To find the steady-state vector of a $2 \times 2$ matrix, you're generally dealing with a Markov chain or a system of linear equations. The steady-state vector represents the long-term behavior of the system, which remains unchanged after multiple transitions.

Given a matrix $A$, the steady-state vector $\mathbf{v}$ is a vector that satisfies:

$A \mathbf{v} = \mathbf{v}$

This means that $\mathbf{v}$ is an eigenvector of matrix $A$ corresponding to the eigenvalue $1$. To find $\mathbf{v}$, you can follow these steps:

Steps to Find the Steady-State Vector:

1. Write the matrix equation: Start by writing the equation $A \mathbf{v} = \mathbf{v}$, which simplifies to: $(A - I) \mathbf{v} = 0$ where $I$ is the identity matrix.

2. Set up the system of equations: The equation $(A - I) \mathbf{v} = 0$ corresponds to a system of linear equations. The matrix $A - I$ is: $A - I = \begin{pmatrix} a_{11} - 1 & a_{12} \\ a_{21} & a_{22} - 1 \end{pmatrix}$ This will be a system of equations that you need to solve for the steady-state vector $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$.

3. Solve the system: Solve the system of equations to find the values of $v_1$ and $v_2$. You can use methods like substitution, elimination, or matrix row reduction.

4. Normalize the vector: Since steady-state vectors are typically normalized (especially in Markov chains) to sum to 1 (i.e., $v_1 + v_2 = 1$), you can use this condition to get the final steady-state vector.

Example

Suppose you have a matrix:

$A = \begin{pmatrix} 0.5 & 0.5 \\ 0.2 & 0.8 \end{pmatrix}$

To find the steady-state vector $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$:

1. Set up the equation $A \mathbf{v} = \mathbf{v}$, which becomes: $\begin{pmatrix} 0.5 & 0.5 \\ 0.2 & 0.8 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$ This simplifies to: $\begin{pmatrix} 0.5v_1 + 0.5v_2 \\ 0.2v_1 + 0.8v_2 \end{pmatrix} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$

2. Subtract the right-hand side from both sides: $\begin{pmatrix} -0.5v_1 + 0.5v_2 \\ -0.2v_1 + 0.8v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

3. This gives the system of equations: $-0.5v_1 + 0.5v_2 = 0$ $-0.2v_1 + 0.8v_2 = 0$

4. Solve the system: From the first equation, we have $v_1 = v_2$. Substituting into the second equation: $-0.2v_1 + 0.8v_1 = 0 \quad \Rightarrow \quad 0.6v_1 = 0 \quad \Rightarrow \quad v_1 = 0$ Hence, $v_2 = 0$.

5. Normalize $\mathbf{v}$: In this case, the normalization condition is $v_1 + v_2 = 1$. Substitute into your expression for $\mathbf{v}$.

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