Exercise 6.4.1
Let be the linear operator on , the matrix of which in the standard ordered basis is
(a) Prove that the only subspaces of invariant under are and the zero subspace.
(b) If is the linear operator on , the matrix of which in the standard ordered basis is , show that has -dimensional invariant subspaces.
Solution: (a) The characteristic polynomial equals This is a parabola opening upwards with vertex , so it has no real roots. If had an invariant subspace it would have to be -dimensional and would therefore have a characteristic value.
(b) Over the characteristic polynomial factors into two linears. Therefore over , has two characteristic values and therefore has at least one characteristic vector. The subspace generated by a characteristic vector is a -dimensional subspace.
Exercise 6.4.2
Let be an invariant subspace for . Prove that the minimal polynomial for the restriction operator divides the minimal polynomial for , without referring to matrices.
Solution: The minimum polynomial of divides any polynomial where . If is the minimum polynomial for then . Therefore, . So since by definition for . Therefore, . Therefore the minimum polynomial for divides .
Exercise 6.4.3
Let be a characteristic value of and let be the space of characteristic vectors associated with the characteristic value . What is the restriction operator ?
Solution: For the transformation . Thus is diagonalizable with single characteristic value . In other words under which it is represented by the matrix
where there are dim ’s on the diagonal.
Exercise 6.4.4
Let
Is similar over the field of real numbers to a triangular matrix? If so, find such a triangular matrix.
Solution: We haveAnd . Thus the minimal polynomial and the only characteristic value is . We now follow the constructive proof of Theorem 5. , a characteristic vector of is . We need such that . satisfies . Now need such that . satisfies . Thus with respect to the basis the transformation corresponding to is .
Exercise 6.4.5
Every matrix such that is similar to a diagonal matrix.
Solution: satisfies the polynomial . Therefore the minimum polynomial of is either , or . In all three cases the minimum polynomial factors into distinct linears. Therefore, by Theorem 6 is diagonalizable.
Exercise 6.4.6
Let be a diagonalizable linear opeartor on the -dimensional vector space , and let be a subspace which is invariant under . Prove that the restriction operator is diagonalizable.
Solution: By the lemma on page 80 the minimum polynomial for divides the minimum polynomial for . Now diagonalizable implies (by Theorem 6) that the minimum polynomial for factors into distinct linears. Since the minimum polynomial for divides it, it must also factor into distinct linears. Thus by Theorem 6 again is diagonalizable.
Exercise 6.4.7
Let be a linear operator on a finite-dimensional vector space over the field of complex numbers. Prove that is diagonalizable if and only if is annihilated by some polynomial over which has distinct roots.
Solution: If is diagonalizable then its minimum polynomial is a product of distinct linear factors, and the minimal polynomial annihilates . This proves “”. Now suppose is annihilated by a polynomial over with distinct roots. Since the base field is this polynomial factors completely into distinct linear factors. Since the minimum polynomial divides this polynomial the minimum polynomial factors completely into distinct linear factors. Thus by Theorem 6, is diagonalizable.
Exercise 6.4.8
Let be a linear operator on . If every subspace of is invariant under , then is a scalar multiple of the identity operator.
Solution: Let be a basis. The subspace generated by is invariant thus is a multiple of . Thus is a characteristic vector since for some . Suppose such that . Then Since the subspace generated by is invariant under . Thus and since coefficients of linear combinations of basis vectors are unique. Thus . Thus is times the identity operator.
Exercise 6.4.9
Let be the indefinite inntegral operator
on the space of continuous functions on the interval . Is the space of polynomial functions invariant under ? Ths space of differentiable functions? The space of functions which vanish at ?
Solution: The integral from to of a polynomial is again a polynomial, so the space of polynomial functions is invariant under . The integral from to of a differntiable function is differentiable, so the space of differentiable functions is invariant under . Now let . Then vanishes at but which does not vanish at . So the space of functions which vanish at is not invariant under .
Exercise 6.4.10
Let be a matrix with real entries. Prove that, if is not similar over to a triangular matrix, then is similar over to a diagonal matrix.
Solution: If is not similar to a tirangular matrix then the minimum polynomial of must be of the form where has no real roots. The roots of are then two non-real complex conjugates and . Thus over the minimum polynomial factors as . Since is real, , and constintute three distinct numbers. Thus by Theorem 6 is diagonalizable over .
Exercise 6.4.11
True or false? If the triangular matrix is similar to a diagonal matrix, then is already diagonal.
Solution: False. Let . Then is triangular and not diagonal. The characteristic polynomial is which has distinct roots, so the minimum polynomial is . Thus by Theorem 6, is diagonalizable.
Exercise 6.4.12
Let be a linear operator on a finite-dimensional vector space over an algebraically closed field . Let be a polynomial over . Prove that is a characteristic value of if and only if , where is a characteristic value of .
Solution: Since is algebraically closed, the corollary at the bottom of page 203 implies there’s a basis under which is represented by a triangular matrix . where if and the , are the characteristic values of . Now where if and for all . Thus the characteristic values of are exactly the ’s where is a characteristic value of . Since is a matrix representative of in the same basis, we conclude the same thing about the transformation .
Exercise 6.4.13
Let be the space of matrices over . Let be a fixed matrix over . Let and be the linear operators on defined by
(a) True or false? If is a diagonalizable (over ), then is diagonalizable.
(b) True or false? If is diagonalizable, then is diagonalizable.
Solution: (a) True by Exercise 10 Section 6.3 page 198 since by Theorem 6 diagonalizability depends entirely on the minimum polynomial.
(b) True. Find a basis so that is diagonal
Let . Then . The matrices such that and all other entries equal zero form a basis for . For any , where Thus only when and . Thus . So is a characteristic value and is a characteristic vector for all . Thus has a basis of characteristic vectors for . Thus is diagonalizable.