Iteration Method
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 T v = 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) = {0}, 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 cT (x). Since the degree of cT (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 cA (x) = det(xIn − 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 eigenvalue 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 geometric 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 T be linear mappings induced by T on W and V /W. Then cT (x) = cT (x)cT (x). (ii). Let V = W1 ⊕ W2 , where W1 and W2 are T -invariant subspaces, and let T1 and T2 be linear mappings induced by T on W1 and W2 . Then cT (x) = cT1 (x)cT2 (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
1
Eigenvalues and Eigenvectors
A scalar λ ∈ K is an eigenvalue of T if there is a nonzero v ∈ V such that T v = 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) = {0}, 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 cT (x). Since the degree of cT (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 cA (x) = det(xIn − 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 eigenvalue 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 geometric 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 T be linear mappings induced by T on W and V /W. Then cT (x) = cT (x)cT (x). (ii). Let V = W1 ⊕ W2 , where W1 and W2 are T -invariant subspaces, and let T1 and T2 be linear mappings induced by T on W1 and W2 . Then cT (x) = cT1 (x)cT2 (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