Coupled Systems and Eigenmodes

Definition A linear update model for $n$ real quantities is a rule

where $x_k \in \mathbb{R}^n$ is the state at step $k$ and $B \in \mathbb{R}^{n \times n}$ is a fixed matrix, the coupling matrix. Then $x_k = B^k x_0$ for every $k \ge 0$. The entry $b_{ij}$ is the factor by which quantity $j$ at the current step contributes to quantity $i$ at the next step, so the contribution is $b_{ij}$ times the current value of quantity $j$. The diagonal entries record what share each quantity keeps, and the entries off the diagonal record how the quantities are coupled. The model assumes that the rule is linear, the same at every step and free of outside input. The total $s_k = x_{k,1} + \cdots + x_{k,n}$ satisfies $s_{k+1} = \mathbf{1}^{\mathsf{T}}Bx_k$, where $\mathbf{1}$ is the vector of ones, so the total is conserved for every starting state exactly when every column of $B$ sums to $1$.

Definition For $x = (x_1, x_2) \in \mathbb{R}^2$, the mean and the half difference are

so that $x_1 = m + d$, $x_2 = m - d$ and $x = m(1, 1) + d(1, -1)$. The orthonormal version uses $q_+ = (1, 1)/\sqrt{2}$ and $q_- = (1, -1)/\sqrt{2}$, the columns of an orthogonal matrix $Q$ with $Q^{\mathsf{T}}Q = I$. The coefficients $c = Q^{\mathsf{T}}x$ are

with $x = Qc = c_+q_+ + c_-q_-$ and $\lVert x \rVert^2 = c_+^2 + c_-^2 = 2(m^2 + d^2)$. For a symmetric coupling matrix with equal diagonal entries $a$ and off-diagonal entries $b$, adding and subtracting the two update equations gives

Changing coordinates does not change the state. The same vector $x$ is described by two signal values or by two pattern coefficients.

Definition In a linear update model with coupling matrix $B \in \mathbb{R}^{n \times n}$, an eigenmode is a nonzero vector $q \in \mathbb{R}^n$ with $Bq = \lambda q$ for a real number $\lambda$, the factor of the mode. The line spanned by $q$ is invariant, since every vector $cq$ on it goes to $\lambda cq$. So a state that starts in the mode stays in it, with $x_k = \lambda^k x_0$. The mode decays when $\lvert \lambda \rvert < 1$, keeps its size when $\lvert \lambda \rvert = 1$ and grows when $\lvert \lambda \rvert > 1$, and a negative factor flips the sign of the state at every step. A real matrix can also have nonreal eigenvalues, which come in conjugate pairs and describe oscillating patterns, but the models in this lesson have real ones.

Definition Let $B \in \mathbb{R}^{n \times n}$ be symmetric. By the spectral theorem $\mathbb{R}^n$ has an orthonormal basis $q_1, \dots, q_n$ with $Bq_i = \lambda_i q_i$ and every $\lambda_i$ real. Let $Q$ be the orthogonal matrix with these columns and $\Lambda = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$. Then $B = Q\Lambda Q^{\mathsf{T}}$, and for every $x_0$ and every $k \ge 0$

which is the matrix identity $B^k = Q\Lambda^kQ^{\mathsf{T}}$. For the proof, induction gives $B^kq_i = \lambda_i^kq_i$. Orthonormality gives $x_0 = \sum_i c_iq_i$ with $c_i = q_i^{\mathsf{T}}x_0$, and linearity finishes the argument. So each mode coefficient evolves on its own, multiplied by its factor at every step. Only the formula for the coefficients uses orthonormality. For a diagonalizable matrix $B = P\Lambda P^{-1}$ the same expansion holds along the columns of $P$ with $c = P^{-1}x_0$.

Definition For a symmetric $H \in \mathbb{R}^{n \times n}$ the quadratic form is $E(x) = x^{\mathsf{T}}Hx$. It is positive definite when $E(x) > 0$ for every nonzero $x$, which happens exactly when every eigenvalue of $H$ is positive. In the two-signal model take

Then

which is positive for $x \ne 0$ and measures both the disagreement and the size of the signals. Since $H = 4(I - B)$, it has the same eigenvectors $q_\pm$, with eigenvalues $4(1 - 3/4) = 1$ and $4(1 - 1/4) = 3$. In mode coordinates $E = c_+^2 + 3c_-^2$, and one step gives

so at each step the energy of a state is multiplied by at most $9/16$. Watch for this. Under a change of coordinates $x = Pc$ with $P$ invertible, the update matrix becomes $P^{-1}BP$, a similarity, while the energy becomes $c^{\mathsf{T}}(P^{\mathsf{T}}HP)c$, a congruence. The two rules agree only when $P^{-1} = P^{\mathsf{T}}$, that is when $P$ is orthogonal.

Definition When the coupling matrix $B \in \mathbb{R}^{n \times n}$ is not symmetric, the real spectral theorem does not apply, and $\mathbb{R}^n$ has no orthonormal basis of eigenvectors of $B$. If $B$ has $n$ linearly independent eigenvectors $p_1, \dots, p_n$ in $\mathbb{C}^n$, for example because it has $n$ distinct eigenvalues in $\mathbb{C}$, then $B = P\Lambda P^{-1}$ with $P = [p_1 \ \cdots \ p_n]$, the mode coordinates are $c = P^{-1}x_0$, and

The eigenvectors need not be orthogonal, so in general $c_i$ is not $p_i^{*}x_0 / \lVert p_i \rVert^2$, and arguments that used an orthonormal mode basis, such as the energy computation above, do not carry over unchanged. When every eigenvalue is real, the eigenvectors can be taken in $\mathbb{R}^n$. Otherwise $P$ and $c$ are complex, and the sum is still the real vector $B^kx_0$. If $B$ is defective there is no eigenbasis at all. Jordan chains then supply the missing directions and add factors polynomial in $k$. A Schur form $B = UTU^{*}$, with $U$ unitary and $T$ upper triangular, always exists over $\mathbb{C}$. Its change of basis is always orthonormal, and its factor $T$ is diagonal exactly when $B$ is normal. Only then do the Schur coordinates evolve independently as modes.