The Eigenvalue One in a Transition Matrix

Why Eigenvalue One Is Guaranteed

A transition matrix describes one-step probabilities, but it is also a linear transformation. Its rows sum to one because, from each state, the probabilities of all possible next states add to one. That simple condition has a direct algebraic consequence: the column vector whose entries are all one is unchanged by the matrix. Thus every finite transition matrix has eigenvalue one. A second fact is equally useful: no eigenvalue, including a complex one, can have modulus greater than one.

Let $(X_n)_{n\ge0}$ be a discrete-time Markov chain on a finite, nonempty state space $S=\{1,\ldots,m\}$, defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Its transition matrix is $P=(p_{ij})$, where $p_{ij}\ge0$ and $\sum_{j=1}^{m}p_{ij}=1$ for every $i$. We regard $P$ as a real matrix but allow complex vectors when discussing its eigenvalues.

Definition: A complex number $\lambda$ is an eigenvalue of $P$ if there is a nonzero vector $v\in\mathbb{C}^m$ such that $Pv=\lambda v$. Such a vector $v$ is an eigenvector. The vector $\mathbf{1}=(1,\ldots,1)^{\mathsf T}$ is the all-ones column vector.

The column-vector convention matters here. Probability distributions are written as row vectors, so an initial distribution propagates as $\mu P$. The vector $\mathbf{1}$, in contrast, is a right eigenvector: it is multiplied on the left by $P$. It is not itself a probability distribution.

Theorem: If $P$ is a finite row-stochastic matrix, then $P\mathbf{1}=\mathbf{1}$, so $1$ is an eigenvalue of $P$. Moreover, every complex eigenvalue $\lambda$ of $P$ satisfies $|\lambda|\le1$.

Proof. For each row $i$, the $i$th entry of $P\mathbf{1}$ is $\sum_{j=1}^{m}p_{ij}=1$. Therefore $P\mathbf{1}=\mathbf{1}$, and since $\mathbf{1}\ne0$, the number $1$ is an eigenvalue. Now let $\lambda$ be any eigenvalue and let $v\ne0$ be a corresponding complex eigenvector. Since $v$ has finitely many coordinates, there is an index $i$ for which $|v_i|=M=\max_{1\le j\le m}|v_j|$. The number $M$ is positive because $v\ne0$. Taking the $i$th coordinate of $Pv=\lambda v$, then using nonnegativity and the row sum, gives $$ |\lambda|M =|\lambda v_i| =\left|\sum_{j=1}^{m}p_{ij}v_j\right| \le\sum_{j=1}^{m}p_{ij}|v_j| \le\sum_{j=1}^{m}p_{ij}M =M. $$ Dividing by $M>0$ gives $|\lambda|\le1$. Both claims follow. $\square$

The bound uses only the defining properties of a row-stochastic matrix: nonnegative entries and row sums equal to one. In particular, it does not assume that $P$ is symmetric, irreducible, or diagonalizable. It also applies to complex eigenvectors, which is important because real matrices can have nonreal eigenvalues.

Worked Examples

Worked Example: The Identity Matrix Has a Repeated Eigenvalue One

Consider the two-state transition matrix $$ P=\begin{bmatrix}1&0\\0&1\end{bmatrix}. $$ Its first row sums to $1+0=1$, and its second row sums to $0+1=1$, so it is row-stochastic. Find its eigenvalues and check the all-ones eigenvector.

The characteristic determinant is $$ \det(\lambda I-P) =\det\begin{bmatrix}\lambda-1&0\\0&\lambda-1\end{bmatrix} =(\lambda-1)^2. $$ Thus the only eigenvalue is $1$, with algebraic multiplicity two. Direct multiplication gives $$ P\begin{bmatrix}1\\1\end{bmatrix} =\begin{bmatrix}1\\1\end{bmatrix}. $$ In fact, $Pv=v$ for every $v\in\mathbb{C}^2$, since $P$ is the identity matrix. This verifies the eigenvalue calculation and shows that eigenvalue one need not be simple.

Worked Example: A Nonsymmetric Two-State Matrix

Let $$ P=\begin{bmatrix}\tfrac12&\tfrac12\\[2pt]\tfrac14&\tfrac34\end{bmatrix}. $$ The first row sums to $\tfrac12+\tfrac12=1$, and the second sums to $\tfrac14+\tfrac34=1$. Its characteristic determinant is $$ \det(\lambda I-P) =(\lambda-\tfrac12)(\lambda-\tfrac34)-\tfrac18 =\lambda^2-\tfrac54\lambda+\tfrac14 =(\lambda-1)(\lambda-\tfrac14). $$ So the eigenvalues are $1$ and $\tfrac14$. For eigenvalue $1$, direct multiplication gives $P(1,1)^{\mathsf T}=(1,1)^{\mathsf T}$. For the other eigenvalue, take $v=(2,-1)^{\mathsf T}$. Then $$ Pv =\begin{bmatrix}\tfrac12(2)+\tfrac12(-1)\\[2pt]\tfrac14(2)+\tfrac34(-1)\end{bmatrix} =\begin{bmatrix}\tfrac12\\-\tfrac14\end{bmatrix} =\tfrac14\begin{bmatrix}2\\-1\end{bmatrix}. $$ This verifies both eigenpairs. Their moduli are $1$ and $\tfrac14$, respectively, as permitted by the theorem.

