TITLE: About the remark of the proof that every operator on non-zero finite dimensional complex vector space has an eigenvalue in Linear Algebra Done Right
QUESTION [8 upvotes]: In the 2nd Edition of Linear algebra Done Right, theorem 5.10 (Every operator on a finite-dimensional, nonzero, complex vector space has an eigenvalue), the author writes as a side note:

Compare the simple proof of this theorem given here with the standard
proof using determinants. With the standard proof, first the difficult
concept of determinants must be defined, then an operator
with 0 determinant must be shown to be not invertible, then
the characteristic polynomial needs to be defined, and by the time the
proof of this theorem is reached, no insight remains about why it is
true.

However, the approach given by the author does not seem to give me any more insight for why this theorem is true than the standard proof using determinants. Both proofs seem to just find some trick to transform this into a problem about roots of polynomials, in order to apply FTA. Am i missing some intuition about this
Here's the proof for reference:

REPLY [3 votes]: I was thinking about the same a few months ago, I think Axler's point is that the proof using determinants needs a theoretical framework that's rather more difficult to develop because you need to define the entities that he mentions in his comment of the proof (determinant and characteristic polynomial ) and show that an operator with $0$ determinant is not invertible. That could be rather painful for a pure mathematician trying to write an elegant text (this is my opinion). Now, if you look at his proof carefully you see that it is rather based on linear independence, a pretty basic concept that he defines at the beginning of the text. Following his approach also has the benefit that you can define the determinant later on as a product of eigenvalues, which to me makes the concept very intuitive.