Let V be an n-dimensional complex vector space where and let be a linear map.
show that has an eigenvector .
i can't solve this question . how can ?
There is a nice proof of this without the use of determinants/charcteristic polynomials. It's here, page 3.
The existence of an eigenvalue implies the existence of an eigenvector. With the definitions from my earlier post it's immediate. What does it mean that has an eigenvalue? This:
What does it mean that has an eigenvector? This:
These formulas are equivalent. The order of the existential quantifiers doesn't matter.
However, Sheldon Axler uses a different definition of an eigenvalue. For him, that has an eigenvalue means this:
3) There exists such that is not injective.
We want to see that 3) implies 2). What does it mean that is not injective? It means that we have two vectors such that and
the last formula being equvalent to
and further equivalent to
Since we have that the vector is non-zero. Therefore, we have found a non-zero vector such that
just as we wanted.