Worked Example: A Transition Matrix with Nonreal Eigenvalues

Consider the deterministic three-state cycle $$ P=\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}. $$ The three row sums are $0+1+0=1$, $0+0+1=1$, and $1+0+0=1$. The characteristic determinant is $$ \det(\lambda I-P) =\det\begin{bmatrix}\lambda&-1&0\\0&\lambda&-1\\-1&0&\lambda\end{bmatrix} =\lambda^3-1 =(\lambda-1)(\lambda^2+\lambda+1). $$ The roots are $1$ and $\lambda_{\pm}=\frac{-1\pm i\sqrt{3}}{2}$. Their moduli are $$ |\lambda_{\pm}| =\sqrt{\left(-\tfrac12\right)^2+\left(\tfrac{\sqrt3}{2}\right)^2} =\sqrt{\tfrac14+\tfrac34} =1. $$ For a direct eigenvector check, each root $\lambda$ satisfies $\lambda^3=1$, and $$ P\begin{bmatrix}1\\\lambda\\\lambda^2\end{bmatrix} =\begin{bmatrix}\lambda\\\lambda^2\\1\end{bmatrix} =\lambda\begin{bmatrix}1\\\lambda\\\lambda^2\end{bmatrix}. $$ For $\lambda=1$ this is the all-ones eigenvector; for the other two roots it verifies the nonreal eigenpairs. Thus eigenvalues can lie on the unit circle without being equal to one.

What Strict Positivity Says at the Boundary

The bound $|\lambda|\le1$ allows equality. The cycle above shows that equality can hold for a nonreal eigenvalue when some transition probabilities are zero. If, instead, every entry of the transition matrix is strictly positive, equality is more restricted.

Proposition: Suppose $P$ is a finite row-stochastic matrix with $p_{ij}>0$ for every pair $i,j$. If $Pv=\lambda v$ for a nonzero complex vector $v$ and $|\lambda|=1$, then $\lambda=1$ and $v$ is a constant multiple of $\mathbf{1}$.

Proof. Choose $i$ such that $|v_i|=M=\max_j|v_j|>0$. The eigenvector equation and the bound used in the theorem give $$ M=|\lambda v_i| =\left|\sum_jp_{ij}v_j\right| \le\sum_jp_{ij}|v_j| \le M. $$ The first and last quantities are equal, so equality holds in both inequalities. Since every $p_{ij}>0$, equality in $\sum_jp_{ij}|v_j|\le M\sum_jp_{ij}=M$ requires $|v_j|=M$ for every $j$: if one coordinate had modulus less than $M$, its positive weight would make the weighted sum strictly less than $M$. Put $u_j=v_j/M$ and $a_j=p_{ij}$. Then $|u_j|=1$, $a_j>0$, $\sum_j a_j=1$, and $\left|\sum_j a_ju_j\right|=1$. Expanding both sides gives $$ 1-\left|\sum_j a_ju_j\right|^2 =\sum_{j<k}a_ja_k|u_j-u_k|^2. $$ The left side is zero. Every term on the right is nonnegative and each coefficient $a_ja_k$ is positive, so $u_j=u_k$ for every pair $j<k$. If there is only one state, the vector already has one coordinate; otherwise this argument shows all coordinates are equal. Hence $v=c\mathbf{1}$ for a nonzero constant $c$. Since $P\mathbf{1}=\mathbf{1}$, the equation $Pv=\lambda v$ becomes $v=\lambda v$, which implies $\lambda=1$ because $v\ne0$. $\square$

This proposition is an eigenvector statement; it does not assert that eigenvalue one has algebraic multiplicity one. Nor does the main theorem assert that the distribution of a Markov chain converges. Eigenvalues are useful in studying matrix powers, but conclusions about convergence require additional arguments and hypotheses.

A Common Mistake: Using Column Sums

A frequent error is to use column sums when checking the right eigenvector $\mathbf{1}$. For the nonsymmetric matrix in the second example, the column sums are $\tfrac12+\tfrac14=\tfrac34$ and $\tfrac12+\tfrac34=\tfrac54$. A mistaken calculation might therefore claim that $P\mathbf{1}=(\tfrac34,\tfrac54)^{\mathsf T}$, and conclude that $\mathbf{1}$ is not an eigenvector. That calculation sums down columns; the entries of $P\mathbf{1}$ are row sums. The correct multiplication is $$ P\mathbf{1} =\begin{bmatrix}\tfrac12+\tfrac12\\[2pt]\tfrac14+\tfrac34\end{bmatrix} =\begin{bmatrix}1\\1\end{bmatrix}. $$ Column sums instead determine the entries of $\mathbf{1}^{\mathsf T}P$, which is a different multiplication. Checking the orientation before using a stochasticity argument prevents this mistake.

The main theorem gives a basic diagnostic for a proposed transition matrix: eigenvalue one must occur, and every eigenvalue must lie in the closed unit disk. These are necessary conditions, not a complete test that an arbitrary matrix is a valid transition matrix. The identity example also warns against assuming that eigenvalue one is unique, while the cycle shows why the bound must include complex eigenvalues and permit equality.

Check Your Understanding

Use the row-sum condition and the eigenvector equation in each response.

For a finite row-stochastic matrix, compute $P\mathbf{1}$ coordinate by coordinate. Why does this prove that one is an eigenvalue?
In the eigenvalue-bound proof, why must the maximum coordinate modulus $M$ be strictly positive?
For the three-state cycle, verify that $\lambda^3=1$ for each root of $\lambda^3-1=0$, and use this to check its stated eigenvector.
Why does the identity matrix show that eigenvalue one need not have multiplicity one?
Which assumption in the strict-positivity proposition rules out the nonreal unit-modulus eigenvalues of the three-state cycle?

The Eigenvalue One in a Transition Matrix

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