Chapter 5 Inner Product Spaces
5-1 Inner Products and Norms
Inner product <x,y>: V is a vector space over F, and <x,y> has the following characteristics:
(a) <ax+bz,y>=a<x,y>+b<z,y>, (b) <x,y>=<y,x>*, (c) <x,x> is positive if x≠0.
Eg. Let u=(u1,u2,u3) and v=(v1,v2,v3). Determine which of the following are inner products on R3.
(i) <u,v>=u12v12+u22v22+u32v32, (ii) <u,v>=2u1v1+u2v2+3u3v3, (iii) <u,v>=u1v1-u2v2+u3v3. [2003交大電
信所]
(Sol.) Only (ii) is the inner product. (i) does not fulfill (a), (iii) does not fulfill (c) if u1v1 +u3v3<u2v2.
Eg. The following calculations fulfill the definition of the inner product:
n
(a) x=(a1,a2,…,an), y=(b1,b2,…,bn), <x,y>=  ai bi , (b) A,B  Mnxn(F), <A,B>=tr(B*A),
i 1
(c) f, g  V, <f,g>=  f x g ( x)dx .
a
0
In Matlab language, we can use the following instructions to obtain the inner product and the outer
product of two vectors:
>>A=[4,-1,3]; B=[-2,5,2]; C=dot(A,B); D=cross(A,B)
A=
4
-1
3
B=
-2
5
2
-14
18
C=
-7
D=
-17
Norm ||．||: A real-valued function on V and satisfies the following conditions for all x, y  V and c  F.
(a) ||cx||=|c|．||x||, (b) ||x||≧0, and ||x||=0 if x=0, (c) ||x+y||≦||x||+||y||.
Eg. Show that ||-x||=||x||. [1993 台大電研]
(Proof) ||-x||=|-1|．||x||=||x||
Eg. The following calculations fulfill the definition of the norm:
(a) V=Mmxn(F), ||A||= max |Aij|, (b) V=C[0,a >0] , ||f||= max f (t ) , (c)
t[ 0 ,a ]
i, j
Euclidean norm of a vector: x  V, ||x||=
V=Fn,
n
2
x   a i 
 i 1

1
2
x, x
Theorem (a)| <x,y>|≦||x||．||y|| (Cauchy-Schwartz Inequality), (b) ||x+y||2 + ||x-y||2 =2||x||2 +2||y||2.
(Remark) By definition of the Euclidean norm.
In Matlab language, we can use the following instructions to obtain the Euclidean norm of a vector:
>> x=[3 4 5]; c=norm(x)
c=
7.0711
Orthogonal vectors x and y: x⊥y  <x,y>=0.
Orthonormal vectors x and y: ||x||=||y||=1 and <x,y>=0 if x≠y.
Eg. Show that if {v1, v2, v3, …, vn} is an orthogonal set of non-zero vectors in an inner product
space, then v1, v2, v3, …, and vn are linearly independent. [2015 中央電研固態組、生醫電子組]
(Proof) Let y=
be a linear combination of v1, v2, v3, …, and vn.
Since {v1, v2, v3, …, vn} is an orthogonal set of non-zero vectors in an inner product space, we have <
vi, vj>=0 if j≠i but < vi, vi>≠0 in case of j=i.
Consider an inner product <y,vj>=<
, vj >=
< vi, vj >=ai< vi, vi>
Set y=0 and then we have ai=0 because < vi, vi>≠0 for each i. It implies that v1, v2, v3, …, and vn are
linearly independent.
m
Theorem V is an inner product space and S={x1,…, xm}. If y=  a i xi , and then (a) aj=<y,xj>/<xj,xj>
i 1
for all j if S is orthogonal, and (b) aj=<y,xj> for all j if S is orthonormal.
m
m
 y, x j 
i 1
i 1
 xj, xj 
(Proof) (a)  y , x j     ai xi , x j    ai  xi , x j   a j  x j , x j  a j 
Eg. (8,-7)=8(1,0)-7(0,1) where 8=<(8,-7),(1,0)> and -7=<(8,-7),(0,1)>
Theorem A linear operator T: V→W, let β={x1,…,xm} and orthogonal set β’={y1,…,ym} are the bases
 T ( x j ), y i 
of V and W, respectively. Then the ij-entry Tij=
. If β’ is an orthonormal basis, then
 yi , yi 
Tij=  T ( x j ), y i  .
m
m
m
k 1
k 1
k 1
(Proof) ∵ T ( x j )   Tkj y k , ∴  T ( x j ), y i     Tkj y k , y i    Tkj  y k , y i   Tij  y i , y i 
 Tij 
 T ( x j ), y i 
 yi , yi 
Eg. For a linear operator T(a1,a2)=(a1-a2,a1,2a1+a2), let β={(1,2),(2,3)} be a basis for R2, and
β’={(1,0,0),(0,1,0),(0,0,1)} be an
orthonormal
basis
for R3, then we have
T(1,2)=(-1,1,4)=-1(1,0,0)+1(0,1,0)+4(0,0,1) and T(2,3)=(-1,2,7)=-1(1,0,0)+2(0,1,0)+7(0,0,1)
 T 
'

