《高等代数》第二学期期末考试卷参考答案(2)

√λ1,λn}=diag{µ1,µ2,···,µn}U,B=C.

C,A=C2,BB=CC,C 1B=CB 1,

D,C 1=DDT,C 1B=DDTBDD 1,

B,C 1BBBB2=C2,

C 1B(C 1B)T=C 1BBT(C 1)T=C 1BBC 1=C 1CCC 1=E,

1,C 1B

P,C 1B=P 1EP=E,B=C.()(11)C,A=C2.Bα1,α2,···,αnn

2

0=(B2 C2)αi=µ2Ciαi Cµi=(µiE+C)(µiE C)αi.(µiE C)αi=0,Cαi=µiαi,i=1,2,···,n

C 1B

C 1B

1,

B

µ1,µ2,···,µn

A=B2=C2,B2 C2=0,µi>0,µiE+C

C(α1,α2,···,αn)=(α1,α2,···,αn)

µ1

µ2

...µn

5

=B(α1,α2,···,αn),

《高等代数》第二学期期末考试卷参考答案

λ1,

B

√

λn}Q 1

B

0,

B

(√

Qdiag{

2)

√

λ2,···,

A

Q,

Q 1AQ=diag{λ1,λ2,···,λn},

B=

√λ1,λn}=

T

QTCQ.B2=C2,PTB2P=diag{λ1,λ2,···,λn}=QTC2Q, U=PQ , diag{ λ1 ,λ2,···,λn}=

11 1 =Udiag{λ1,λ2,···,λn}UT,U=E,P=Q,B=C.

121

1 1 , =E.

21

(6

)

b,(

A,B,ABλAB

λ=ab.)(1011A,BABPP 1AP=D1,P 1BP=D2,D1,D2

P 1ABP=D1D2,D1D2

A

a,B

)

A,B,

AB=BA,

A,B

A

B

AB

A,B

()(11)ABλVλ.A,B,ABAB=(AB)T=BTAT=BA,A(AB)=A(BA)=(AB)A.A,BRnAAB

VλAAVλA|Vλa,Va.ABVaBBVab,Vb.0=α∈Vb Va Vλ,AB=λα,Aα=aα,Bα=bα,

λα=ABα=A(bα)=abα,α=0,λ=ab.

(1)A,BP,P 1AP,P 1BPA,B

P,QP 1AP,Q 1BQP,Q

AB=BA,A,BP.AB=BA

(

BX=bX,

λ

2)A,B,ABABX=A(bX)=abX,a,b.

AB=BA,λ=ab.

A,B

λ

X,

AX=aX,

(10

)

A

n

V={X∈Rn|XTAX=0}.

(1)

(2)

V

Rn

V(

Rn)(11

r(A)=r<n

)

dimV.

6

《高等代数》第二学期期末考试卷参考答案

1λ1,Y3TDY3

λ1,

1

>0,

QY1,QY2∈V,

(2)(1)A

R

QY3= QY1+QY2

E0 r

00

V, Er

T

λ1,0,0,···,0),

Y1TDY1=0,Y2TDY2=0,

V 0 ,0

C,

A=±CTC,

V={X∈Rn|CX=0}={CX=0

},

r(C)=r(CTC)=r(A)=r,

D=

Ep

dimV=n r(C)=n r.

,p,q

(

)

A,

P,

PAP=D,

T

Eq

A

XTAX=XT(PT) 1DP 1X=YTDY,

2222

XTAX=0 YTDY=0 y1+···+yp yp+1 ··· yr=0.

(I)

p=r

22

XTAX=0 y1+···+yp=0 y1=···=yp=0 X=P(0,···,0,yp+1,···,yn)T.

V={P(0,···,0,bp+1,···,bn)T|bi∈R(i=p+1,···,n)},

V

(II)

Rnq=r

dimV=n p=n r.

V={P(0,···,0,bq+1,···,bn)T|bi∈R(i=q+1,···,n)},

V(III)

RndimV=n q=n r.p>0q>0VRn

α=P(ε1+εr),β=P(ε1 εr)∈V,

TT (α+β)A(α+β)=(2ε1)

Ep

Eq

(2ε1)=4=0.

(1)VR(2)VRn()

n

p=rq=r.

dimV=n r.

X,Y∈V,XTAY=0

7