Recipe

Find a Markov chain's exact steady state

Use when: you have a column-stochastic transition matrix $P$ and want the exact steady-state distribution $\mathbf x^\ast$ , not just its approximate behavior after many steps

Goal: the exact fractions making up $\mathbf x^\ast$ , satisfying both $P\mathbf x^\ast=\mathbf x^\ast$ and that its entries sum to 1

Taught in Markov chains and steady states

Quick recipe

Set up $(P-I)\mathbf x^\ast=\mathbf 0$ , dropping one redundant row
Add the normalization equation $\sum x_i = 1$
Solve the resulting system for the exact fractions
Check the result against both defining conditions

1
Write $P\mathbf x^\ast=\mathbf x^\ast$ out one equation per row and move everything to one side; then drop any one of those rows — it's guaranteed to be redundant once the rest hold together with the normalization equation.

Example

For $P=\begin{bmatrix} 0.7 & 0.4 \\ 0.3 & 0.6 \end{bmatrix}$ with $\mathbf x^\ast = (a,b)$ : row 1 gives $a=0.7a+0.4b \implies -0.3a+0.4b=0$ . Row 2 would give $b=0.3a+0.6b \implies 0.3a-0.4b=0$ — literally the negative of row 1, so it's dropped as redundant.

Why: every column of $P$ sums to $1$ , which forces the $n$ equations of $(P-I)\mathbf x=\mathbf 0$ to always sum to the zero equation — one of them is never independent information.
2
Since $\mathbf x^\ast$ has to be an actual probability distribution, add the equation that its entries sum to $1$ — this replaces the row just dropped.

Example

$a+b=1$ .

Why: without this equation, the system only pins $\mathbf x^\ast$ down up to a scalar multiple — any eigenvector for $\lambda=1$ solves $(P-I)\mathbf x=\mathbf 0$ — and only one specific multiple is a valid distribution.
3
Solve the surviving equation from Step 1 together with the normalization equation from Step 2, by substitution or elimination.

Example

From $-0.3a+0.4b=0$ : $a=\tfrac43b$ . Substitute into $a+b=1$ : $\tfrac43b+b=1 \implies \tfrac73b=1 \implies b=\tfrac37$ , so $a=\tfrac47$ .

Why: this is an ordinary linear system once Steps 1 and 2 are set up — the Markov structure only decided which equations to use, not how to solve them.
4
Confirm the entries sum to $1$ , and substitute back into $P\mathbf x^\ast$ to confirm it reproduces $\mathbf x^\ast$ exactly.

Example

$\tfrac47+\tfrac37=1$ ✓, and $P\mathbf x^\ast = \begin{bmatrix} 0.7(\tfrac47)+0.4(\tfrac37) \\ 0.3(\tfrac47)+0.6(\tfrac37) \end{bmatrix} = \begin{bmatrix} \tfrac47 \\ \tfrac37 \end{bmatrix} = \mathbf x^\ast \ \checkmark$

Why: this verifies both defining properties of a steady state at once — being an eigenvector for $\lambda=1$ , and actually summing to $1$ as a valid probability distribution.

Common mistakes

Using all $n$ rows of $(P-I)\mathbf x=\mathbf 0$ without dropping the redundant one, turning an easy system into a needlessly tangled one.
Forgetting the normalization equation and reporting some eigenvector for $\lambda=1$ that doesn't actually sum to $1$ as if it were the steady state.
Solving for an eigenvector of $P$ at the wrong eigenvalue — the steady state is specifically the eigenvector for $\lambda=1$ , not just any eigenvector of $P$ .
A sign slip moving terms across the equation, especially easy here since every row of $P-I$ only subtracts $1$ from one diagonal entry.