Let $A$ be a square matrix such that $A^T$ is a positive transition matrix. Show that $\lambda = -1$ fails to be an eigenvalue of $A$.

Question

Question states:

Consider a square matrix $A$ such that all entries are positive and the sum of the entries in each row is $1$, i.e. $A^T$ is a positive transition matrix. Show that $\lambda = -1$ fails to be an eigenvalue of $A$.

I am not sure how to prove this. I know that all eigenvalues of $A$ has their absolute values less than or equal to 1, and that 1 is an eigenvalue.

Answer 1

Perron-Frobenius would imply all eigenvalues (except $\lambda=1$) have absolute value strictly less than $1$.

Answer 2

Let $S$ be the set of $n\,{\times}\,n$ matrices with positive real entries such that the sum of the entries in each row is $1$.

Suppose $A \in S$ has $\lambda = -1$ as an eigenvalue.

Out goal is to derive a contradiction.

Let $x = (x_1,...,x_n)$ be a nonzero real $n$-vector such that $Ax = -x$.

Let $m = \text{min}(x_1,...,x_n)$, and let $M = \text{max}(x_1,...,x_n)$.

If $m = M$, then $x_1 = \cdots = x_n$, but then from $A\in S$, we would get $Ax = x$, contrary to the assumption that $Ax = -x$.

Thus, we must have $m < M$.

Let $A^2x = y = (y_1,...,y_n)$. Then

$$y = A^2x = A(Ax) = A(-x) = -Ax = -(-x) = x$$

But $A \in S$ implies $A^2 \in S$, hence $A^2x = y$ implies that each of $y_1,...,y_n$ is a strict convex combination of $x_1,...,x_n$. It follows that $m < y_1,...,y_n < M$, contrary to $y=x$.