $e^{-it\lambda }(e^{-i\lambda h}-1)\mathrm{/}h$ by [F6]. If $\lambda \mathrm{\ne }0$, set $z=-i\lambda h$ and use [F8] at $0$, together with $e^0=1$ from step 2.1, to obtain the limit $-i\lambda e^{-it\lambda }$; if $\lambda =0$, the quotient is identically zero. The spectral sum is finite, so differentiating it term by term gives $U_H^{\mathrm{\prime }}(t)={\sum }_{\lambda }(-i\lambda )e^{-it\lambda }{P}_{\lambda }=-iH{U}_H(t)$.

Let $Y$ be any differentiable solution with $Y(0)=I_V$. Write its entries in the orthonormal eigenbasis from step 1.1 as $y_{jk}(t)$. Taking each matrix coordinate of the difference quotient for $Y^{\mathrm{\prime }}(t)=-iHY(t)$ gives $y_{jk}^{\mathrm{\prime }}(t)=-i{\lambda }_j{y}_{jk}(t)$. Put $q_j(t):=e^{i{\lambda }_{j}t}$. By [F6], for $h\mathrm{\ne }0$, $\frac{q_j(t+h)-q_j(t)}{h}=e^{i{\lambda }_{j}t}\frac{e^{i{\lambda }_{j}h}-1}{h}\mathrm{.}$ If ${\lambda }_j=0$, this quotient is zero. Otherwise set $z=i{\lambda }_{j}h$; [F8] at $0$, with $e^0=1$ from step 2.1, gives ${\lim }_{z\to 0}(e^z-1)\mathrm{/}z=1$, and hence $q_j^{\mathrm{\prime }}(t)=i{\lambda }_j{q}_j(t)$. The real and imaginary parts of $y_{jk}$ and $q_j$ are differentiable. Since $\mathrm{Re}(q_j{y}_{jk})=(\mathrm{Re}{q}_j)(\mathrm{Re}{y}_{jk})-(\mathrm{Im}{q}_j)(\mathrm{Im}{y}_{jk})$ and $\mathrm{Im}(q_j{y}_{jk})=(\mathrm{Re}{q}_j)(\mathrm{Im}{y}_{jk})+(\mathrm{Im}{q}_j)(\mathrm{Re}{y}_{jk})$, the real product rule [F10] gives the complex product rule. Thus $g_{jk}(t):=q_j(t)y_{jk}(t)$ is differentiable and $g_{jk}^{\mathrm{\prime }}(t)=q_j^{\mathrm{\prime }}(t)y_{jk}(t)+q_j(t)y_{jk}^{\mathrm{\prime }}(t)=0.$ Each real-valued component $h$ of $g_{jk}$ is therefore differentiable on $\mathbb{R}$ with derivative zero. It is continuous at every $c$: for $x\mathrm{\ne }c$, $h(x)-h(c)=\frac{h(x)-h(c)}{x-c}(x-c),$ and the difference quotient tends to the finite value $h^{\mathrm{\prime }}(c)$, so it is bounded near $c$ and the product tends to zero. Thus both components are continuous on every $[a,b]$ and differentiable on $(a,b)$. Applying [F9] to each component on any $a<b$ shows both are constant on $\mathbb{R}$. Since $q_j(0)=1$ by step 2.1 and $Y(0)=I_V$, $g_{jk}(0)={\delta }_{jk}$; therefore $y_{jk}(t)={\delta }_{jk}{e}^{-i{\lambda }_{j}t}$ by [F6]. Thus $Y(t)=U_H(t)$. The zero-dimensional case has only one endomorphism and is unique as well.