Finding general solution for Markov Matrix Eigenvalues.

Question

I am being asked to solve a Markov Matrix in general terms. Essentially I need to prove an eigenvalue with always be $1$.

I have set up the following matrix:

\begin{pmatrix}p&1-p\\q&1-q\end{pmatrix}

I think my issue is either with my matrix setup, or my algebra. But I cannot figure out how to get that $1$ value when using $\det (A - \lambda I) = 0$.

Answer 1

With your matrix:

\begin{equation} P = \begin{pmatrix} p && 1 - p \\ q && 1 - q \\ \end{pmatrix} \end{equation}

Now, based on the definition of eigenvalues $Px = λx$, we can now write $(P - λ)x = 0$ which is essentially equivalent to $det(P - λE) = 0$.

No matter whether you use the canonical approach or the Cayley-Hamilton, you can write the eigen equation as below:

\begin{equation} λ^2 - (p-q+1)λ + (p - q) = 0 \end{equation}

You can use the formula for solving quadratic equations to get the roots of λ, and clearly, if you replace $λ = 1$, the equation holds.