 1
  1
 4
 1
2 
7 
,
where
T11=-1=<T(1,2),(1,0,0)>,
T21=1=<T(1,2),(0,1,0)>,
T31=4=<T(1,2),(0,0,1)>, T12=-1<T(2,3),(1,0,0>, T22=2=<T(2,3),(0,1,0)>, and T32=7=<T(2,3),(0,0,1)>
5-2 Gram-Schmidt Orthogonalization Process
Gram-Schmidt Orthogonalization Process: Let S={y1,y2,y3,…,yn} be a linearly independent subset
 y 2 , x1 
of V and S’={x1,x2,x3,…,xn} be an orthogonal subset of V. Set x1=y1, x2= y2 x1 , x3=
 x1 , x1 
y3-
k 1  y , x 
 y3 , x1 
 y3 , x2 
k
j
 x j . Then Span(S)=Span(S’).
 x1  x2 , …, and xk=yk- 
 x1 , x1 
 x2 , x2 
j 1  x j , x j 
Eg. For V=R3, β={(1,1,0),(2,0,1),(2,2,1)}, find an orthogonal basis for V by the Gram-Schmidt
orthogonalization process. [2007 台科大電研]
(Sol.) Let x1  (1,1,0), x2  (2,0,1) 
x3  (2,2,1) 
 (2,0,1), (1,1,0) 
 (1,1,0)  (2,0,1)  22 (1,1,0)  (1,1,1)
 (1,1,0), (1,1,0) 
 (2,2,1), (1,1,0) 
 (2,2,1), (1,1,1) 
 (1,1,0) 
 (1,1,1)  ( 13 , 13 , 23 ) .
 (1,1,0), (1,1,0) 
 (1,1,1), (1,1,1) 
Then β’={x1,x2,x3} is an orthogonal basis.
Eg. Let the vector space P2 have the inner product <p,q>=
1

1
p( x)q( x)dx . Apply the
Gram-Schmidt process to transform the standard basis S={1,x,x2} into an orthonormal basis.
[2005 北科大電腦通訊所]
1
 y 2 , x1 
 x 1dx  1 =x-0=x,
(Sol.) S={1,x,x2}={ y1, y2, y3}. Let x1= y1=1, x2= y2 x1 =x- 11
 x1 , x1 
1  1dx

1
1
1


x 2  1dx
x 2  xdx
 y3 , x1 
 y3 , x2 
1
1



1

1
x3= y3 x1  x2 =x2- 1
1 - 1
 x =x2- -0= x2- .
3
3
 x1 , x1 
 x2 , x2 
1  1dx
x  xdx
1
1
1
45 2 1
1
3
∴ Orthogonal basis: {1,x,x2- }  Orthonormal basis: {
,
x,
(x - )}
3
3
8
2 2
Sperp: S⊥={x  V: <x,y>=0 for all y  S} is an orthogonal set of S.
Eg. For V=C3, S=Span{(1,0,i),(1,2,1)}, compute S⊥.
(Sol.) Suppose S⊥=Span{(a,b,c)},
 a, b, c , 1,0, i   0  a  ci

