University:
California State University, Northridge
Course:
MATH 462 | Advanced Linear Algebra
Academic year:

2024
Views:

392

Pages:

1
Author:

calb10e8i

Math 462 1. Let V be a finite dimensional vector space over F and S, T ∈ L(V ). Moreover, let (v1 , . . . , vn ) be a basis such that both S and T have upper triangular matrics in this basis. Show that ST also has an upper triangular matrix in this basis. Solution: Since S and T have upper triangular matrices in the given bases we have that for any k ∈ {1, . . . , n} Uk = span(v1 , . . . , vk ) is an invariant subspace for both S and T . Since T Uk ∈ uK and SUK ∈ Uk it follows that ST Uk ∈ UK , so uk is an invariant subspace of ST for all k, and ST has an upper triangular matrix in the given basis. 2. Repeat the previous problem with diagonal instead of upper triangular matrices for S and T . Solution: Since both S and T have diagonal matrices in the given basis there are numbers τ1 , . . . , τn and σ1 , . . . , σn such that Svj = σj vj , T v j = τj vj , ∀j ∈ {1, . . . , n} But then ST vj = Sτj vj = τj Svj = τj σj vj , ∀j ∈ {1, . . . , n} Therefore ST has a diagonal matrix. 3. Give an example of an operator whose matrix with respect to some basis contains only zeros on the main diagonal, but the operator is invertible. Solution: Consider the operator T : R2 → R2 which maps T : (x, y) 7→ (y, x). This operator is clearly invertible, infact T −1 = T and its matrix in the standard basis is µ ¶ 0 1 1 0 4. Give an example of an operator whose matrix with respect to some basis contains only non-zero numbers on the main diagonal, but the operator is not invertible. Solution: Consider the operator T : R2 → R2 which maps T : (x, y) 7→ (x + y, x + y). This operator is not invertible, since every vector of the form (a, −a) is in the kernel of T , and ker T 6= {0}. Its matrix in the standard basis is µ ¶ 1 1 1 1 5. Suppose S, T ∈ L(V ) and S is invetible. Prove that if p is a polynomial, then p(ST S −1 ) = Sp(T )S −1 . Solution: Observe that for any k ∈ {0, . . . n} ST k S −1 = ST (SS −1 )T · · · (SS −1 )T S −1 = (ST S −1 )k . The coefficients of p are numbers which commute with any linear operator. The result follows immediately.