Diagnosing Misleading Long-Run Eigenvalue Conclusions

Why Eigenvalue Lists Can Mislead

Eigenvalues describe possible modes of a discrete system, but they do not by themselves describe the long-run behavior of every initial state. A mode may be absent from the initial state, two modes of equal magnitude may keep a sequence oscillating, and a repeated eigenvalue may belong to a matrix whose powers grow. Markov models add another source of error: the correct multiplication side depends on whether probabilities are stored as rows or columns.

Consider a discrete system \(\mathbf{x}_{k+1}=\mathbf{A}\mathbf{x}_k\), where \(\mathbf{A}\in\mathbb{R}^{n\times n}\), \(\mathbf{x}_0\in\mathbb{R}^n\), and \(k=0,1,2,\ldots\). Its state after \(k\) steps is \(\mathbf{x}_k=\mathbf{A}^k\mathbf{x}_0\). An eigenvalue \(\lambda\) of \(\mathbf{A}\) has a corresponding eigenvector \(\mathbf{v}\ne\mathbf{0}\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Consequently, if the initial state is an eigenvector, then \(\mathbf{A}^k\mathbf{v}=\lambda^k\mathbf{v}\). This explains why the magnitude \(|\lambda|\) is informative: powers decay when \(|\lambda|<1\), grow when \(|\lambda|>1\), and do not decay when \(|\lambda|=1\). But a general initial state may combine several eigenvectors, and that combination matters.

The Initial State Determines Which Modes Appear

This example rules out the claim that an eigenvalue outside the unit circle forces every initial state to grow. The correct question is whether that state has a nonzero component in a growing mode. For diagonalizable matrices, the following criterion makes the distinction exact, including when eigenvalues are complex.

Proof. For each eigenvalue \(\lambda\), diagonalizability gives \(\mathbf{A}^k\mathbf{x}_0=\sum_{\mu}\mu^k\mathbf{x}_{\mu}\). Define the polynomial \(q_{\lambda}(t)=\prod_{\mu\ne\lambda}(t-\mu)/(\lambda-\mu)\), where the product is over the distinct eigenvalues other than \(\lambda\). If there are no other eigenvalues, the product is \(1\). On the eigenspace for \(\lambda\), \(q_{\lambda}(\mathbf{A})\) acts as multiplication by \(q_{\lambda}(\lambda)=1\); on every other eigenspace it acts as multiplication by \(0\). Therefore \(q_{\lambda}(\mathbf{A})\mathbf{A}^k\mathbf{x}_0=\lambda^k\mathbf{x}_{\lambda}\). If \(\mathbf{A}^k\mathbf{x}_0\) converges, applying the fixed linear map \(q_{\lambda}(\mathbf{A})\) shows that \(\lambda^k\mathbf{x}_{\lambda}\) converges for each \(\lambda\). If \(\mathbf{x}_{\lambda}\ne\mathbf{0}\), choose a nonzero coordinate \(c\) of that vector. The corresponding coordinate of the sequence is \(c\lambda^k\). For \(|\lambda|>1\) its magnitude grows without bound. For \(|\lambda|=1\) and \(\lambda\ne1\), convergence is impossible: if \(\lambda^k\) converged, its successive differences would tend to zero, but \(|\lambda^{k+1}-\lambda^k|=|\lambda-1|>0\) for every \(k\). Thus every nonzero component must satisfy the stated condition. Conversely, if those conditions hold, each term with \(|\lambda|<1\) tends to zero, and the term for \(\lambda=1\) stays equal to \(\mathbf{x}_1\). There are only finitely many eigenvalues, so their sum converges to \(\mathbf{x}_1\). \(\square\)

Equal Magnitudes and Complex Modes

The criterion shows why looking only for a “largest” eigenvalue magnitude can miss persistent oscillation. When two eigenvalues have the same magnitude, neither mode necessarily fades relative to the other. In particular, an eigenvalue of magnitude \(1\) other than \(1\) can prevent convergence when its component is present.

Complex eigenvalues require the same care. A real matrix can have complex eigenvalues even though its real states remain real. A complex pair with magnitude less than \(1\) produces a rotating or alternating component that shrinks to zero. A pair on the unit circle can keep a component moving without settling.

