Stationary Distributions as Eigenvectors

From a Transition Matrix to a Stationary Distribution

A Markov transition matrix describes how a probability state changes in one step. A stationary distribution is a probability state that is unchanged by that step. Finding one is therefore an eigenvector problem, but with an extra requirement: an eigenvector must be converted into a vector of probabilities. Solving the eigenvalue equation without checking nonnegativity and normalization can produce a vector that is not a probability distribution.

We use the column-vector convention. For a finite state space with \(n\) states, a probability state is a column vector \(\mathbf{p}\in\mathbb{R}^n\). Its entries give the probabilities of the individual states. A transition matrix \(\mathbf{P}\in\mathbb{R}^{n\times n}\) is column-stochastic when its entries are nonnegative and the entries in each column sum to \(1\). The \(j\)-th column records the probabilities of moving to each state when the current state is \(j\). The update rule is \(\mathbf{p}_{\text{new}}=\mathbf{P}\mathbf{p}_{\text{old}}\).

Let \(\mathbf{1}\in\mathbb{R}^n\) denote the column vector whose entries are all \(1\). The normalization condition is \(\mathbf{1}^{\top}\mathbf{p}=1\). Nonnegativity is a separate condition: it is not implied by either the eigenvector equation or normalization alone. A nonzero eigenvector can have negative entries, and even an eigenvector with positive entries might not yet be normalized.

Why Eigenvalue One Is the Right Equation

Under the column-vector convention, stationarity means that applying the transition matrix leaves the state exactly unchanged. This is precisely the equation \(\mathbf{P}\mathbf{p}=\mathbf{p}\), or equivalently \((\mathbf{P}-\mathbf{I}_n)\mathbf{p}=\mathbf{0}\). The solution vectors of the latter homogeneous system form the eigenspace for eigenvalue \(1\). To find stationary distributions, solve that system and then retain only the solutions that satisfy the probability conditions.

Proof. Suppose \(\mathbf{p}\) is a stationary distribution. By definition it is a probability vector and \(\mathbf{P}\mathbf{p}=\mathbf{p}\). Rearranging the latter equality gives \((\mathbf{P}-\mathbf{I}_n)\mathbf{p}=\mathbf{0}\), while being a probability vector gives \(\mathbf{1}^{\top}\mathbf{p}=1\) and \(p_i\geq0\) for every \(i\). Conversely, suppose all three displayed conditions hold. The first gives \(\mathbf{P}\mathbf{p}=\mathbf{p}\). The second and third say exactly that \(\mathbf{p}\) is a probability vector. Therefore \(\mathbf{p}\) is a stationary distribution. \(\square\)

The theorem gives a direct calculation strategy: row-reduce \(\mathbf{P}-\mathbf{I}_n\), describe its null space, and impose the sum and sign conditions. A convenient uniqueness test is available when that null space is one-dimensional.

Proof. Let \(\mathbf{p}\) and \(\mathbf{q}\) be stationary distributions. Both satisfy \((\mathbf{P}-\mathbf{I}_n)\mathbf{x}=\mathbf{0}\), so both lie in the one-dimensional null space. Since each is a probability vector, neither is zero. In a one-dimensional space, one nonzero vector is a scalar multiple of any other, so \(\mathbf{q}=c\mathbf{p}\) for some real \(c\). Taking sums of entries gives \(1=\mathbf{1}^{\top}\mathbf{q}=c\mathbf{1}^{\top}\mathbf{p}=c\). Hence \(\mathbf{q}=\mathbf{p}\), proving uniqueness. \(\square\)

This is a sufficient uniqueness test: a one-dimensional eigenspace for eigenvalue \(1\), together with an existing stationary distribution, guarantees uniqueness. If the eigenspace has larger dimension, that test does not decide uniqueness by itself. Instead, impose the probability conditions and determine whether they leave one solution or several.

Solving and Checking the Equations

The normalization step is not merely a convenient final adjustment. It selects the scale of a nonzero eigenvector, while nonnegativity checks whether that scaled vector can represent probabilities. Both checks are needed even after the eigenvalue equation has been solved.

When There Is More Than One Stationary Distribution

The eigenvalue-one equation need not determine a single probability vector. If its solution space contains several directions, the probability constraints can leave a whole family of stationary distributions. The following example makes this distinction explicit.

This example also shows why finding one valid eigenvector does not settle uniqueness. To check uniqueness in a particular problem, identify the full solution set of the eigenvalue-one system and then apply the probability restrictions to that set. If the null space is one-dimensional and a stationary distribution has been found, the uniqueness theorem gives a quick conclusion. If it has more than one dimension, inspect the normalized, nonnegative solutions rather than assuming either uniqueness or nonuniqueness.

Common Mistake: Switching the Probability Convention

With column probability states, the stationary equation is \(\mathbf{P}\mathbf{p}=\mathbf{p}\), not \(\mathbf{p}^{\top}\mathbf{P}=\mathbf{p}^{\top}\). Confusing these conventions can give an answer that satisfies the wrong equation. For instance, consider the column-stochastic matrix

A mistaken row-vector calculation gives \((\frac12,\frac12)\mathbf{P}=(\frac12,\frac12)\), which might seem to make \((\frac12,\frac12)^{\top}\) stationary. But under the column convention the required product is

The correct equation gives \(p_1+\frac12p_2=p_1\), so \(p_2=0\). Normalization then gives \(p_1=1\). Indeed, \(\mathbf{P}(1,0)^{\top}=(1,0)^{\top}\). Always match the multiplication order to the stated convention before interpreting a solution as stationary.

What the Calculation Establishes

A stationary distribution records a probability state that one transition leaves unchanged; finding it is not a claim that every starting state will approach it. The equation \(\mathbf{P}\mathbf{p}=\mathbf{p}\) answers the fixed-state question, while questions about repeated transitions and long-term convergence require additional analysis. For the present calculation, keep three checks together: the vector solves the eigenvalue-one equation, its entries sum to \(1\), and none of its entries is negative. Then check uniqueness by analyzing the full set of solutions, not just one candidate.