Generalized Eigenvectors Definition: eigenspace, E(λ, T) Suppose T 2 L(V) and λ 2 F. The eigenspace of T corresponding to λ, denoted E(λ, T), is defined by E(λ, T) = null(T − λI): Definition: eigenvector Clearly Suppose T 2 L(V) and λ 2 F is an v 2 null(T − λI) () Tv = λv: eigenvalue of T. A vector v 2 V is called Thus E(λ, T) is the set of eigenvectors an eigenvector of T corresponding to λ of T corresponding to λ, along with the if v 6= 0 and Tv = λv. 0 vector. Eigenvalue and Eigenvector Review Definition: eigenvalue Suppose T 2 L(V). A number λ 2 F is called an eigenvalue of T if there exists v 2 V such that v 6= 0 and Tv = λv. Definition: eigenspace, E(λ, T) Suppose T 2 L(V) and λ 2 F. The eigenspace of T corresponding to λ, denoted E(λ, T), is defined by E(λ, T) = null(T − λI): Clearly v 2 null(T − λI) () Tv = λv: Thus E(λ, T) is the set of eigenvectors of T corresponding to λ, along with the 0 vector. Eigenvalue and Eigenvector Review Definition: eigenvalue Suppose T 2 L(V). A number λ 2 F is called an eigenvalue of T if there exists v 2 V such that v 6= 0 and Tv = λv. Definition: eigenvector Suppose T 2 L(V) and λ 2 F is an eigenvalue of T. A vector v 2 V is called an eigenvector of T corresponding to λ if v 6= 0 and Tv = λv. Clearly v 2 null(T − λI) () Tv = λv: Thus E(λ, T) is the set of eigenvectors of T corresponding to λ, along with the 0 vector. Eigenvalue and Eigenvector Review Definition: eigenvalue Definition: eigenspace, E(λ, T) Suppose T 2 L(V). A number λ 2 F is Suppose T 2 L(V) and λ 2 F. The called an eigenvalue of T if there exists eigenspace of T corresponding to λ, v 2 V such that v 6= 0 and Tv = λv. denoted E(λ, T), is defined by E(λ, T) = null(T − λI): Definition: eigenvector Suppose T 2 L(V) and λ 2 F is an eigenvalue of T. A vector v 2 V is called an eigenvector of T corresponding to λ if v 6= 0 and Tv = λv. Thus E(λ, T) is the set of eigenvectors of T corresponding to λ, along with the 0 vector. Eigenvalue and Eigenvector Review Definition: eigenvalue Definition: eigenspace, E(λ, T) Suppose T 2 L(V). A number λ 2 F is Suppose T 2 L(V) and λ 2 F. The called an eigenvalue of T if there exists eigenspace of T corresponding to λ, v 2 V such that v 6= 0 and Tv = λv. denoted E(λ, T), is defined by E(λ, T) = null(T − λI): Definition: eigenvector Clearly Suppose T 2 L(V) and λ 2 F is an v 2 null(T − λI) () Tv = λv: eigenvalue of T. A vector v 2 V is called an eigenvector of T corresponding to λ if v 6= 0 and Tv = λv. Eigenvalue and Eigenvector Review Definition: eigenvalue Definition: eigenspace, E(λ, T) Suppose T 2 L(V). A number λ 2 F is Suppose T 2 L(V) and λ 2 F. The called an eigenvalue of T if there exists eigenspace of T corresponding to λ, v 2 V such that v 6= 0 and Tv = λv. denoted E(λ, T), is defined by E(λ, T) = null(T − λI): Definition: eigenvector Clearly Suppose T 2 L(V) and λ 2 F is an v 2 null(T − λI) () Tv = λv: eigenvalue of T. A vector v 2 V is called Thus E(λ, T) is the set of eigenvectors an eigenvector of T corresponding to λ of T corresponding to λ, along with the if v 6= 0 and Tv = λv. 0 vector. Definition: generalized eigenspace, G(λ, T) Suppose T 2 L(V) and λ 2 F. The gen- eralized eigenspace of T corresponding to λ, denoted G(λ, T), is defined to be the set of all generalized eigenvectors of T corresponding to λ, along with the 0 vector. Description of generalized eigenspaces Suppose T 2 L(V) and λ 2 F. Then G(λ, T) = null(T − λI)dim V : Generalized Eigenvectors Definition: generalized eigenvector Suppose T 2 L(V) and λ is an eigen- value of T. A vector v 2 V is called a gen- eralized eigenvector of T corresponding to λ if v 6= 0 and (T − λI)jv = 0 for some positive integer j. Description of generalized eigenspaces