1 i
⊥

 a, b, c , 1,2,1  0  b   1  i c , S = Span{( i, 2 ,1)}

2

Eg. Show that (a) {x1,x2,…,xk} is an ordered basis for W (subspace of V) and {x1,x2,…,xk,xk+1,…,xn}
is an ordered basis V, then {xk+1,…,xn} is an order basis of for W⊥.
(b) W is a subspace of V, then dim(V)=dim(W)+dim(W⊥). [台大電研]
n
(Proof) (a) x  V  x    x, xi  xi
i 1
If x W  , then  x, xi  0 for 1  i  k . Therefore, x 
n
  x, x
i  k 1
(b) According to (a), dim( V )=n=k+(n-k)=dim( W )+dim( W  )
i
 xi  Spanx k 1 ,  , x n 
Eg. Show that (a) W1  W2   W1  W2


and (b) W1  W2   W1  W2  . [1990, 1999 台大電

研]
(Proof) W1  W2   W1  W2   W1  W1  W2  W2  V



W1 and W2  W1  W2 ,  W1  W2   W1 and W2  W1  W2   W1  W2







x  W1  W2 , then y  W1 and z  W2  y , x  z, x  0
 cy  dz W1  W2 ,  cy  dz, x  c  y, x  d  z, x  0
 x  W1  W2   W1  W2  (W1  W2 )  , ∴ W1  W2   W1  W2





On the other hand, U 1  U 2   U 1  U 2  U 1  U 2  U 1  U 2



Set U 1  W1 , U 2  W2  W1  W2  W1  W2 








 


5-3 Various Matrices
Adjoint matrix: A* is the complex conjugate transpose of A. Note: A*≠adj(A) and (AB)*=B*A*.
1  i 2
1  i 3i 
Eg. For A= 
, then A*= 
.

4 
  3i 4
 2
Eg. For T: C2→C2 by T(a1,a2)=(2ia1+3a2,a1-a2), if β is the standard ordered basis, find T*(a1,a2).
[1998 台大電研]
(Sol.) T 1,0  2i,1  2i  (1,0)  1 (0,1) and T 0,1  3,1  3  (1,0)  (1)  (0,1)
2i 3 
 2i 1 
 T   
 T*   

 , ∴ T*(a1,a2)=( -2ia1+a2,3a1-a2).
 1  1
 3  1
 
Theorem For A  Mm×n(F), x  Fn, y  Fm, then <Ax,y>=<x,A*y>.
(Note:  T x , y  x, T *  y   for x, y  V , T : V  V )
Eg.
 1 1  x  u 
 x  y  u 
 x  1 2i  u 
 x  u  2iv 
 2i 3  y , v    2ix  3 y , v   xu  yu  2ixv  3 yv   y , 1 3  v    y ,  u  3v 

   

  
  
 
  

Theorem (a) A  Mm×n(F), Rank(A*A)=Rank(A). (b) A  Mm×n(F), if Rank(A)=n, then A*A is
invertible.
(Proof) (a) A*Ax=0  Ax=0
1. " ": Ax  0 implies A* Ax  0
2. "": 0  A* Ax, x
n
 Ax, A** x
m
 Ax, Ax m , ∴ Ax=0
(b) A  M mn F   A* A  M nn F 
Rank(A*A)=Rank(A)=n, ∴ A*A is invertible
Orthogonal matrix A: AAt=AtA=I. Unitary matrix A:. AA*=A*A=I.
1 / 2
Eg. A= 
1 / 2
 1/ 2  t  1/ 2 1/ 2 
t
t
, A =
 , AA =A A=I, ∴ A is orthogonal.
1/ 2 

1
/
2
1
/
2


i / 2
Eg. A= 
i / 2
 i / 2
 i / 2
 , A*= 
i/ 2 
 i/ 2
 i / 2
 , AA*=A*A=I, ∴ A is unitary.
 i / 2
Normal Matrix A: AA*=A*A.
i  i 
 i  i 
Eg. A= 
, A*= 

 , AA*=A*A, ∴ A is normal.
i i 
 i  i
Eg. T is a normal operator on V. Show that (a) ||T(x)||= ||T*(x)||, (b) T(x)=λx  T *  x    x , (c) λ1
and λ2 are distinct eigenvalues of T with corresponding eigenvector x1 and x2. Then x1 and x2 are
orthogonal. [2012 台大電研]
(Proof) (a) T x   T x , T x   T *T x , x  TT * x , x  T * x , T * x   T * x 
2
2
(b) Suppose T x   x for some x V . Let U  T  I ,
0  U x   U * x   (T *  I )x  , ∴ T *  x    x .
(c) Let T x1   1 x1 , T x2   2 x2 ,
1 x1 , x2  1 x1 , x2  T x1 , x2  x1 , T * x2   x1 , 2 x2  2 x1 , x2
 1  2 ,  x1 , x2  0 .
Self-adjoint (Hermitian) matrix A: A=A*.
2  5i 
 1
Eg. 
 is a self-adjoint matrix.
2  5i  3 
(Note: A self-adjoint matrix is also a normal matrix, but a normal matrix may not be self-adjoint)
Theorem If A is a self-adjoint matrix, all eigenvalues of A are real. [2000 台大電研]
(Proof) x  Ax  A* x  x for some x≠0,     λ is real.
Gramian matrix A:  B  Mm×n(F) such that A=BBt then A is called the Gramian matrix.
A is symmetric.

