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Solution to Linear Algebra Hoffman & Kunze Chapter 6.3


Exercise 6.3.1

Solution: The minimal polynomial for the identity operator is x1. It annihilates the identity operator and the monic zero degree polynomial p(x)=1 does not, so it must be the minimal polynomial. The minimal polynomial for the zero operator is x. It is a monic polynomial that annihilates the zero operator and again the monic zero degree polynomial p(x)=1 does not, so it must be the minimal polynomial.


Exercise 6.3.2

Solution: The characteristic polynomial is
|x0c1xb01xa|=|x0c10x2axb01xa|=1|xc1x2axb|=x3ax2bxc.Now for any r,sF
A2+rA+s=[0cac0bc+ba1ab+a2]+[00rcr0rb0rra]+[s000s000s]=[scac+rcrb+sc+ba+br1a+rb+a2+ra+s]0.Thus f(A)0 for all fF[x] such that deg(F)=2. Thus the minimum polynomial cannot have degree two, it must therefore have degree three. Since it divides x3ax2bxc it must equal x3ax2bxc.


Exercise 6.3.3

Solution: The characteristic polynomial equals
|x11001x+10022x21111x|=|x111x+1||x211x|,=x2(x22x+1)=x2(x1)2.Here we used (5-20) page 158. The minimum polynomial is clearly not linear, thus the minimal polynomial is one of x2(x1)2, x2(x1), x(x1)2 or x(x1). We will plug A in to the first three and show it is not zero. It will follow that the minimum polynomial must be x2(x1)2.
A2=[0000000033322221]AI=[0100120022111111]and
(AI)2=[1200230011000000]Thus
A2(AI)=[0000000011111111]0A(AI)2=[1100110000000000]0and
A(AI)=[1100110011111122]0.Thus the minimal polynomial must be x2(x1)2.


Exercise 6.3.4

Solution: Not diagonalizable, because for characteristic value c=0 the matrix AcI=A and A is row equivalent to
[1100001000010000]which has rank three. So the null space has dimension one. So if W is the null space for AcI then W has dimension one, which is less than the power of x in the characteristic polynomial. So by Theorem 2, page 187, A is not diagonalizable.


Exercise 6.3.5

Solution: Tk=0 the only characteristic value is zero. We know the minimal polynomial divides this so the minimal polynomial is of the form tr for some 1rn. Thus by Theorem 3, page 193, the characteristic polynomial’s only root is zero, and the characteristic polynomial has degree n. So the characteristic polynomial equals tn. By Theorem 4 (Caley-Hamilton) Tn=0.


Exercise 6.3.6

Solution: If A2=0 and A0 then the minimal polynomial is x or x2. So any A0 such that A2=0 has minimal polynomial x2. E.g.
A=[000100000].


Exercise 6.3.7

Solution: 1,x,x2,,xn is a basis.
10x1x22xxnnxn1The matrix for D is therefore
[0100000200000300000n00000]Suppose A is a matrix such that aij=0 except when j=i+1. Then A2 has aij=0 except when j=i+2. A3 has aij=0 except when j=i+3. Etc., where finally An+1=0. Thus if aij0 j=i+1 then Ak0 for k<n+1 and An+1=0. Thus the minimum polynomial divides xn+1 and cannot be xk for k<n+1. Thus the minimum polynomial is xn+1.


Exercise 6.3.8

Solution: P can be given in the standard basis by left multiplication by A=[1000]. Since P is given by left multiplication by a matrix, P is clearly linear. Since A is diagonal, the characteristic values are the diagonal values. Thus the characteristic values of A are 0 and 1. The characteristic polynomial is a degree two monic polynomial for which both 0 and 1 are roots. Therefore the characteristic polynomial is x(x1). If the characteristic polynomial is a product of distinct linear terms then it must equal the minimal polynomial. Thus the minimal polynomial is also x(x1).


Exercise 6.3.9

Solution: Suppose A is n×n. Claim: |xIA|=xn+trace(A)xn1+. Proof by induction: case n=2. A=[abcd]. |xIA|=x2(a+d)x+(adbc). The trace of A is a+d so we have established the claim for the case n=2. Suppose true for up to n1. Let r=a22+a33++ann. Then
|xa11a12a1na21xa22a2nan1an2xann|Now expanding by minors using the first column, and using induction, we get that this equals
(xa11)(xn1rxn2+)a21(polynomial of degree n2)+a31(polynomial of degree n2)+=xn+(r+a11)xn1+polynomial of degree at most n2=xntr(A)xn1+Now if f(x)=(xc1)d1(xck)dk then the coefficient of xn1 is c1d1+ckdk so it must be that c1d1+ckdk=tr(A).


Exercise 6.3.10

Solution: If we represent a n×n matrix as a column vector by stacking the columns of the matrix on top of each other, with the first column on the top, then the transformation T is represented in the standard basis by the matrix
M=[AA00A].And since
f(M)=[f(A)f(A)00f(A)]it is evident that f(M)=0 f(A)=0.


Exercise 6.3.11

Solution: We have
|InABxIn|= |InABxIn||InA0In|= |In0BxInBA|= |xInBA|Similarly,|InABxIn|= |InABxIn||In0Bx1In|= |Inx1ABA0xIn|= xn|Inx1AB|=|xInAB|.So AB and BA have the same characteristic polynomials.

But the minimum polynomials need not be equal. To see this let A=[0010] and B=[1000]. Then AB=[0010] and BA=[0000] so the minimal polynomial of BA is x and the minimal polynomial of AB is clearly not x (it is in fact x2).

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
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