must see it has some proofs of kunze + more

must see it has some proofs of kunze + more - Throughout in...

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Unformatted text preview: Throughout in this text V will be a vector space of finite dimension n over a field K and T : V → V will be a linear transformation. 1 Eigenvalues and Eigenvectors A scalar λ ∈ K is an eigenvalue of T if there is a nonzero v ∈ V such that Tv = λv. In this case v is called an eigenvector of T corresponding to λ. Thus λ ∈ K is an eigenvalue of T if and only if ker( T- λI ) 6 = { } , and any nonzero element of this subspace is an eigenvector of T corresponding to λ. Here I denotes the identity mapping from V to itself. Equivalently, λ is an eigenvalue of T if and only if det( T- λI ) = 0 . Therefore all eigenvectors are actually the roots of the monic polynomial det( xI- T ) in K. This polynomial is called the characteristic polynomial of T and is denoted by c T ( x ) . Since the degree of c T ( x ) is n, the dimension of V, T cannot have more than n eigenvalues counted with multiplicities. If A ∈ K n × n , then A can be regarded as a linear mapping from K n to itself, and so the polynomial c A ( x ) = det( xI n- A ) is the characteristic polynomial of the matrix A, and its roots in K are the eigenvalues of A. A subspace W of V is T-invariant if T ( W ) ⊆ W. The zero subspace and the full space are trivial examples of T-invariant subspaces. For an eigen- value λ of T the subspace E ( λ ) = ker( T- λI ) is T-invariant and is called the eigenspace corresponding to λ. The dimension of E ( λ ) is the geomet- ric multiplicity of eigenvalue λ, and the multiplicity of λ as a root of the characteristic polynomial of T is the algebraic multiplicity of λ 1.1 (i). Let W be a T-invariant subspace of V. and let T and e T be linear mappings induced by T on W and V/W. Then c T ( x ) = c T ( x ) c e T ( x ) . (ii). Let V = W 1 ⊕ W 2 , where W 1 and W 2 are T-invariant subspaces, and let T 1 and T 2 be linear mappings induced by T on W 1 and W 2 . Then c T ( x ) = c T 1 ( x ) c T 2 ( x ) . 1.2 The geometric multiplicity of eigenvalue λ cannot exceed its algebraic multiplicity. Proof. Let the geometric multiplicity of λ be m. Then λ will have m lin- early independent eigenvectors u 1 ,...,u m , that is, Tu i = λu i for i ∈ 1(1) m. Extend this to a basis of V : B = { u 1 ,...,u m ,u m +1 ,...,u n } . Then [ T ] B = 1 λI m B A . Therefore c T ( x ) = ( x- λ ) m c A ( x ) and the algebraic multiplicity of λ is at least m. Recall that L ( V ) , the set of linear operators on V is a vector space over K of dimension n 2 . Thus, if T is a linear operator on V , then I = T ,T,...,T n 2 , are linearly dependent, that is, there are scalars α ,α 1 ,...,α n 2 not all of which are zero such that α I + α 1 T + ··· + α n 2 T n 2 = 0 ( ∈ L ( V )) . Therefore, if f ( x ) = α + α 1 x + ··· + α n 2 x n 2 , then f ( T ) = 0 . This shows that S = { p ( x ) ∈ K [ x ] | p ( T ) = 0 } is a nonzero principal ideal of K [ x ] . The monic polynomial which generates this ideal is called the minimal polynomial of T . We will denote the minimal polynomial of....
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This note was uploaded on 03/24/2012 for the course MATH 203 taught by Professor Ankit during the Spring '12 term at Evergreen.

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must see it has some proofs of kunze + more - Throughout in...

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