Theorem A is a Gramian matrix  
all eigenvalues of A are nonnegative.
(Proof) (1) At=(BBt)t=BBt=A, ∴ A is symmetric. (2) For some x  0, and Ax=λx,
then
Ax, x  x, x   x, x  BB t x, x  B t x, B t x  0,  x, x  0,    0
Eg. Which of the following matrices is Hermitian? Which is normal?
 i

2  i
 1
B 2
A

2  i  1 
 1
 2
1 
i
1 
 3 1 i i
0

2  C   i
0
 2  i  D  1  i 1 3 [交大電子所]

i 

  i
 1 2  i
3 1
0 
2 
(Sol.) A=A* and D=D*, ∴ A and D are both Hermitian matrices,
 1 i
1  i 
CC*=C*C, BB*= 
, B*B= 

 , ∴ C is normal but B is not normal.
 i 1
i 1 
5-4 Special Characteristics of Matrices
Ax, x >0 for all x≠0, then A is positive definite.
Positive definite: If
Eg. If A is a positive-definite matrix, then all the eigenvalues of A are positive.
(Proof) For some x≠0 and Ax=λx, positive definite:
Ax, x  x, x   x, x  0    0 .
Eg. Let A be a complex normal (or real symmetric) n×n matrix with eigenvalues: λ1, λ2, …, λn,
n
n
i 1
i 1
2
n
and then show that (a) tr(A)=   i and tr(A*A)=  i , (b) det(A)=  i . [1998 台大電研]
i 1
n
(Proof) Let A=QDQ-1 and D be diagonal, tr(A)=tr(QDQ-1)=tr(DQQ-1)=tr(D)=   i
i 1


 
n
det  A  det QDQ 1  det Q  det D  det Q 1  det Q  det D det Q 1  det D    i .
i 1
2
1
Eg. Let λi be the eigenvalues of A= 
- 1

0
1 -1 0 
3 4 2 
, 1≦i≦4, (a)
4 1 2 

2 21 
4

i
i 1
 ? (b)
4

i
?
i 1
( c ) I s A p o s i t i v e definite? [台大機研]
4
(So1.) ∵ A is a real symmetric matrix, ∴ (a)
 i  tr(A) = 2+3+1+1=7, (b)
i 1
4
  =det(A) = 10,
i
i 1
(c) Method 1:  λ＜0  A is not positive definite.
 2 1 -1 
 2 1


Method 2: det([2])=2＞0 and det( 
 )=5＞0, but det(  1 3 4  )=-38＜0  Not positive definite.
1
3


- 1 4 1 
A is unitarily equivalent to B:  P fulfills B=P*AP and P-1=P*.
Theorem (a) A is complex normal  A is unitary equivalent to a diagonal matrix. (b) A is real
symmetric  A is orthogonally equivalent to a real diagonal matrix.


(Proof) AA*  PDP * PDP *

*
 PDP * PD * P *  PDD * P *
A  PDP * , A* A  PDP *  PDP *   PD * DP * .  DD *  D * D,  A* A  AA*
*
Projection: V=W1⊕W2, x1  W1, and x2  W2, x=x1+x2. If T(x)=x1, then T is projection on W1. That is,
R(T)=W1={x: T(x)=x} and N(T)= W2. (Note: T2=T if T is a projection.)
Orthogonal Projection: If R(T)⊥=N(T) and N(T)⊥= R(T) for T: V→V be a projection.
Theorem T is a linear operator on V. Then T is an orthogonal projection.  T2=T=T*.
Eg. Let T: R2→R2 be a linear operator. For any z  R2, T(z)=p, where p is the projection of z on the
line x=y. Find [T]. [交大電信所]
1 
2  } and the standard basis of R2 is { 1  , 0  }.
(Sol.) The orthonormal basis of line x=y is { 
 0  1 
1 
   

