How many eigenvalues does an $n\times n$ matrix have, and how does this relate to the algebraic multiplicity of the dominant eigenvalue?

Question

I'm having to reevaluate my understanding of eigenvalues and how many eigenvalues an $n\times n$ matrix possesses. Previously, I had thought that such a matrix $A$ possessed $d\leq n$ complex eigenvalues, and that this number $d$ was determined by the number of distinct roots of the matrix's characteristic polynomial. Furthermore, the sum of the complex eigenvalues' algebraic multiplicities equals $n$, since the characteristic polynomial of $A$ necessarily has degree $n$ and therefore has $n$ complex roots (where a root of multiplicity $m$ is counted $m$ times). However, I came across a problem involving dominant eigenvalues that prompted me to question this interpretation:

If $A$ has a dominant eigenvalue $\lambda_1$, prove that the eigenspace $E_{\lambda_1}$ is one-dimensional.

Solution: If $\lambda$ is dominant, then $|\lambda|>|\gamma|$ for all other eigenvalues $\gamma$. But this means that the algebraic multiplicity of $\lambda$ is $1$, since it appears only once in the list of eigenvalues listed with multiplicity, so its geometric multiplicity is $1$ and thus its eigenspace is one-dimensional.

So if I'm reading this explanation correctly, a list of all of the eigenvalues of $A$ should include $i$ instances of an eigenvalue with algebraic multiplicity $i$. In other words, every $n \times n$ matrix has exactly $n$ complex eigenvalues, and there is a distinction between the number of eigenvalues that a matrix possesses and the number of distinct eigenvalues that a matrix possesses. This subtle distinction seemed arbitrary until I considered the solution to this problem, which seems to require that all eigenvalues be treated as separate entities, even if they possess the same scalar values. For example, an eigenvalue $\lambda =2$ with algebraic multiplicity $2$ should actually be thought of as two eigenvalues $\lambda _1 = \lambda _2 = 2$. With this understanding, it is clear that neither $\lambda _1$ nor $\lambda _2$ can be a dominant eigenvalue, since it is not true that $|\lambda _1|>|\lambda _2|$, nor is it true that $|\lambda _2|>|\lambda _1|$. In fact, it is impossible for any eigenvalue with algebraic multiplicity greater than $1$ to be dominant. From here, I am comfortable with the fact that a dominant eigenvalue (if it exists) must also have geometric multiplicity $1$, since the geometric multiplicity of an eigenvalue is always less than or equal to the corresponding algebraic multiplicity.

Is this the correct way to interpret the preceding proof/eigenvalues in general? Hopefully I've articulated my thought process clearly, and thank you for taking the time to help!

Answer 1

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.

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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.