Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 22/refcontrol
- Relations between eigenspaces
We have seen in Lemma 21.7 that the eigenspace to is the kernel of the endomorphism. More general, the following characterization holds.
Let be a field,MDLD/field a -vector spaceMDLD/vector space and
a linear mapping.MDLD/linear mapping Let . Then
Let . Then if and only if , and this is the case if and only if holds, which means .
In particular,
is an eigenvalue of if and only if is not injective. For a given , this property can be checked with the help of a linear system
(or the determinant),
and the eigenspace can be determined. However, it is not a linear problem to decide whether has eigenvalues at all and how those can be determined. We will continue to study a linear mapping
by considering the differences to homotheties for various .
For an -matrix , we have to determine the kernel of the matrix . If, for example, we want t know whether the matrix has the eigenvalue , then
shows that this is not the case.
Let be a field,MDLD/field a -vector spaceMDLD/vector space and
a linear mapping.MDLD/linear mapping Let be elements in . Then
Let . Then
Therefore,
and this implies, because of , .
Let be a field,MDLD/field a -vector spaceMDLD/vector space and
a linear mapping.MDLD/linear mapping Let be eigenvectorsMDLD/eigenvectors for (pairwise) different eigenvaluesMDLD/eigenvalues . Then are
linearly independent.MDLD/linearly independentWe 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,MDLD/field a finite-dimensionalMDLD/finite-dimensional (vs) -vector spaceMDLD/vector space and
a linear mapping.MDLD/linear mapping Then there exist at most many eigenvaluesMDLD/eigenvalues
for .Proof
In particular, an endomorphism on a finite-dimensional vector space has only finitely many eigenvalues.
- Geometric multiplicity
The restriction of a linear mapping to an eigenspace is the homothety with the corresponding eigenvalue, thus a very simple linear mapping. For a diagonal matrix
the standard basis has the property that every basis vector is an eigenvector for the linear mapping given by the matrix. In this case, it is easy to describe the eigenspaces, see Example 21.4 ; the eigenspace for consists of all linear combinations of the standard vectors , for which equals . In particular, the dimension of the eigenspace equals the number how often occurs as an diagonal element. In general, the dimensions of the eigenspaces are important invariants of an endomorphism.
Let be a field,MDLD/field a -vector spaceMDLD/vector space and
a linear mapping.MDLD/linear mapping For we call
the geometric multiplicity
of .In particular, a number is an eigenvalue of if and only if its geometric multiplicity is at least . It is easy to give examples where the geometric multiplicity of an eigenvalue is any number between and the dimension of the space.
Let denote a field,MDLD/field and let denote a -vector spaceMDLD/vector space of finite dimension. Let
be a linear mapping.MDLD/linear mapping Then the sumMDLD/sum (linear subspaces) of the eigenspacesMDLD/eigenspaces is direct,MDLD/direct (sum) and we have
This follows directly from Lemma 22.3 .
- Diagonalizability
Let denote a field,MDLD/field let denote a vector space,MDLD/vector space and let
denote a linear mapping.MDLD/linear mapping Then is called diagonalizable, if has a basisMDLD/basis (vs) consisting of eigenvectorsMDLD/eigenvectors
for .Let denote a field,MDLD/field and let denote a finite-dimensionalMDLD/finite-dimensional (fgvs) vector space.MDLD/vector space Let
denote a
linear mapping.MDLD/linear mapping Then the following statements are equivalent.- is diagonalizable.MDLD/diagonalizable (vs)
- There exists a basis of such that the describing matrixMDLD/describing matrix is a diagonal matrix.MDLD/diagonal matrix
- For every describing matrix
with respect to a basis , there exists an
invertible matrixMDLD/invertible matrix
such that
is a diagonal matrix.
The equivalence between (1) and (2) follows from the definition, from Example 21.4 , and the correspondence between linear mappings and matrices. The equivalence between (2) and (3) follows from Corollary 11.12 .
If is diagonalizable and if the eigenvalues together with their geometric multiplicities are known, then we can simply write down a corresponding diagonal matrix: just take the diagonal matrix such that, in the diagonal, the eigenvalues occur as often as the geometric multiplicities. In particular, the corresponding diagonal matrix of a diagonalizable mapping is, up to the ordering of the diagonal entries, uniquely determined.
We continue with Example 21.5 . There exists the two eigenvectorsMDLD/eigenvectors and for the different eigenvaluesMDLD/eigenvalues and , so that the mapping is diagonalizable,MDLD/diagonalizable due to Corollary 22.10 . With respect to the basisMDLD/basis (vs) , consisting of these eigenvectors, the linear mapping is described by the diagonal matrix
The transformation matrix,MDLD/transformation matrix from the basis to the standard basis , consisting of and , is simply
The inverse matrixMDLD/inverse matrix is
Because of Corollary 11.12 , we have the relation
Let denote a field,MDLD/field and let denote a finite-dimensionalMDLD/finite-dimensional (fgvs) vector space.MDLD/vector space Let
denote a linear mapping.MDLD/linear mapping Suppose that there exists different eigenvalues.MDLD/eigenvalues Then is
diagonalizable.MDLD/diagonalizableBecause of Lemma 22.3 , there exist linearly independentMDLD/linearly independent eigenvectors.MDLD/eigenvectors These form, due to Corollary 8.10 , a basis.MDLD/basis (vs)
Let denote a field,MDLD/field and let denote a -vector spaceMDLD/vector space of finite dimension. Let
be a linear mapping.MDLD/linear mapping Then is diagonalizableMDLD/diagonalizable (ev) if and only if is the direct sumMDLD/direct sum (vs) of the
eigenspaces.MDLD/eigenspacesIf is diagonalizable, then there exists a basisMDLD/basis (vs) of consisting of eigenvectors.MDLD/eigenvectors Hence,
Therefore,
That the sum is direct follows from Lemma 22.2 . If, the other way round,
holds, then we can choose in every eigenspace a basis. These bases consist of eigenvectors and yield together a basis of .
== Example Example 22.12
change==
We consider -shearing matrices
with . The condition for some to be an eigenvalueMDLD/eigenvalue means
This yields the equations
For , we get and hence also , that is, only can be an eigenvalue. In this case, the second equation is fulfilled, and the first equation becomes
For , we get and thus is the eigenspace for the eigenvalue , and is an eigenvector which spans this eigenspace. For , we have the identity matrix, and the eigenspaceMDLD/eigenspace for the eigenvalue is the total plane. For , there is a one-dimensional eigenspace, and the mapping is not diagonalizable.MDLD/diagonalizable
The product of two diagonal matrices is again a diagonal matrix. The following example shows that the product of two diagonalizable matrices is not necessarily diagonalizable.
Let and denote two lines in through the origin, and let and denote the reflections at these axes. A reflection at an axis is always diagonalizable,MDLD/diagonalizable the axis and the line orthogonal to the axis are eigenlines (with eigenvalues and ). The compositionMDLD/composition
of the reflections is a plane rotation,MDLD/plane rotation the angle of rotation being twice the angle between the two lines. However, a rotation is only diagonalizable if the angle of rotation is or degree. If the angle between the axes is different from degree, then does not have any eigenvector.
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