Thread: Question about definition of the term "diagonally dominant eigenvalue"

I need help. Consider a matrix that has only one eigenvalue (say 3) but this value has algebraic multiplicity bigger than 1.
Is 3 the dominant eigenvalue of the matrix or not? When determining if there is dominant eigenvalue,
how do we "count" eigenvalues? As 3, 3 or just one 3?

Edit: Such a matrix can be for example:

| 3 1 |
| 0 3 |

Edit: The topic of the thread - "diagonally dominant eigenvalue" is a mistake and cannot be edited. Please disregard "diagonally"

um,... yes. an eigenvalue is just an eigenvalue. the multiplicity of an eigenvalue only comes into it when you are considering which eigenvectors belong to that eigenvalue. eigenvalues with algebraic multiplicity may split into multiple linearly independent eigenvectors (or may not). if a matrix has but a single eigenvalue, yes, that eigenvalue is the dominant one, by default.

one caveat, however: a real nxn matrix may have complex eigenvalues which dominate the real ones. for practical applications, it is often assumed the matrix is diagonalizable over the field being used.

So, the matrix

| 3 0 |
| 0 3 |

has dominant eigenvalue. That is 3.

yes.