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.

Eigenvalues of a transition probability matrix
Can we link another hash table for each key, in a hash table? 

My target is to have a very fast 2-d data structure to store cells in a spreadsheet.

I'll store "all columns within a row" in a hash table. Then do this to all 'r' number of rows. Next make the final hash table for rows and store all the 'r' number of hash tables in the new hash table. is this way efficient? or is there a better way?
