The geometric multiplicity of an eigenvalue is defined to be the number of linearly independent eigenvectors associated with that eigenvalue.
The algebraic multiplicity of an eigenvalue is defined as the eigenvalue's multiplicity as a root of the characteristic polynomial.
I don't think a convention is well-established: in some contexts, I see "different eigenvalues" refer to a set of distinct values with associated algebraic multiplicities, while in other contexts, I see "different eigenvalues" refer to the set of $n$ eigenvalues, possibly with repetitions due to multiplicity. Typically one can either discern which convention is being used, or the author should take care to clarify what is meant.
In your case, I think you just have to read the definition of "dominant eigenvalue" carefully. Based on the problem writing "dominant eigenvalue $\lambda_1$," I suspect the definition is written as
if $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A$, then $\lambda_1$ is considered dominant if $|\lambda_1| > |\lambda_i|$ for all $i \ne 1$
or something like that, which is unambiguous compared to
$|\lambda| > |\gamma|$ for all other eigenvalues $\gamma$
which is very ambiguous for the reasons you raise.
Now that we know that the context of your question is the power method, then my above guess on what "dominant eigenvalue" means is incorrect.
Let $\lambda_1, \ldots, \lambda_m$ be the distinct eigenvalues of $A$ with multiplicities $n_1, \ldots, n_m$. If $|\lambda_1| > |\lambda_i|$ for all $i \ne 1$, then $\lambda_1$ is said to be the dominant eigenvalue. The power method will converge to something in the eigenspace corresponding to $\lambda_1$. To ensure that it does not converge to zero, the initial vector must not be orthogonal to the eigenspace.
Best Answer
Hints. (I am more accustomed to the convention that a vector in $\mathbb R^n$ denoted by a single letter is a column vector. So, I will write $B=A+uu^TA^{-1}$ instead, where $u=\alpha^T\in\mathbb R^{n\times1}$ is a column vector.)