Eigenmodes and Repeated Updates
A discrete linear system updates its state repeatedly by the same matrix. Even if the matrix has eigenvalues with very different magnitudes, not all of them necessarily affect a particular trajectory: the initial state may have zero component in some eigendirections. To predict long-term behavior, we therefore need two pieces of information: the eigenvalues and the initial state’s components along their eigenvectors.
Consider \(\mathbf{x}_{k+1}=\mathbf{A}\mathbf{x}_k\), where \(k=0,1,2,\dots\), \(\mathbf{x}_k\in\mathbb{C}^n\), and \(\mathbf{A}\in\mathbb{C}^{n\times n}\). Real systems are included by taking \(\mathbf{A}\) and \(\mathbf{x}_0\) real. Assume that \(\mathbf{A}\) is diagonalizable over \(\mathbb{C}\): there is an invertible matrix \(\mathbf{P}\) whose columns \(\mathbf{v}_1,\dots,\mathbf{v}_n\) are eigenvectors, with corresponding eigenvalues \(\lambda_1,\dots,\lambda_n\). Write the initial state uniquely as \(\mathbf{x}_0=\sum_{j=1}^n c_j\mathbf{v}_j\). Each term \(c_j\mathbf{v}_j\) is the initial contribution of one eigenmode.
To justify the formula, let \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\). Then \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\), so \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). Since \(\mathbf{x}_0=\mathbf{P}\mathbf{c}\), where \(\mathbf{c}=(c_1,\dots,c_n)^\top\), repeated multiplication gives \(\mathbf{x}_k=\mathbf{P}\mathbf{D}^k\mathbf{c}\), which is the displayed sum. Equivalently, each update multiplies the \(j\)th eigenvector contribution by \(\lambda_j\).
What the Magnitude of an Eigenvalue Tells Us
For one active mode \(c\lambda^k\mathbf{v}\), its scalar multiplier has magnitude \(|c|\,|\lambda|^k\). Thus it is the modulus \(|\lambda|\), not just the sign or the numerical value of \(\lambda\), that controls the mode’s size. When \(|\lambda|<1\), the mode decays to zero. When \(|\lambda|>1\), it grows without bound. When \(|\lambda|=1\), its magnitude stays constant; its direction or sign may still change.
Some boundary cases are important. If \(\lambda=1\), the mode is constant. If \(\lambda=-1\), it alternates sign at each update. A complex eigenvalue \(\lambda\) with \(|\lambda|=1\) also preserves the magnitude of its complex mode while its phase changes; in a real system, conjugate modes can combine to give oscillatory behavior. If \(\lambda=0\), the mode is present in \(\mathbf{x}_0\), but it becomes zero after one update.
Worked Example: Active Growth and Decay in One System
Let \(\mathbf{A}=\begin{bmatrix}\tfrac54&\tfrac34\\[2pt]\tfrac34&\tfrac54\end{bmatrix}\). Find its eigenmodes and describe the trajectory starting at \(\mathbf{x}_0=(3,-1)^\top\).
Take \(\mathbf{v}_1=(1,1)^\top\) and \(\mathbf{v}_2=(1,-1)^\top\). Direct multiplication gives
The two eigenvectors are independent: if \(a\mathbf{v}_1+b\mathbf{v}_2=\mathbf{0}\), then \(a+b=0\) and \(a-b=0\), so \(a=b=0\). They form a basis of \(\mathbb{R}^2\). The initial state decomposes as \(\mathbf{x}_0=\mathbf{v}_1+2\mathbf{v}_2=(3,-1)^\top\). Therefore
The first active mode grows because its eigenvalue has modulus \(2>1\); the second decays because its eigenvalue has modulus \(\tfrac12<1\). As a check at \(k=1\), the formula gives \((2,2)^\top+(1,-1)^\top=(3,1)^\top\), while direct multiplication gives
In particular, the trajectory does not tend to zero or stay bounded: the growing mode is active.