2 
1 
1  
2  >= 1 , so the orthogonal projection of 1  onto the line x=y is
Inner product: <   ,
0 
2
0   1 
 
2

 1  1 
1 
2  =  2 = 1
1

 1  2
2

2   2 
Inner

1 
2 

1  1
0  + 2
 
0 
1 
 
1 
0  
2  >= 1 , so the orthogonal projection of 0  onto the line x=y is
product: <   ,
1 
2
1   1 
 
2

1  1 
1 
1
2  =  2  = 1 1  + 1 0  , ∴ [T]=  2
2  , and we have [T]2=[T]=[T]*
1
1
1   1  2 0  2 1 
 2
2 
2   2 
1 
 0  0 
Eg. Find the orthogonal projection of the vector v=  0  onto the subspace S=Span(  1 , 1  ).
 
   
  2
 1  1 
[交大電子所]

 

 0   0 
 0  0 
(Sol.) Transform the orthogonal basis {  1 , 1  } into the orthonormal one: {  1  ,  1  }.

   
2 
2
 1
 1 
 1  1 

2  
2 
1 
< 0 
 
  2


 0 
1 
0




1
, 
>=-√2, so the orthogonal projection of the vector v= 0
onto  1 is

 
 
2
 1

  2
 1 

2 


 0  0
-√2  1  =  1  .

2  
 1
 1

2 


1   0 
1 
0 






1
< 0 ,
>=-√2, so the orthogonal projection of the vector v= 0
onto 1  is
  
 
 
2
  2  1 
  2
1 

2 


 0  0
-√2  1  =  1 .

2  
 1   1

2 




 0 
 0   1   0   0  1 
1 
 1  1   0   0 














1
1
∵ 0 -(-√2 
)-(-√2
)= 0 - 1 -  1 = 0 , ∴ v=  0  = 0  +  1  +  1


 
       
2
2        
 1

 1    2 1  1 0 
  2
  2 0  1  1


2 
2 
1 
 0  0 


and { 0 }⊥S=Span(  1 , 1  )
 
   
0 
 1  1 
Rotation:  orthogonal basis β={x1,x2} for W, and  θ is real and fulfills T(x1)=x1cosθ+x2sinθ,
T(x2)=-x1sinθ+x2cosθ, and T(y)=y,  y  W⊥, where W is a 2-dimensional subspace. And then T is a
⊥
⊥
rotation of W about W , where W is the axis of rotation.
Reflection: T(x)=-x,  x  W and T(y)=y,  y  W⊥. And then T is a reflection of W about W⊥, where
W is a 1-dimensional subspace.
Eg. T: R2→R2 by T(a,b) = (-a,b). Let W=Span({e1}) and W⊥=Span({e2}). It is reflection of R2 about
W⊥ (the y-axis).
Theorem T is an orthogonal operator on a 2-dimensional real inner product space. If det(T)=1,
then T is a rotation. If det(T)=-1, then T is a reflection.
cos  sin  
(Proof) By definition, T   
 is a rotation, ∴ det(T) = 1.
 sin  cos 
Choose   z1 , z 2 ,... be an orthogonal basis for V and z1  W,
T
  1 0   0
0 1 
 

         det(T) = -1.


