An affine itemtive system has the form 𝐮^(k+1)=T 𝐮^(k)+𝐛, 𝐮^(0)=𝐜. (a) Under what conditions doe

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An affine itemtive system has the form 𝐮^(k+1)=T 𝐮^(k)+𝐛,...

An affine itemtive system has the form $\mathbf{u}^{(k+1)}=T \mathbf{u}^{(k)}+\mathbf{b}, \mathbf{u}^{(0)}=\mathbf{c}$. (a) Under what conditions does the system have an equilibrium solution $\mathbf{u}^{(k)} \equiv \mathbf{u}^{\star}$ ? (b) In such cases, find a formula for the general solution. Hint: Look at $\mathbf{v}^{(k)}=\mathbf{u}^{(k)}-\mathbf{u}^*$. (c) Solve the following affine iterative systems: (i) $\mathbf{u}^{(k+1)}=\left(\begin{array}{rr}6 & 3 \\ -3 & -4\end{array}\right) \mathbf{u}^{(k)}+\left(\begin{array}{l}1 \\ 2\end{array}\right), \quad \mathbf{u}^{(0)}=\left(\begin{array}{r}4 \\ -3\end{array}\right)$, (ii) $\mathbf{u}^{(k+1)}=\left(\begin{array}{rr}-1 & 2 \\ 1 & -1\end{array}\right) \mathbf{u}^{(k)}+\left(\begin{array}{l}1 \\ 0\end{array}\right), \quad \mathbf{u}^{(0)}=\left(\begin{array}{l}0 \\ 1\end{array}\right)$. (iki) $\mathbf{u}^{(k+1)}=\left(\begin{array}{rrr}-3 & 2 & -2 \\ -6 & 4 & -3 \\ 12 & -6 & -5\end{array}\right) \mathbf{u}^{(k)}+\left(\begin{array}{r}1 \\ -3 \\ 0\end{array}\right), \quad \mathbf{u}^{(0)}=\left(\begin{array}{r}1 \\ 0 \\ -1\end{array}\right)$, (iv) $\mathbf{u}^{(k+1)}=\left(\begin{array}{rrr}-\frac{5}{6} & \frac{1}{3} & -\frac{1}{6} \\ 0 & -\frac{1}{2} & \frac{1}{3} \\ 1 & -1 & \frac{2}{3}\end{array}\right) \mathbf{u}^{(k)}+\left(\begin{array}{r}\frac{1}{6} \\ -\frac{1}{3} \\ -\frac{1}{2}\end{array}\right), \quad \mathbf{u}^{(0)}=\left(\begin{array}{r}\frac{1}{6} \\ -\frac{2}{3} \\ \frac{1}{3}\end{array}\right)$. (d) Discuss what happens in cases in which there is no fixed point, assuming that $T$ is complete.

An affine itemtive system has the form $\mathbf{u}^{(k+1)}=T \mathbf{u}^{(k)}+\mathbf{b}, \mathbf{u}^{(0)}=\mathbf{c}$.
(a) Under what conditions does the system have an equilibrium solution $\mathbf{u}^{(k)} \equiv \mathbf{u}^{\star}$ ?
(b) In such cases, find a formula for the general solution. Hint: Look at $\mathbf{v}^{(k)}=\mathbf{u}^{(k)}-\mathbf{u}^*$.
(c) Solve the following affine iterative systems:
(i) $\mathbf{u}^{(k+1)}=\left(\begin{array}{rr}6 & 3 \\ -3 & -4\end{array}\right) \mathbf{u}^{(k)}+\left(\begin{array}{l}1 \\ 2\end{array}\right), \quad \mathbf{u}^{(0)}=\left(\begin{array}{r}4 \\ -3\end{array}\right)$,
(ii) $\mathbf{u}^{(k+1)}=\left(\begin{array}{rr}-1 & 2 \\ 1 & -1\end{array}\right) \mathbf{u}^{(k)}+\left(\begin{array}{l}1 \\ 0\end{array}\right), \quad \mathbf{u}^{(0)}=\left(\begin{array}{l}0 \\ 1\end{array}\right)$.
(iki) $\mathbf{u}^{(k+1)}=\left(\begin{array}{rrr}-3 & 2 & -2 \\ -6 & 4 & -3 \\ 12 & -6 & -5\end{array}\right) \mathbf{u}^{(k)}+\left(\begin{array}{r}1 \\ -3 \\ 0\end{array}\right), \quad \mathbf{u}^{(0)}=\left(\begin{array}{r}1 \\ 0 \\ -1\end{array}\right)$,
(iv) $\mathbf{u}^{(k+1)}=\left(\begin{array}{rrr}-\frac{5}{6} & \frac{1}{3} & -\frac{1}{6} \\ 0 & -\frac{1}{2} & \frac{1}{3} \\ 1 & -1 & \frac{2}{3}\end{array}\right) \mathbf{u}^{(k)}+\left(\begin{array}{r}\frac{1}{6} \\ -\frac{1}{3} \\ -\frac{1}{2}\end{array}\right), \quad \mathbf{u}^{(0)}=\left(\begin{array}{r}\frac{1}{6} \\ -\frac{2}{3} \\ \frac{1}{3}\end{array}\right)$.
(d) Discuss what happens in cases in which there is no fixed point, assuming that $T$ is complete.

Key Concepts

General Solution of Nonhomogeneous Linear Recurrences

The general solution to an affine iterative system combines the particular solution (the fixed point u*) with the solution to the homogeneous transformation. Specifically, once u* is found, the general solution is given by u^(k) = u* + T^k (c - u*), where c is the initial condition. This construction highlights the superposition of steady-state and transient dynamics, a key methodological step in solving nonhomogeneous linear recurrences.

Stability and Convergence Analysis in the Absence of a Fixed Point

When the matrix (I - T) is singular and no fixed point exists, the system's behavior is determined solely by the properties of T. In these cases, the analysis focuses on the eigenvalues, the structure (such as completeness) of the matrix T, and the corresponding modes of the homogeneous system. The long-term behavior, including potential divergence, oscillations, or other dynamics, is inferred from the spectral properties of T.

Transformation to a Homogeneous System

By introducing a change of variables such as v^(k) = u^(k) - u*, the affine system is transformed into a homogeneous one: v^(k+1) = T v^(k). This transformation separates the dynamics of deviations from the steady state, allowing analysis of the convergence and transient behavior of the system using the properties of the linear operator T.

Affine Iterative Systems

An affine iterative system is a discrete dynamical system where the next state is defined as a linear transformation of the current state plus a constant vector. This type of system, written in the form u^(k+1) = T u^(k) + b, captures both the homogeneous (linear) dynamics and an external forcing term, making it a fundamental model for many processes in linear algebra and numerical analysis.

Equilibrium Solution (Fixed Point)

The equilibrium or fixed point of an affine iterative system is a state u* that remains unchanged by the iterative process, meaning u* = T u* + b. To find this fixed point, one typically rearranges the equation to (I - T) u* = b, and a unique equilibrium exists when the matrix (I - T) is invertible. This concept is crucial in understanding steady-state behaviors in dynamical systems.

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