Generalized Eigenvectors Deﬁnition: Eigenspace, E(Λ, T)

Generalized Eigenvectors Deﬁnition: eigenspace, E(λ, T)

Suppose T ∈ L(V) and λ ∈ 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 ∈ L(V) and λ ∈ F is an v ∈ null(T − λI) ⇐⇒ Tv = λv. eigenvalue of T. A vector v ∈ 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 ∈ L(V). A number λ ∈ F is called an eigenvalue of T if there exists v ∈ V such that v 6= 0 and Tv = λv. Deﬁnition: eigenspace, E(λ, T)

Suppose T ∈ L(V) and λ ∈ F. The eigenspace of T corresponding to λ, denoted E(λ, T), is deﬁned by E(λ, T) = null(T − λI).

Clearly v ∈ 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 ∈ L(V). A number λ ∈ F is called an eigenvalue of T if there exists v ∈ V such that v 6= 0 and Tv = λv.

Deﬁnition: eigenvector

Suppose T ∈ L(V) and λ ∈ F is an eigenvalue of T. A vector v ∈ V is called an eigenvector of T corresponding to λ if v 6= 0 and Tv = λv. Clearly v ∈ 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 ∈ L(V). A number λ ∈ F is Suppose T ∈ L(V) and λ ∈ F. The called an eigenvalue of T if there exists eigenspace of T corresponding to λ, v ∈ 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 ∈ L(V) and λ ∈ F is an eigenvalue of T. A vector v ∈ 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)

Deﬁnition: eigenvector Clearly Suppose T ∈ L(V) and λ ∈ F is an v ∈ null(T − λI) ⇐⇒ Tv = λv. eigenvalue of T. A vector v ∈ 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 ∈ L(V) and λ ∈ F. The generalized 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 ∈ L(V) and λ ∈ F. Then G(λ, T) = null(T − λI)dim V .

Generalized Eigenvectors

Deﬁnition: generalized eigenvector

Suppose T ∈ L(V) and λ is an eigenvalue of T. A vector v ∈ V is called a generalized 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 ∈ L(V) and λ ∈ F. Then G(λ, T) = null(T − λI)dim V .

Generalized Eigenvectors

Definition: generalized eigenvector Definition: generalized eigenspace, G(λ, T) Suppose T ∈ L(V) and λ is an eigenvalue of T. A vector v ∈ V is called a gen- Suppose T ∈ L(V) and λ ∈ F. The generalized 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

Description of generalized eigenspaces

Suppose T ∈ L(V) and λ ∈ F. Then G(λ, T) = null(T − λI)dim V . We have 3 T (z1, z2, z3) = (0, 0, 125z3)

Example of Generalized Eigenspaces

Deﬁne T ∈ 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 ∈ C. Thus The corresponding eigenspaces are G(0, T) = {(z , z , 0): z , z ∈ C}. easily seen to be 1 2 1 2 E(0, T) = {(z , 0, 0): z ∈ C} 1 1 We have and 3 (T−5I) (z1, z2, z3) = (−125z1+300z2, −125z2, 0). E(5, T) = {(0, 0, z ): z ∈ C}. 3 3 Thus Note that this operator T does not G(5, T) = {(0, 0, z ): z ∈ C}. have enough eigenvectors to span its 3 3 domain C3. Now C3 = G(0, T) ⊕ G(5, T).

Example of Generalized Eigenspaces

Deﬁne T ∈ 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 ∈ C. Thus

G(0, T) = {(z1, z2, 0): z1, z2 ∈ C}.

We have 3 (T−5I) (z1, z2, z3) = (−125z1+300z2, −125z2, 0). Thus Note that this operator T does not G(5, T) = {(0, 0, z ): z ∈ C}. have enough eigenvectors to span its 3 3 domain C3. Now C3 = G(0, T) ⊕ G(5, T).

Example of Generalized Eigenspaces

Deﬁne T ∈ 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) = {(z1, 0, 0): z1 ∈ C} and

E(5, T) = {(0, 0, z3): z3 ∈ C}. We have 3 T (z1, z2, z3) = (0, 0, 125z3)

for all z1, z2, z3 ∈ C. Thus

G(0, T) = {(z1, z2, 0): z1, z2 ∈ C}.

