How Is the Steady-State Solution Related to Eigenvalues and Eigenvectors?

A steady-state solution is an eigenvector of the transition matrix corresponding to the eigenvalue $\lambda=1$. After this eigenvector is normalized so that its entries sum to $1$, its entries give the long-term probabilities of the states.

For a Markov chain using the IB convention that columns represent the state being left, the state vector after $n$ transitions is

$$\mathbf{s}_n=T^n\mathbf{s}_0.$$

At steady state, another transition does not change the probability distribution. Therefore,

$$T\mathbf{s}=\mathbf{s}=1\mathbf{s}.$$

This is exactly the eigenvector equation $T\mathbf{v}=\lambda\mathbf{v}$, with $\lambda=1$ and $\mathbf{v}=\mathbf{s}$.

For example, let

$$T=\begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix},\qquad \mathbf{s}=\begin{pmatrix}x\\y\end{pmatrix}.$$

Solving $T\mathbf{s}=\mathbf{s}$ gives

$$0.8x+0.3y=x \quad\Rightarrow\quad 0.2x=0.3y,$$

so $x:y=3:2$. Normalizing using $x+y=1$ gives

$$\mathbf{s}=\begin{pmatrix}0.6\\0.4\end{pmatrix}.$$

Thus, the long-term probabilities are $0.6$ and $0.4$. For a regular Markov chain, repeated multiplication by $T$ approaches this distribution regardless of the initial state.

This also explains the long-term behaviour. Powers of $T$ suppress components associated with eigenvalues whose absolute values are less than $1$, while the component in the direction of the eigenvector for $\lambda=1$ remains.

An equivalent method is to solve $(T-I)\mathbf{s}=\mathbf{0}$, together with $x+y=1$. The matrix equation determines only a ratio because every non-zero scalar multiple of an eigenvector is also an eigenvector; normalization selects the required probability vector.

A common misconception is that any eigenvector for $\lambda=1$ is already a probability distribution. It must be scaled so all entries are non-negative and sum to $1$.

On a GDC, you may obtain the steady state by calculating a large power of $T$ and multiplying by an initial state. However, if an exact solution is required, solve the simultaneous equations instead. In either method, check that the resulting entries are valid probabilities.

Exam technique: This is HL-only content from AHL 4.19: Transition matrices and Markov chains. Show both $T\mathbf{s}=\mathbf{s}$ and the normalization condition, then interpret the probabilities in context.