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Solution to Linear Algebra Hoffman & Kunze Chapter 6.4


Exercise 6.4.1

Let T be the linear operator on R2, the matrix of which in the standard ordered basis is
A=[1122].(a) Prove that the only subspaces of R2 invariant under T are R2 and the zero subspace.

(b) If U is the linear operator on C2, the matrix of which in the standard ordered basis is A, show that U has 1-dimensional invariant subspaces.

Solution: (a) The characteristic polynomial equals |x112x2|=(x1)(x2)+2=x23x+4. This is a parabola opening upwards with vertex (3/2,7/4), so it has no real roots. If T had an invariant subspace it would have to be 1-dimensional and T would therefore have a characteristic value.

(b) Over C the characteristic polynomial factors into two linears. Therefore over C, T has two characteristic values and therefore has at least one characteristic vector. The subspace generated by a characteristic vector is a 1-dimensional subspace.


Exercise 6.4.2

Let W be an invariant subspace for T. Prove that the minimal polynomial for the restriction operator TW divides the minimal polynomial for T, without referring to matrices.

Solution: The minimum polynomial of TW divides any polynomial f(t) where f(TW)=0. If f is the minimum polynomial for T then F(T)v=0 vV. Therefore, f(T)w=0 wW. So f(TW)w=0 wW since by definition f(TW)w=f(T)w for wW. Therefore, f(TW)=0. Therefore the minimum polynomial for TW divides f.


Exercise 6.4.3

Let c be a characteristic value of T and let W be the space of characteristic vectors associated with the characteristic value c. What is the restriction operator TW?

Solution: For wW the transformation T(w)=cw. Thus TW is diagonalizable with single characteristic value c. In other words under which it is represented by the matrix
[cc00c]where there are dim(W) c’s on the diagonal.


Exercise 6.4.4

Let
A=[010222232].Is A similar over the field of real numbers to a triangular matrix? If so, find such a triangular matrix.

Solution: We haveA2=[222000222].And A3=0. Thus the minimal polynomial x3 and the only characteristic value is 0. We now follow the constructive proof of Theorem 5. W={0}, α1 a characteristic vector of A is [101]. We need α2 such that Aα2{α1}. α2=[112] satisfies Aα2=α1. Now need α3 such that Aα3{α1,α2}. α3=[001] satisfies Aα3=2α1+2α2. Thus with respect to the basis {α1,α2,α3} the transformation corresponding to A is [012002000].


Exercise 6.4.5

Every matrix A such that A2=A is similar to a diagonal matrix.

Solution: A2=A A satisfies the polynomial x2x=x(x1). Therefore the minimum polynomial of A is either x, x1 or x(x1). In all three cases the minimum polynomial factors into distinct linears. Therefore, by Theorem 6 A is diagonalizable.


Exercise 6.4.6

Let T be a diagonalizable linear opeartor on the n-dimensional vector space V, and let W be a subspace which is invariant under T. Prove that the restriction operator TW is diagonalizable.

Solution: By the lemma on page 80 the minimum polynomial for TW divides the minimum polynomial for T. Now T diagonalizable implies (by Theorem 6) that the minimum polynomial for T factors into distinct linears. Since the minimum polynomial for TW divides it, it must also factor into distinct linears. Thus by Theorem 6 again TW is diagonalizable.


Exercise 6.4.7

Let T be a linear operator on a finite-dimensional vector space over the field of complex numbers. Prove that T is diagonalizable if and only if T is annihilated by some polynomial over C which has distinct roots.

Solution: If T is diagonalizable then its minimum polynomial is a product of distinct linear factors, and the minimal polynomial annihilates T. This proves “”. Now suppose T is annihilated by a polynomial over C with distinct roots. Since the base field is C 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, T is diagonalizable.


Exercise 6.4.8

Let T be a linear operator on V. If every subspace of V is invariant under T, then T is a scalar multiple of the identity operator.

Solution: Let {αi} be a basis. The subspace generated by αi is invariant thus Tαi is a multiple of αi. Thus αi is a characteristic vector since Tαi=ciαi for some ci. Suppose i,j such that cicj. Then T(αi+αj)=Tαi+Tαj=ciαi+cjαj=c(αi+αj). Since the subspace generated by {αi,αj} is invariant under T. Thus ci=c and cj=c since coefficients of linear combinations of basis vectors are unique. Thus Tαi=cαi i. Thus T is c times the identity operator.


Exercise 6.4.9

Let T be the indefinite inntegral operator
(Tf)(x)=0xf(t)dton the space of continuous functions on the interval [0,1]. Is the space of polynomial functions invariant under T? Ths space of differentiable functions? The space of functions which vanish at x=1/2?

Solution: The integral from 0 to x of a polynomial is again a polynomial, so the space of polynomial functions is invariant under T. The integral from 0 to x of a differntiable function is differentiable, so the space of differentiable functions is invariant under T. Now let f(x)=x1/2. Then f vanishes at 1/2 but 0xf(t)dt=12x212x which does not vanish at x=1/2. So the space of functions which vanish at x=1/2 is not invariant under T.


Exercise 6.4.10

Let A be a 3×3 matrix with real entries. Prove that, if A is not similar over R to a triangular matrix, then A is similar over C to a diagonal matrix.

Solution: If A is not similar to a tirangular matrix then the minimum polynomial of A must be of the form (xc)(x2+ax+b) where x2+ax+b has no real roots. The roots of x2+ax+b are then two non-real complex conjugates z and z¯. Thus over C the minimum polynomial factors as (xc)(xz)(xz¯). Since c is real, c, z and z¯ constintute three distinct numbers. Thus by Theorem 6 A is diagonalizable over C.


Exercise 6.4.11

True or false? If the triangular matrix A is similar to a diagonal matrix, then A is already diagonal.

Solution: False. Let A=[1100]. Then A is triangular and not diagonal. The characteristic polynomial is x(x1) which has distinct roots, so the minimum polynomial is x(x1). Thus by Theorem 6, A is diagonalizable.


Exercise 6.4.12

Let T be a linear operator on a finite-dimensional vector space over an algebraically closed field F. Let f be a polynomial over F. Prove that c is a characteristic value of f(T) if and only if c=f(t), where t is a characteristic value of T.

Solution: Since F is algebraically closed, the corollary at the bottom of page 203 implies there’s a basis under which T is represented by a triangular matrix A. A=[aij] where aij=0 if i>j and the aii, i=1,,n are the characteristic values of T. Now f(A)=[bij] where bij=0 if i>j and bii=f(aii) for all i=1,,n. Thus the characteristic values of f(A) are exactly the f(c)’s where c is a characteristic value of A. Since f(A) is a matrix representative of f(T) in the same basis, we conclude the same thing about the transformation T.


Exercise 6.4.13

Let V be the space of n×n matrices over F. Let A be a fixed n×n matrix over F. Let T and U be the linear operators on V defined by
T(B)=ABU(B)=ABBA(a) True or false? If A is a diagonalizable (over F), then T is diagonalizable.

(b) True or false? If A is diagonalizable, then U 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 A is diagonal
A=[c1c200cn].Let B=[bij]. Then U(B)=ABBA. The n2 matrices Bij such that bij0 and all other entries equal zero form a basis for V. For any Bij, U(Bij)=ABijBijA=[dij] where dij=cibijcjbij=(cicj)bij. Thus dij0 only when i=i and j=j. Thus U(Bij)=(cicj)Bij. So cicj is a characteristic value and Bij is a characteristic vector for all i,j. Thus V has a basis of characteristic vectors for U. Thus U is diagonalizable.

Linearity

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