Exercises 5D

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1 Suppose π is a finite-dimensional complex vector space and π β β(π).
(a) Prove that if $π^4$ = πΌ, then π is diagonalizable.
(b) Prove that if $π^4$ = π, then π is diagonalizable.
(c) Give an example of an operator π β β(π 2 ) such that $π^4$ = $π^2$and π is not diagonalizable.
2 Suppose π β β(π) has a diagonal matrix π΄ with respect to some basis of π.
Prove that if π β π
, then π appears on the diagonal of π΄ precisely dim πΈ(π, π) times.
3 Suppose π is finite-dimensional and π β β(π).
Prove that if the operator π is diagonalizable, then π = null π β range π.
4 Suppose π is finite-dimensional and π β β(π). Prove that the following are equivalent. (a) π = null π β range π. (b) π = null π + range π. (c) null π β© range π = {0}.
5 Suppose π is a finite-dimensional complex vector space and π β β(π). Prove that π is diagonalizable if and only if π = null(π β ππΌ) β range(π β ππΌ) for every π β π.
6 Suppose π β β(π
5 ) and dim πΈ(8, π) = 4. Prove that π β 2πΌ or π β 6πΌ is invertible.
7 Suppose π β β(π) is invertible. Prove that πΈ(π, π) = πΈ( 1 π , π β1) for every π β π
with π β 0.
8 Suppose π is finite-dimensional and π β β(π). Let π1 , β¦, ππ denote the distinct nonzero eigenvalues of π. Prove that dim πΈ(π1 , π) + β― + dim πΈ(ππ, π) β€ dim range π.
9 Suppose π
, π β β(π
3 ) each have 2, 6, 7 as eigenvalues. Prove that there exists an invertible operator π β β(π
3 ) such that π
= πβ1ππ.
10 Find π
, π β β(π
4 ) such that π
and π each have 2, 6, 7 as eigenvalues, π
and π have no other eigenvalues, and there does not exist an invertible operator π β β(π
4 ) such that π
= πβ1ππ.
11 Find π β β(π 3 ) such that 6 and 7 are eigenvalues of π and such that π does not have a diagonal matrix with respect to any basis of π 3 .
12 Suppose π β β(π 3 ) is such that 6 and 7 are eigenvalues of π. Furthermore, suppose π does not have a diagonal matrix with respect to any basis of π 3 . Prove that there exists (π§1 , π§2 , π§3 ) β π3 such that π(π§1 , π§2 , π§3 ) = (6 + 8π§1 , 7 + 8π§2 , 13 + 8π§3 ).
13 Suppose π΄ is a diagonal matrix with distinct entries on the diagonal and π΅ is a matrix of the same size as π΄. Show that π΄π΅ = π΅π΄ if and only if π΅ is a diagonal matrix.
14 (a) Give an example of a finite-dimensional complex vector space and an operator π on that vector space such that π 2 is diagonalizable but π is not diagonalizable. (b) Suppose π
= π, π is a positive integer, and π β β(π) is invertible. Prove that π is diagonalizable if and only if π π is diagonalizable.
15 Suppose π is a finite-dimensional complex vector space, π β β(π), and π is the minimal polynomial of π. Prove that the following are equivalent. (a) π is diagonalizable. (b) There does not exist π β π such that π is a polynomial multiple of (π§ β π)2 . (c) π and its derivative π β² have no zeros in common. (d) The greatest common divisor of π and π β² is the constant polynomial 1. The greatest common divisor of π and π β² is the monic polynomial π of largest degree such that π and π β² are both polynomial multiples of π. The Euclidean algorithm for polynomials (look it up) can quickly determine the greatest common divisor of two polynomials, without requiring any information about the zeros of the polynomials. Thus the equivalence of (a) and (d) above shows that we can determine whether π is diagonalizable without knowing anything about the zeros of π.
16 Suppose that π β β(π) is diagonalizable. Let π1 , β¦, ππ denote the distinct eigenvalues of π. Prove that a subspace π of π is invariant under π if and only if there exist subspaces π1 , β¦, ππ of π such that ππ β πΈ(ππ , π) for each π and π = π1 β β― β ππ.
17 Suppose π is finite-dimensional. Prove that β(π) has a basis consisting of diagonalizable operators.
18 Suppose that π β β(π) is diagonalizable and π is a subspace of π that is invariant under π. Prove that the quotient operator π/π is a diagonalizable operator on π/π. The quotient operator π/π was defined in Exercise 38 in Section 5A. Linear Algebra Done Right, fourth edition, by Sheldon Axler 174 Chapter 5 Eigenvalues and Eigenvectors
19 Prove or give a counterexample: If π β β(π) and there exists a subspace π of π that is invariant under π such that π|π and π/π are both diagonalizable, then π is diagonalizable. See Exercise 13 in Section 5C for an analogous statement about uppertriangular matrices.
20 Suppose π is finite-dimensional and π β β(π).
Prove that π is diagonalizable if and only if the dual operator π β² is diagonalizable.
21 The Fibonacci sequence πΉ0 , πΉ1 , πΉ2 , β¦ is defined by πΉ0 = 0, πΉ1 = 1, and πΉπ = πΉπβ 2 + πΉπβ 1 for π β₯ 2. Define π β β(π 2 ) by π(π₯, π¦) = (π¦, π₯ + π¦).
(a) Show that π π (0, 1) = (πΉπ , πΉπ+1 ) for each nonnegative integer π.
(b) Find the eigenvalues of π.
(c) Find a basis of π 2 consisting of eigenvectors of π.
(d) Use the solution to (c) to compute π π (0, 1).
Conclude that πΉπ = 1 β5 [(1 + β5 2 ) π β ( 1 β β5 2 ) π ] for each nonnegative integer π. (e) Use (d) to conclude that if π is a nonnegative integer, then the Fibonacci number πΉπ is the integer that is closest to 1 β5 ( 1 + β5 2 ) π . Each πΉπ is a nonnegative integer, even though the right side of the formula in (d) does not look like an integer. The number 1 + β5 2 is called the golden ratio.
22 Suppose π β β(π) and π΄ is an π-by-π matrix that is the matrix of π with respect to some basis of π. Prove that if |π΄π,π | > π β π = 1 π β π |π΄π,π | for each π β {1, β¦, π}, then π is invertible. This exercise states that if the diagonal entries of the matrix of π are large compared to the nondiagonal entries, then π is invertible. 23 Suppose the definition of the Gershgorin disks is changed so that the radius of the π th disk is the sum of the absolute values of the entries in column (instead of row) π of π΄, excluding the diagonal entry. Show that the Gershgorin disk theorem (5.67) still holds with this changed definition.