Say I have a F - vector space V and a subspace U given by U={va : a is in F}. Now suppose I have an operator defined by $Tv=av$. Clearly, U is invariant under T, since for any element of U, say bv, I can write $T(bv)=bT(v)=bav$ which is in U.
Here is my question, how many eigenvalues does T have?
It seems to me that there are infinitely many, since for different elements of U, i can write $T(u)=cbv$, where c is such that $u=cv$, in which case cb would be an eigen value. This would mean that there are infinitely many eigenvectors, but i know this does not make sense.
Where is the mistake in my thought process?
In the case in which $V$ has finite dimension, such an operator has $n$ eigenvalues, where $n$ is the dimension of $V$. Eigenvalues are the roots of the characteristic equation, that is a polynomial equation over F. Such roots cannot be more than n (but can be less) in F itself. In this case they are n and are all equal. See it also from a geometrical point of view: if it were possible to have infinite many eigenvalues you would also have infinite many eigenvectors. But eigenvectors are linearly indipendent and so V would have infinite dimension.