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Eigenspace/Kernel/Linearly independent/Section

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Let be a field, a -vector space and

a linear mapping. Then

In particular, is an

eigenvalue of if and only if is not

injective.

Proof


More general, we have the following characterization.


Let be a field, a -vector space and

a linear mapping. Let . Then

Let . Then if and only if , and this is the case if and only if holds, which means .



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 fixed space. 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.


Let be a field, a -vector space and

a linear mapping. Let be elements in . Then

Proof



Let be a field, a -vector space and

a linear mapping. Let be eigenvectors for (pairwise) different eigenvalues . Then are

linearly independent.

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 .



Let be a field, a finite-dimensional -vector space and

a linear mapping. Then there exist at most many eigenvalues

for .

Proof