1.3 Diagonalizable Operators
5
Theorem 1.3.1 Let V be a n-dimensional vector space over K and λ ∈ K . Let
T : V −→ V be a linear operator on V and I n be the identity matrix of order n.
Then the following are equivalents:
(i) λ is an eigenvalue of T ;
(ii) the operator T − λI n is not invertible;
(iii) det(T − λI n ) = 0.
Exercise 1.3.1 Prove the equivalences given of Theorem 1.3.1.
From Theorem 1.2.2, there exists a bijection between the set of all linear transformations T : V −→ W and the set of all matrices m × n with entries in K . In
particular, if V = W and if B V is an basis of V , by considering M = [T ]
B V
B V
, it follows that T − λI n is invertible if and only if M − λI n is invertible. Thus, if A is
a square matrix of order n over the field K , an eigenvalue of A in K is a scalar
λ ∈ K such that det(A − λI n ) = 0. The polynomial p(λ) = det(A − λI n ) is called
characteristic polynomial of A.
Definition 1.3.2 Let V be a finite-dimensional vector space and let T : V −→ V
be a linear operator on V . Then T is said to be diagonalizable if there exists a basis
of eigenvectors for V .
The following theorem characterizes diagonalizable operators.
Theorem 1.3.2 Let T : V −→ V be a linear operator on V , where V is n
-dimensional. Assume that λ 1 , λ 2 , . . . , λ r are distinct eigenvalues of T and W i =
ker(T − λ i I n ). Then the following conditions are equivalent:
(i) T is diagonalizable;
(ii) the characteristic polynomial for T is p(λ) = (λ − λ 1 )
d 1 · . . . · (λ − λ r )
d r ,
where dim(W i ) = d i , for all i = 1, 2, . . . , r;
(iii) dim(W 1 ) + dim(W 2 ) + · · · + dim(W r ) = dim(V ).
Exercise 1.3.2 Show Theorem 1.3.2.
1.4 Tensor Products
Let us now consider that U, V, W are vector spaces over K . A function B : U ×
V −→ W is called bilinear if for satisfies two conditions:
(1) for each u ∈ U , the function B u : V −→ W defined as B u (v) = B(u, v) is a
linear transformation;
(2) for each v ∈ V , the function B v : U −→ W defined as B v (u) = B(u, v) is a
linear transformation.
Keeping this concept in mind, we can define tensor product of vector spaces.
5
Theorem 1.3.1 Let V be a n-dimensional vector space over K and λ ∈ K . Let
T : V −→ V be a linear operator on V and I n be the identity matrix of order n.
Then the following are equivalents:
(i) λ is an eigenvalue of T ;
(ii) the operator T − λI n is not invertible;
(iii) det(T − λI n ) = 0.
Exercise 1.3.1 Prove the equivalences given of Theorem 1.3.1.
From Theorem 1.2.2, there exists a bijection between the set of all linear transformations T : V −→ W and the set of all matrices m × n with entries in K . In
particular, if V = W and if B V is an basis of V , by considering M = [T ]
B V
B V
, it follows that T − λI n is invertible if and only if M − λI n is invertible. Thus, if A is
a square matrix of order n over the field K , an eigenvalue of A in K is a scalar
λ ∈ K such that det(A − λI n ) = 0. The polynomial p(λ) = det(A − λI n ) is called
characteristic polynomial of A.
Definition 1.3.2 Let V be a finite-dimensional vector space and let T : V −→ V
be a linear operator on V . Then T is said to be diagonalizable if there exists a basis
of eigenvectors for V .
The following theorem characterizes diagonalizable operators.
Theorem 1.3.2 Let T : V −→ V be a linear operator on V , where V is n
-dimensional. Assume that λ 1 , λ 2 , . . . , λ r are distinct eigenvalues of T and W i =
ker(T − λ i I n ). Then the following conditions are equivalent:
(i) T is diagonalizable;
(ii) the characteristic polynomial for T is p(λ) = (λ − λ 1 )
d 1 · . . . · (λ − λ r )
d r ,
where dim(W i ) = d i , for all i = 1, 2, . . . , r;
(iii) dim(W 1 ) + dim(W 2 ) + · · · + dim(W r ) = dim(V ).
Exercise 1.3.2 Show Theorem 1.3.2.
1.4 Tensor Products
Let us now consider that U, V, W are vector spaces over K . A function B : U ×
V −→ W is called bilinear if for satisfies two conditions:
(1) for each u ∈ U , the function B u : V −→ W defined as B u (v) = B(u, v) is a
linear transformation;
(2) for each v ∈ V , the function B v : U −→ W defined as B v (u) = B(u, v) is a
linear transformation.
Keeping this concept in mind, we can define tensor product of vector spaces.