Why the Initial State Matters
The eigenvalues describe what the matrix can do, but the coefficients \(c_j\) determine which of those behaviors occur in a given trajectory. An eigenvalue larger than \(1\) in modulus does not cause growth for an initial state whose coefficient in that eigendirection is zero. Similarly, a persistent eigenvalue matters only if its mode is active. This is why long-term conclusions must be based on the active eigenvalues, not on the full spectrum alone.
Worked Example: A Growing Eigenvalue That Is Not Active
Consider \(\mathbf{A}=\operatorname{diag}(3,1,\tfrac12)\) and \(\mathbf{x}_0=(0,2,4)^\top\). The coordinate eigenvectors are \(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\), with eigenvalues \(3,1,\tfrac12\), respectively. Since \(\mathbf{x}_0=0\mathbf{e}_1+2\mathbf{e}_2+4\mathbf{e}_3\), the mode with eigenvalue \(3\) is inactive. The trajectory is
Thus the state converges to \((0,2,0)^\top\): the first coordinate stays zero, the second persists, and the third decays. As a direct check, \(\mathbf{x}_1=(0,2,2)^\top\), and \(\mathbf{A}\mathbf{x}_0=(0,2,2)^\top\) as well. The inactive growing eigenvalue does not change this trajectory.
Worked Example: Persistence and Alternation
Let \(\mathbf{A}=\operatorname{diag}(-1,\tfrac12)\) and \(\mathbf{x}_0=(3,-2)^\top\). Its eigenvectors are \(\mathbf{e}_1,\mathbf{e}_2\), and the initial state has both modes active. Hence
The first coordinate has constant magnitude \(3\) and alternates between \(3\) and \(-3\); the second tends to zero. For example, \(\mathbf{x}_1=(-3,-1)^\top\), and multiplication verifies this: \(\mathbf{A}\mathbf{x}_0=(-3,-1)^\top\). The trajectory is bounded, but it does not converge because its first coordinate alternates.
Criteria for Boundedness and Convergence
The mode-by-mode description gives exact criteria for the whole state. Call an eigenvalue active for \(\mathbf{x}_0\) when its coefficient in the eigenvector expansion of \(\mathbf{x}_0\) is nonzero. If an eigenvalue is repeated, it is best to group together the components in its eigenspace: its mode is active when that total eigenspace component is nonzero. In an eigenvector basis, this grouping avoids making the conclusion depend on a particular choice of basis inside the eigenspace.
Proof. Choose an eigenvector basis and write \(\mathbf{x}_k=\mathbf{P}\mathbf{D}^k\mathbf{c}\), where \(\mathbf{P}\) is invertible and \(\mathbf{c}\) contains the initial coefficients. Use the Euclidean norm and its induced matrix norm. The inequalities \(\|\mathbf{P}\mathbf{y}\|\le\|\mathbf{P}\|\|\mathbf{y}\|\) and \(\|\mathbf{y}\|\le\|\mathbf{P}^{-1}\|\|\mathbf{P}\mathbf{y}\|\) hold for every vector \(\mathbf{y}\).
If all active eigenvalues have modulus at most \(1\), every coordinate \(c_j\lambda_j^k\) is bounded, so \(\|\mathbf{D}^k\mathbf{c}\|\) and then \(\|\mathbf{x}_k\|\) are bounded. Conversely, if an active eigenvalue satisfies \(|\lambda_j|>1\), then \(|c_j\lambda_j^k|=|c_j||\lambda_j|^k\) tends to infinity. Consequently \(\|\mathbf{D}^k\mathbf{c}\|\) tends to infinity, and the inequality \(\|\mathbf{D}^k\mathbf{c}\|\le\|\mathbf{P}^{-1}\|\|\mathbf{x}_k\|\) forces \(\|\mathbf{x}_k\|\) to be unbounded.
