Proving a Primitive Matrix Has 1 Eigenvalue

[johnjohn1]

Well, I'm writing project on Marcov chains, and I've stumbled upon the statement that a Primitive matrix only has one eigenvalue on its Spectral circle.

Primitive Matrix = A matrix (lets call it U) where U^n > 0 for n > 1

That's nice, but how am i supposed to prove that?

 

[jvicens]

Hello,

Thank you for reaching out with your question. I would like to clarify and provide some insight on the statement you have stumbled upon.

First, let's define what a primitive matrix is. A primitive matrix is a square matrix where all elements are non-negative and there exists a positive integer n where the matrix raised to the nth power has all positive elements. This is represented as U^n > 0 for n > 1.

Now, to prove that a primitive matrix only has one eigenvalue on its spectral circle, we need to first understand the spectral circle. The spectral circle of a matrix is the set of all complex numbers λ such that the matrix minus λ times the identity matrix is not invertible. In other words, these are the values of λ that make the determinant of the matrix equal to 0.

Next, we need to understand the concept of eigenvalues. Eigenvalues are the values λ that satisfy the equation Ax = λx, where A is the matrix and x is a non-zero vector. In simpler terms, eigenvalues are the values that when multiplied by a vector, result in a scaled version of that same vector.

Now, let's put these two concepts together. A primitive matrix has only one eigenvalue on its spectral circle because if there were more than one eigenvalue on the spectral circle, then there would be more than one value of λ that makes the determinant of the matrix equal to 0. This would mean that there are multiple values of λ that satisfy the equation Ax = λx, which goes against the definition of eigenvalues.

In conclusion, a primitive matrix only has one eigenvalue on its spectral circle because having more than one would contradict the definition of eigenvalues. I hope this explanation helps in your project on Markov chains. If you have any further questions, please don't hesitate to reach out.