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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 $\det (A - \lambda I) = 0$.

Sonamu
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Dogjus
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  • Your matrix setup is OK. The row sums are all 1, so it's a right stochastic (vs. left stochastic) matrix. What algebraic approach have you attempted? And what eigenvalues do you obtain using that approach? – brentertainer Jun 17 '23 at 15:40
  • Sum of eigenvalues is trace, and product of eigenvalues is determinant. – Andrew Jun 18 '23 at 07:16

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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.

Sonamu
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  • If you would like to know why the largest eigenvalues of a Markov matrix is 1, you can refer to the answer to this post https://math.stackexchange.com/questions/2822190/proof-that-markov-matrix-has-eigenvalue-of-1 – Sonamu Jun 18 '23 at 06:04