Eigenvalues of a transition probability matrix I have read that, for
$$I - \alpha P$$
where $I$ is the $n\times n$ identity matrix, $\alpha \in (0,1]$, and $P$ is the transition probability matrix with dimensions $n \times n$, $I - \alpha P$ is an invertible matrix because the eigenvalues of any transition probability matrix lie within the unit circle of the complex plane. "Thus no eigenvalue of $\alpha P$ can be equal to 1, which is a necessary and sufficient condition for $I - \alpha P$ to be invertible."
Why is that?
 A: For any scalar $\alpha$ and square matrix $P$, the following are equivalent:


*

*$\alpha P$ has an eigenvalue $1$

*$\alpha Pv = v$ for some non-zero vector $v$

*$(I - \alpha P)v = 0$ for some non-zero vector $v$

*$I - \alpha P$ is not invertible

A: As stated, the result is FALSE.  Presuming P is indeed a Markov Chain transition matrix, i.e., is row stochastic, i.e., all entries are non-negative with all row sums equal to 1, then the largest eigenvalue of P is exactly 1. Given that α = 1 satisfies α ∈ (0,1],  I - α * P in this case has an eigenvalue = 0 (i.e., corresponding to P's eigenvalue of 1).  Therefore, such an I - α * P is not invertible; and in this case,  α * P = P, which has largest eigenvalue = 1.  The result would be true if the restriction on α were changed to α ∈(0,1).  
The statement, "eigenvalues of any transition probability matrix lie within the unit circle of the complex plane" is true only if "within" is interpreted to mean inside or on the boundary of the unit circle, as is the case for the largest eigenvalue, 1.
Proof that P has an eigenvalue = 1.  All row sums of P = 1, therefore,
P  * (vector of ones) = (vector of ones)

Therefore, 1 is an eigenvalue of P with right eigenvector = (vector of ones), thereby establishing the falsehood of the claim made in the opening post of the thread.