Repeated Eigenvalues Do Not Settle the Question

A repeated eigenvalue is not itself a prediction about convergence. One must also know how the matrix acts on the associated directions. In particular, diagonalizability matters: a matrix with a repeated eigenvalue may have a full basis of eigenvectors, or it may fail to have one. The next proposition exhibits two matrices with the same repeated eigenvalue and different long-run behavior.

Proof. Let \(\mathbf{C}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\) and \(\mathbf{D}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\). Both have characteristic polynomial \((1-\lambda)^2\), so both have the same repeated eigenvalue \(1\). For \(\mathbf{x}_0=(0,1)^\top\), \(\mathbf{C}^k\mathbf{x}_0=(0,1)^\top\) for every \(k\), which converges. Write \(\mathbf{D}=\mathbf{I}_2+\mathbf{N}\), where \(\mathbf{N}=\begin{bmatrix}0&1\\0&0\end{bmatrix}\) and \(\mathbf{N}^2=\mathbf{O}\). Induction gives \(\mathbf{D}^k=\mathbf{I}_2+k\mathbf{N}=\begin{bmatrix}1&k\\0&1\end{bmatrix}\): it holds for \(k=0\), and multiplying \(\mathbf{I}_2+k\mathbf{N}\) by \(\mathbf{I}_2+\mathbf{N}\) gives \(\mathbf{I}_2+(k+1)\mathbf{N}\) because \(\mathbf{N}^2=\mathbf{O}\). Therefore \(\mathbf{D}^k\mathbf{x}_0=(k,1)^\top\), which does not converge. The same repeated eigenvalue is compatible with different outcomes, so the eigenvalue list alone is insufficient. \(\square\)

This example also warns against applying the diagonalizable-matrix criterion without checking its hypothesis. The matrix \(\mathbf{D}\) is not diagonalizable: its eigenvector equation \((\mathbf{D}-\mathbf{I}_2)\mathbf{v}=\mathbf{0}\) forces the second coordinate of \(\mathbf{v}\) to be zero, so its eigenspace has only one independent direction and cannot form a basis of \(\mathbb{C}^2\).

Markov Matrices: Match the State Convention

A Markov transition matrix encodes probabilities, so the convention for storing a probability state must match the matrix. A matrix is row-stochastic if its entries are nonnegative and each row sums to \(1\); it naturally updates row probability vectors by multiplication on the right. A matrix is column-stochastic if its entries are nonnegative and each column sums to \(1\); it naturally updates column probability vectors by multiplication on the left. Using the wrong side may fail even to preserve the sum of the probabilities.

There is a related eigenvector trap. The right eigenvector \((1,1)^\top\) of this row-stochastic \(\mathbf{P}\) has eigenvalue \(1\), but it is not a probability vector: its entries sum to \(2\). The stationary probability state in the row convention is a left eigenvector, \((1,0)\mathbf{P}=(1,0)\); in the column convention it is a right eigenvector of \(\mathbf{P}^{\top}\). An eigenvector calculation is meaningful for a Markov limit only after the state orientation and probability normalization have been checked.

A Practical Diagnostic

Before predicting a long-run limit from eigenvalues, check the following in order:

1. Confirm the state equation and whether states are rows or columns. For Markov matrices, verify that the chosen multiplication preserves the probability sum.
2. Check whether the matrix is diagonalizable before using a sum of ordinary eigenmodes as a complete description.
3. Decompose the initial state into eigenvector components. A growing or oscillating eigenvalue is relevant only if its component is nonzero.
4. Inspect every present mode: \(|\lambda|<1\) decays, \(\lambda=1\) stays fixed, \(|\lambda|>1\) grows, and \(|\lambda|=1\) with \(\lambda\ne1\) does not settle.
5. For a repeated eigenvalue, do not infer behavior from its value alone; check the eigenspace and whether the matrix is diagonalizable.

The distinction is between information about the matrix and information about a particular trajectory. Eigenvalues constrain possible dynamics, while initial components determine which modes participate. In Markov systems, even a correct eigenvalue computation can answer the wrong question if the transition convention is mismatched.