For convergence, if every active eigenvalue has modulus less than \(1\) or equals \(1\), then each coordinate \(c_j\lambda_j^k\) tends to zero or to \(c_j\), respectively. Thus \(\mathbf{D}^k\mathbf{c}\) converges, and multiplication by \(\mathbf{P}\) gives the stated limit. Conversely, suppose \(\mathbf{x}_k\) converges. Then \(\mathbf{D}^k\mathbf{c}=\mathbf{P}^{-1}\mathbf{x}_k\) converges, so every coordinate \(c_j\lambda_j^k\) converges. An active coordinate with \(|\lambda_j|>1\) cannot converge because its magnitude tends to infinity. An active coordinate with \(|\lambda_j|=1\) and \(\lambda_j\ne1\) cannot converge: the difference between consecutive terms has magnitude \(|c_j||\lambda_j-1|\), a fixed positive number, whereas consecutive terms of any convergent sequence must have difference tending to zero. Therefore each active eigenvalue must satisfy \(|\lambda_j|<1\) or \(\lambda_j=1\). \(\square\)
This theorem distinguishes boundedness from convergence. A mode on the unit circle keeps a bounded magnitude, but unless its eigenvalue is \(1\), it prevents convergence when active. For real eigenvalues, the simplest example is \(\lambda=-1\), which produces sign alternation. For complex unit-modulus eigenvalues, a changing phase can likewise prevent convergence.
Common Pitfall: Looking Only at the Largest Eigenvalue
A common mistake is to conclude that every trajectory grows whenever the matrix has an eigenvalue with modulus greater than \(1\). For \(\mathbf{A}=\operatorname{diag}(3,\tfrac12)\) and \(\mathbf{x}_0=(0,4)^\top\), that reasoning would predict growth from the eigenvalue \(3\). But its coefficient is zero, and the actual calculation is
For a direct update check, \(\mathbf{A}\mathbf{x}_0=(0,2)^\top=\mathbf{x}_1\). The correction is to inspect the initial state’s eigenvector components and classify only the active modes. The converse error is also possible: an eigenvalue of modulus less than \(1\) does not imply that every trajectory decays to zero, because an active \(\lambda=1\) mode can persist and an active eigenvalue of modulus greater than \(1\) can grow.
For a diagonalizable system, the main checks can be organized as follows:
| Active eigenvalue | Mode behavior | Effect on the full trajectory |
|---|---|---|
| \(|\lambda|<1\) | Decays to zero | Does not prevent convergence |
| \(\lambda=1\) | Constant | Contributes to the limit |
| \(|\lambda|=1,\ \lambda\ne1\) | Persistent, possibly oscillatory | Bounded but prevents convergence |
| \(|\lambda|>1\) | Grows without bound | Makes the trajectory unbounded |
| \(\lambda=0\) | Zero after one update | Does not prevent convergence |
These criteria are specifically for diagonalizable systems, where the state splits into a finite sum of eigenmodes. The practical lesson is to compute the initial eigenvector components before drawing a long-term conclusion: eigenvalue magnitudes describe possible behavior, while active modes determine the behavior that actually occurs.
Check Your Understanding
For each question, state which modes are active and justify the long-term conclusion from their eigenvalues.
- For \(\mathbf{A}=\operatorname{diag}(\tfrac13,2)\) and \(\mathbf{x}_0=(6,0)^\top\), find \(\mathbf{x}_k\) and its limit.
- For \(\mathbf{A}=\operatorname{diag}(1,-1)\) and \(\mathbf{x}_0=(2,3)^\top\), is the trajectory bounded? Does it converge?
- Suppose a diagonalizable system has initial components only in eigenspaces with eigenvalues \(\tfrac12\) and \(1\). Describe its limit.
- Explain why an active eigenvalue with modulus greater than \(1\) makes a diagonalizable trajectory unbounded, even if other active modes decay.
- Can a bounded trajectory fail to converge? Give an eigenvalue and an active mode that demonstrate this.