Suppose T 2 L(V) and λ 2 F. Then G(λ, T) = null(T − λI)dim V : Generalized Eigenvectors Definition: generalized eigenvector Definition: generalized eigenspace, G(λ, T) Suppose T 2 L(V) and λ is an eigen- value of T. A vector v 2 V is called a gen- Suppose T 2 L(V) and λ 2 F. The gen- eralized eigenvector of T corresponding eralized eigenspace of T corresponding λ (λ, ) to λ if v 6= 0 and to , denoted G T , is defined to be the set of all generalized eigenvectors of j (T − λI) v = 0 T corresponding to λ, along with the 0 for some positive integer j. vector. Generalized Eigenvectors Definition: generalized eigenvector Definition: generalized eigenspace, G(λ, T) Suppose T 2 L(V) and λ is an eigen- value of T. A vector v 2 V is called a gen- Suppose T 2 L(V) and λ 2 F. The gen- eralized eigenvector of T corresponding eralized eigenspace of T corresponding λ (λ, ) to λ if v 6= 0 and to , denoted G T , is defined to be the set of all generalized eigenvectors of j (T − λI) v = 0 T corresponding to λ, along with the 0 for some positive integer j. vector. Description of generalized eigenspaces Suppose T 2 L(V) and λ 2 F. Then G(λ, T) = null(T − λI)dim V : We have 3 T (z1; z2; z3) = (0; 0; 125z3) The eigenvalues of T are 0 and 5. for all z1; z2; z3 2 C. Thus The corresponding eigenspaces are G(0; T) = f(z ; z ; 0): z ; z 2 Cg: easily seen to be 1 2 1 2 E(0; T) = f(z ; 0; 0): z 2 Cg 1 1 We have and 3 (T−5I) (z1; z2; z3) = (−125z1+300z2; −125z2; 0): E(5; T) = f(0; 0; z ): z 2 Cg: 3 3 Thus Note that this operator T does not G(5; T) = f(0; 0; z ): z 2 Cg: have enough eigenvectors to span its 3 3 domain C3. Now C3 = G(0; T) ⊕ G(5; T): Example of Generalized Eigenspaces Define T 2 L(C3) by T(z1; z2; z3) = (4z2; 0; 5z3): We have 3 T (z1; z2; z3) = (0; 0; 125z3) for all z1; z2; z3 2 C. Thus The corresponding eigenspaces are G(0; T) = f(z ; z ; 0): z ; z 2 Cg: easily seen to be 1 2 1 2 E(0; T) = f(z ; 0; 0): z 2 Cg 1 1 We have and 3 (T−5I) (z1; z2; z3) = (−125z1+300z2; −125z2; 0): E(5; T) = f(0; 0; z ): z 2 Cg: 3 3 Thus Note that this operator T does not G(5; T) = f(0; 0; z ): z 2 Cg: have enough eigenvectors to span its 3 3 domain C3. Now C3 = G(0; T) ⊕ G(5; T): Example of Generalized Eigenspaces Define T 2 L(C3) by T(z1; z2; z3) = (4z2; 0; 5z3): The eigenvalues of T are 0 and 5. We have 3 T (z1; z2; z3) = (0; 0; 125z3) for all z1; z2; z3 2 C. Thus G(0; T) = f(z1; z2; 0): z1; z2 2 Cg: We have 3 (T−5I) (z1; z2; z3) = (−125z1+300z2; −125z2; 0): Thus Note that this operator T does not G(5; T) = f(0; 0; z ): z 2 Cg: have enough eigenvectors to span its 3 3 domain C3. Now C3 = G(0; T) ⊕ G(5; T): Example of Generalized Eigenspaces Define T 2 L(C3) by T(z1; z2; z3) = (4z2; 0; 5z3): The eigenvalues of T are 0 and 5. The corresponding eigenspaces are easily seen to be E(0; T) = f(z1; 0; 0): z1 2 Cg and E(5; T) = f(0; 0; z3): z3 2 Cg: We have 3 T (z1; z2; z3) = (0; 0; 125z3) for all z1; z2; z3 2 C. Thus G(0; T) = f(z1; z2; 0): z1; z2 2 Cg: We have 3 (T−5I) (z1; z2; z3) = (−125z1+300z2; −125z2; 0): Thus G(5; T) = f(0; 0; z3): z3 2 Cg: Now C3 = G(0; T) ⊕ G(5; T): Example of Generalized Eigenspaces Define T 2 L(C3) by T(z1; z2; z3) = (4z2; 0; 5z3): The eigenvalues of T are 0 and 5. The corresponding eigenspaces are easily seen to be E(0; T) = f(z1; 0; 0): z1 2 Cg and E(5; T) = f(0; 0; z3): z3 2 Cg: Note that this operator T does not have enough eigenvectors to span its domain C3. Thus G(0; T) = f(z1; z2; 0): z1; z2 2 Cg: We have 3 (T−5I) (z1; z2; z3) = (−125z1+300z2; −125z2; 0): Thus G(5; T) = f(0; 0; z3): z3 2 Cg: Now C3 = G(0; T) ⊕ G(5; T): Example of Generalized Eigenspaces Define T 2 L(C3) by We have 3 T(z1; z2; z3) = (4z2; 0; 5z3): T (z1; z2; z3) = (0; 0; 125z3) The eigenvalues of T are 0 and 5.