Toeplitz circulant and displacement structured solvers — Examples

By [F1]–[F2], the $k$th eigenvalue of this generator is ${\lambda }_k=2-{\omega }^k-{\omega }^{k(N-1)}\mathrm{.}$ Since ${\omega }^N=\exp (-2\pi i)=1$ by [F3], ${\omega }^{k(N-1)}={\omega }^{-k}$, giving the stated symbol.

Put $z={\omega }^k$. It is nonzero by [F3], and ${\lambda }_k=0⟺z+z^{-1}=2⟺(z-1{)}^2=0⟺z=1,$ where the middle equivalence follows after multiplying by $z$, and the last uses [F4]. By the exponential kernel characterization in [F3], ${\omega }^k=1$ exactly when $-2\pi ik\mathrm{/}N\in 2\pi i\mathbb{Z}$, equivalently $k\mathrm{/}N$ is an integer. For $0\le k<N$ this means $k=0$. Thus only the constant Fourier mode is a zero mode.

At $k=0$, $({\mathcal{F}}_{N}f{)}_0=N^{-1\mathrm{/}2}{\sum }_j{f}_j$. Hence [F2] gives solvability precisely when ${\sum }_j{f}_j=0$. All nonzero modes then determine ${\mathcal{F}}_{N}x$ uniquely, while the zero-mode coordinate is free; imposing ${\sum }_j{x}_j=0$ fixes that coordinate to zero, proving uniqueness on the zero-mean subspace.

When $N=4$, $\omega =\exp (-\pi i\mathrm{/}2)$, so the addition law gives ${\omega }^2=\exp (-\pi i)$ and $\exp (-\pi i)\exp (\pi i)=1$. Euler's identity gives $\exp (\pi i)=-1$, hence ${\omega }^2=-1$. Since $\omega \mathrm{\ne }0$, this implies ${\omega }^{-1}=-\omega$, and the symbol from step 1.1 gives $({\lambda }_0,{\lambda }_1,{\lambda }_2,{\lambda }_3)=(0,2,4,2)$. The vector $f=(1,-1,1,-1{)}^T$ has sum zero. Directly, each cyclic neighbor of $f_j$ is $-f_j$, so $(C_{4}f{)}_j=2f_j-(-f_j)-(-f_j)=4f_j$. Therefore $x=f\mathrm{/}4$ solves $C_{4}x=f$, and its zero mean gives the unique zero-mean solution.

The compatibility condition is exactly orthogonality to the constant null mode, and division is only performed on the nonzero Fourier modes. The concrete $N=4$ vector verifies the normalization directly.

The matrix is Hermitian Toeplitz with $t_0=1$ and $t_1=t_{-1}=2$. Direct multiplication gives $T(-\frac{1}{3}\left(\begin{array}{cc}1& -2\\ -2& 1\end{array}\right))=I_2,$ so $T$ is invertible. For $z=(1,-1{)}^T$, $z^{\ast }Tz=-2<0$, hence it is not positive definite by [F1].

The solver starts with ${\nu }_0=1$, computes ${\alpha }_1=t_1\mathrm{/}{\nu }_0=2$ and then ${\nu }_1={\nu }_0(1-\mathrm{\mid }{\alpha }_1{\mathrm{\mid }}^2)=-3$. Since this pivot is nonpositive, [F2] returns nonpositive-pivot$(1)$ before constructing an inverse-column denominator.

Centering and weak stationarity give $E[X_t]=0$ and $\gamma (0)=\mathrm{Var}(X_t)$ for every $t$. From the autoregression and innovation orthogonality, $\gamma (0)=\mathrm{Var}(\frac{1}{2}{X}_{t-1}+{\epsilon }_t)=\frac{1}{4}\gamma (0)+\frac{3}{4},$ so $\gamma (0)=1$.

