Beforehand anything, $T: V\rightarrow V$ is an endomorphism of a vector space with dimension $n$.
Theorem:
TFAE equivalent:
i) $c$ is an eigenvalue of $T$
ii) $T-cI$ is not invertible,
iii) $\det(T-cI)=0$.
i) $\iff$ ii) and ii) $\implies$ iii) are obvious for me. When doing iii) $\implies$ ii) I start having doubts.
One argument goes like this:
iii) means that the matrix of $T-cI$(with respect to any basis), call it $A$, is not invertible* which means $T-cI$ is not invertible, then $T-cI$ either is not injective or not surjective. I say it can perfectly be only not surjective, and so $c$ may not be an eingenvalue of $T$ if we just know iii,
The other argument is more "essential", and goes like this:
[Same than first argument until *], then $A$ can't be row reduced to the identity matrix. Suppose any column of $A$ is full of $0$'s(otherwise, we are done). Experience has shown me that A is row-equivalent to a matrix which has, at least, one row of $0$'s(I have seen other post which shows this is true, anyway). As far as I know, if one wishes to know the Null space of a linear transformation represented by a certain matrix, one has to row reduce the matrix and see if it's equivalent to the identity or not(if not, one can find a relation between the coordinates and after a while, know explicitly the Null space), So if I row reduce $A$, I will find that a row is full of zeros and so one variable is independant, which means $T-cI$ has nonzero vectors in its null space, so $c$ is indeed an eigenvalue.
What is the most dubious from the 2nd argument is that I would be actually proving the title of the post, which I tend not to believe. Where am I wrong?
Edit: I've just realised that I've actually proven in the second argument that T-cI is NOT injective(It also serves as a proof to show that if det A=0 then the endomorphism asocciated with A is not injective).
I answer to the question on the title. The answer is no. Because if $detA=0$ ($A$ is a square matrix since it is associated to an endomorphism) then the rank is not maximal, then the kernel of the linear transformation associated to $A$ is not trivial with respect to any basis you choose!
Moreover $f$ is an endomorphism of a vector space $V$ with finite dimension, injectivity is equivalent to surjectivity. So your first argument work: $(T-cI)v=0$ for some vector $v\neq 0$, then $Tv=cv$.