  0

 0   0 1
Theorem The composition of a reflection and a rotation is a reflection.
(Proof) det(T1T2)=det(T1)det(T2)= -1
normal on V over F  C
Spectral Theorem If T is 
, and λ1, …, λk are distinct eigenvalues of
self - adjoint on V over F  R
T. Let Wi be the eigenspace of T corresponding to λi, and Ti be an orthogonal projection on Wi.
(Ti(x)=xi  Wi). Then (a) V=W1⊕W2⊕W3…⊕Wk, (b) Wi⊥=  W j (direct sum), (c) TiTj=δijTi, (d)
ji
I=T1+T2+…+Tk, (e) T=λ1T1+λ2T2+…+λkTk.
 0, i  j
(Proof) (c) Let x  0 and T j ( x)  x j , then Ti Tj ( x)  Ti ( x j )  
 i j Ti ( x)  TiT j   i j Ti
 xi , i  j
(d) x  Ix  x1  ...  xk  T1 ( x)  T2 ( x)  ...  Tk ( x)  (T1  T2  ...  Tk )( x) , ∴ I  T1  T2  ...  Tk .
(e) For x V , then x  x1  ...  xk , where xi  Wi ,
T ( x)  T ( x1 )  ...  T ( xk )  1 x1  ...  k xk  1T1 ( x)  ...  k Tk ( x)  (1T1  ...  k Tk )( x) ,
∴ T  1T1  ...  k Tk
Theorem (a) F=C, T is a unitary operator  T* is normal and |λ|=1 for all eigenvalues. (b)
T=-T*  each λ is pure imaginary. (c) T is a projection  each λ is either 0 or 1.
5-5 Bilinear Forms
Bilinear form, H: If (a) H(ax1+bx2,y)=aH(x1,y)+bH(x2,y) and (b) H(x,ay1+by2)=aH(x,y1)+bH(x,y2) for
x1, y1, x2, y2, x, y  V, a, b  F.
Theorem For each H  B(Fn),  ! A  Mn×n(F) fulfills H(x,y)=xtAy for all x, y  Fn.
a  b 
a  b 
Eg. Define H : R 2  R 2  R by H (  1 ,  1 )  2a1b1  3a1b2  4a 2 b1  a 2 b2 for all  1 ,  1   R 2 ,
a2  b2 
a2  b2 
a1 
b1 
2 3 
t
then let x   , y   , A  
  H ( x, y )  x Ay  [a1 a 2 ]
a
b
4

1


 2
 2
2 3 
4  1


Symmetric bilinear form: H(x,y)=H(y,x) for all x, y  V.
Quadratic form: K: V→F if  H  B(V) fulfills K(x)=H(x,x)=xtAx.
Eg. Let f(t1,t2,t3)= 2t1  t 2
2
 f (t1 , t 2 , t 3 )  t1 t 2
2
t3 
0
2 3
t1 


 6t1t 2  4t 2 t 3  K ( t 2  ) , and then A= 3  1  2
t 3 
0  2 0 
0  t 1 
2 3
3  1  2 t  is a quadratic form.

  2
0  2 0  t 3 
 b1 
b  .
 2
Eg. There are two bases (x1,y1) and (x2,y2) in R2. If the equations of the same ellipse represented
by two distinct bases are described as follows, respectively: 2x12-4x1y1+5y12-36=0 and
x22+6y22-36=0. Please find the transformation matrix between these two bases. [2004 台大電研]
x 
2
(Sol.) Let K  1   2 x1 2  5 y1 2  4 x1 y1 , A  
 2
 y1 
 2  a1 
a 
2
    0   1   a  , a  0,

4  a2 
1
 a2 
a 
1
 2

b 
 4  2  b 
1  1  A  1I    1   
a
2
 2    2  
 2
, det 


  2 5        1  6,   1, 6
5


b 
1
2  6  A  2 I    1   
  1   0   1   b  , b  0

 2
b2   2  1 b2 
b2 


Q


2
5
1
5


x
∴  1  
y 
 1 

1 
5  fulfills Q 1 AQ  Q t AQ  1 0 for    1 2, 1  1   .

0 6
 
 
2 


 5 1 5  2 

5 
2
5
1
5
1 
5    x 2   K  x2   x 2  6y 2  2 x1 2  4 x1 y1  5 y1 2  36  0  x 2 2  6 y 2 2  36  0
2
2
y 
2   y 2 
 2

5 
Eg. [t1,t2,t3]t R3, 3t12+3t22+3t32-2t1t3+2√2 (t1+t3)+1=0. Find a basis β for R3 such that the above
equation is simplified.
t1 
 3 0  1
2
2
2


(Sol.) Let K t 2   3t1  3t 2  3t 3  2t1t 3 , then A   0 3 0  . It is found that an orthogonal
t 3 
 1 0 3 

0

matrix Q  1
0

1 
 0
3 0 0 
1
  1 
2 
  1   1   


t
0  fulfills Q AQ  0 4 0 for    1 ,
0 ,
0 .
  2   2  
1 
 0

0 0 2

  1
  1 
 

2
1

2
0
1
2

 t1  0

∴ t 2   1
t 3  0

1
2
0
1
2
1 
2   s1 
0    s 2   K ( x)  3s12  4s2 2  2s3 2
1 
 s3 

2 

 3t1  3t 2  3t 3  2t1t3  2 2 (t1  t 3 )  1  0  3s1  4s2  2s3  4s3  1  0 .
2
2
2
2
2
2