We have 3 (T−5I) (z1, z2, z3) = (−125z1+300z2, −125z2, 0). Thus

G(5, T) = {(0, 0, z3): z3 ∈ C}.

Now C3 = G(0, T) ⊕ G(5, T).

Example of Generalized Eigenspaces

Deﬁne T ∈ 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) = {(z1, 0, 0): z1 ∈ C} and

E(5, T) = {(0, 0, z3): z3 ∈ C}. Note that this operator T does not have enough eigenvectors to span its domain C3. Thus

G(0, T) = {(z1, z2, 0): z1, z2 ∈ C}.

We have 3 (T−5I) (z1, z2, z3) = (−125z1+300z2, −125z2, 0). Thus

G(5, T) = {(0, 0, z3): z3 ∈ C}.

Now C3 = G(0, T) ⊕ G(5, T).

Example of Generalized Eigenspaces

Deﬁne T ∈ 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. for all z1, z2, z3 ∈ C. The corresponding eigenspaces are easily seen to be

E(0, T) = {(z1, 0, 0): z1 ∈ C} and

E(5, T) = {(0, 0, z3): z3 ∈ C}. Note that this operator T does not have enough eigenvectors to span its domain C3. We have 3 (T−5I) (z1, z2, z3) = (−125z1+300z2, −125z2, 0). Thus

G(5, T) = {(0, 0, z3): z3 ∈ C}.

Now C3 = G(0, T) ⊕ G(5, T).

Example of Generalized Eigenspaces

Deﬁne T ∈ 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. for all z1, z2, z3 ∈ C. Thus The corresponding eigenspaces are G(0, T) = {(z , z , 0): z , z ∈ C}. easily seen to be 1 2 1 2

E(0, T) = {(z1, 0, 0): z1 ∈ C} and

E(5, T) = {(0, 0, z3): z3 ∈ C}. Note that this operator T does not have enough eigenvectors to span its domain C3. Thus

G(5, T) = {(0, 0, z3): z3 ∈ C}.

Now C3 = G(0, T) ⊕ G(5, T).

Example of Generalized Eigenspaces

Deﬁne T ∈ L(C3) by We have 3 T(z1, z2, z3) = (4z2, 0, 5z3). T (z1, z2, z3) = (0, 0, 125z3)

Example of Generalized Eigenspaces

Deﬁne T ∈ 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. for all z1, z2, z3 ∈ C. Thus The corresponding eigenspaces are G(0, T) = {(z , z , 0): z , z ∈ C}. easily seen to be 1 2 1 2 E(0, T) = {(z , 0, 0): z ∈ C} 1 1 We have and 3 (T−5I) (z1, z2, z3) = (−125z1+300z2, −125z2, 0). E(5, T) = {(0, 0, z ): z ∈ C}. 3 3 Thus Note that this operator T does not G(5, T) = {(0, 0, z ): z ∈ C}. have enough eigenvectors to span its 3 3 domain C3. Now C3 = G(0, T) ⊕ G(5, T). k n n 0 = a1(T − λ1I) (T − λ2I) ··· (T − λmI) v1 n n = a1(T − λ2I) ··· (T − λmI) w n n = a1(λ1 − λ2) ··· (λ1 − λm) w. The equation above implies that a1 = 0. In a similar fashion, aj = 0 for each j. Thus v1,..., vm is linearly independent.

Hence Tw = λ1w. Thus (T − λI)w = (λ1 − λ)w for every λ ∈ F. Hence n n (T − λI) w = (λ1 − λ) w for every λ ∈ F, where n = dim V. Apply the operator k n n (T − λ1I) (T − λ2I) ··· (T − λmI) to both sides, getting Proof Suppose a1,..., am ∈ F are such that 0 = a1v1 + ··· + amvm. Let k be the largest nonnegative integer k such that (T − λ1I) v1 6= 0. Let k w = (T − λ1I) v1. k+1 Thus (T − λ1I)w = (T − λ1I) v1 = 0.