For the separate general-RHS solve, set $T=\left(\begin{array}{cc}1& 1\mathrm{/}2\\ 1\mathrm{/}2& 1\end{array}\right)$ as in the Statement. For each nonzero $z\in {\mathbb{C}}^2$, $z^{\ast }Tz=\mathrm{\mid }{z}_1+z_2\mathrm{/}2{\mathrm{\mid }}^2+3\mathrm{\mid }{z}_2{\mathrm{\mid }}^2\mathrm{/}4>0$, so $T$ is Hermitian positive definite. Its first lag data are $t_0=1,t_1=1\mathrm{/}2$, hence ${\alpha }_1=t_1\mathrm{/}{t}_0=1\mathrm{/}2$, ${\varphi }_1={\alpha }_1=1\mathrm{/}2$, and ${\nu }_1=t_0(1-\mathrm{\mid }{\alpha }_1{\mathrm{\mid }}^2)=3\mathrm{/}4$. The endpoint columns from [F5] are $v=(-2\mathrm{/}3,4\mathrm{/}3{)}^T,u=J_{2}v=(4\mathrm{/}3,-2\mathrm{/}3{)}^T,u_0=4\mathrm{/}3>0.$ The lower triangular Toeplitz products are $L(u)L(u{)}^{\ast }=\left(\begin{array}{cc}16\mathrm{/}9& -8\mathrm{/}9\\ -8\mathrm{/}9& 20\mathrm{/}9\end{array}\right),L(Z_{2}v)L(Z_{2}v{)}^{\ast }=\left(\begin{array}{cc}0& 0\\ 0& 4\mathrm{/}9\end{array}\right)\mathrm{.}$ Therefore the formula in [F5] gives $T^{-1}=\left(\begin{array}{cc}4\mathrm{/}3& -2\mathrm{/}3\\ -2\mathrm{/}3& 4\mathrm{/}3\end{array}\right),T^{-1}b=(2\mathrm{/}3,2\mathrm{/}3{)}^T\mathrm{.}$ Direct multiplication also verifies $T(2\mathrm{/}3,2\mathrm{/}3{)}^T=(1,1{)}^T$.

For each integer $h\ge 1$, covariance bilinearity and $\mathrm{Cov}({\epsilon }_t,X_{t-h})=0$ give $\gamma (h)=\mathrm{Cov}(X_t,X_{t-h})=\frac{1}{2}\gamma (h-1)\mathrm{.}$ Induction yields $\gamma (h)=2^{-h}$ for $h\ge 0$; covariance symmetry gives $\gamma (-h)=\gamma (h)$.

The order-one recursion gives ${\kappa }_1=\gamma (1)\mathrm{/}\gamma (0)=1\mathrm{/}2$, ${\varphi }_1=(1\mathrm{/}2)$ and ${\nu }_1=1(1-1\mathrm{/}4)=3\mathrm{/}4$. At order two, ${\kappa }_2=\frac{\gamma (2)-{\varphi }_{1,1}\gamma (1)}{{\nu }_1}=\frac{1\mathrm{/}4-(1\mathrm{/}2)(1\mathrm{/}2)}{3\mathrm{/}4}=0,$ so ${\varphi }_2=({\varphi }_{1,1}-{\kappa }_2{\varphi }_{1,1},{\kappa }_2{)}^T=(1\mathrm{/}2,0{)}^T$ and ${\nu }_2={\nu }_1(1-{\kappa }_2^2)=3\mathrm{/}4$. By steps 1.1 and 2.1, ${\mathrm{\Gamma }}_1=[1]$ and ${\mathrm{\Gamma }}_2=T$. Step 1.2 proves $T$ Hermitian positive definite, so the real leading covariance blocks are positive definite; [F3] applies and [F4] identifies this vector as the unique prediction solution.

The first calculation reuses the published real covariance prediction recursion, while the second is a distinct arbitrary-right-hand-side Hermitian Toeplitz solve; the computed vector satisfies its stated system.