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Let $P$ be a $2 \times2$ transition matrix. $$P=\begin{pmatrix} 1-p & p\\q & 1-q\end{pmatrix}$$ Show that the eigenvalues of $P$ are $1$ and $1-p-q$.

I am able to show that one of the eigenvalues is $1$, however I am not sure how to show that the second eigenvalue is $1-p-q$.

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    $\begingroup$ Hint: the trace is the sum of the eigenvalues. $\endgroup$ – Robert Israel Feb 13 '18 at 7:30
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We know that the product of all eigenvalues is equal to the $\det$ and the sum of eigenvalues is equal to the trace of a matrix. Hence $\lambda_1 \lambda_2 = 1 - p - q$ and $\lambda_1+\lambda_2 = 2 -p - q$. I think you are able to continue

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It is possible that you have not learnt that the trace of a matrix is the sum of its eigenvalues and the determinant of a matrix is the product of its eigenvalues. Hence, I provide an alternative approach:


Note that the following statements are equivalent:

  • $\lambda$ is an eigenvalue of $P$
  • $(P-\lambda I)\mathbf{x}=\mathbf{0}$ has a nontrivial solution
  • $N(P-\lambda I)\neq \{\mathbf{0}\}$
  • $(P-\lambda I)$ is singular
  • $\det(P-\lambda I)=0$

Hence, all we need to do is check that one of these statements satisfy, then all the others satisfy. I will assume you have correctly shown that $\lambda_1=1$ is an eigenvalue of $P$. I will prove that $\lambda_2=1-p-q$ is an eigenvalue of $P$.


The easiest condition to check, in my opinion is if $\det(P-\lambda_2 I)=0$. Note that: $$\det(P-\lambda_2 I)=\begin{vmatrix} 1-p-(1-p-q) & p \\ q & 1-q-(1-p-q) \end{vmatrix}=\begin{vmatrix} q & p \\ q & p \end{vmatrix}=0$$ Where in the last step, we have used the fact that if a $n\times n$ matrix has two identical rows or columns, its determinant is zero. Thus, $1-p-q$ is an eigenvalue of $P$. $$\tag*{$\square$}$$

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A stochastic matrix always has $1$ as an eigenvalue. If it’s row-stochastic, all of its rows sum to $1$, but summing rows is equivalent to right-multiplying by the column vector that consists entirely of $1$s, hence it’s a right eigenvector with eigenvalue $1$. (If the matrix is column-stochastic, you can apply the same argument to its transpose, and use the fact that transposes have identical eigenvalues.)

For any matrix, once you know all but one eigenvalue (counting multiplicities), you get the last one “for free” because the sum of the eigenvalues is equal to the trace of the matrix. So, for this matrix, you know that the second eigenvalue must be $(1-p)+(1-q)-1 = 1-p-q.$

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