Eigenspace/Kernel/Linearly independent/Section
Lemma
Let be a field, a -vector space and
a linear mapping. Then
eigenvalue of if and only if is not injective.
Proof
More general, we have the following characterization.
Remark
Beside the eigenspace for , which is the kernel of the linear mapping, the eigenvalues and are in particular interesting. The eigenspace for consists of all vectors which are sent to themselves. Restricted to this linear subspace, the mapping is just the identity, it is called the fixspace. The eigenspace for consists in all vector which are sent to their negative. On this linear subspace, the mapping acts like the reflection at the origin.
Lemma
Let be a field, a -vector space and
a linear mapping. Let be eigenvectors for (pairwise) different eigenvalues . Then are linearly independent.
Proof
We prove the statement by induction on . For , the statement is true. Suppose now that the statement is true for less than vectors. We consider a representation of , say
We apply to this and get, on one hand,
On the other hand, we multiply the equation with and get
We look at the difference of the two equations, and get
By the induction hypothesis, we get for the coefficients , . Because of , we get for , and because of , we also get .
Corollary
Let be a field, a finite-dimensional -vector space and
a linear mapping. Then there exist at most many eigenvalues for .