Rewiev Problems:

———————————-
1. Let S = ft + 3; t 4g and T = ft + 4; 2t + 1g be two ordered bases for
P1 :
(a) Find the coordinate vector [p(t)]T for p(t) = t + 14:
We will …nd a; b such that
p(t) = t + 14 = a(t + 4) + b( 2t + 1)
t + 14 = (a 2b)t + (4a + b)
Then
a 2b = 1
+ (2)4a + b = 14
————————
9a = 1 + 28 = 27 : a = 3 and from the second eqn. b = 14 4a = 2
a 3
The coordinate vector [p(t)]T = =
b 2
(b) Find the transition matrix PS T from T basis to S basis.
1 1
Take the S basis S = ft + 3; t 4g : t + 3 ! ; t 4!
3 4
1 2
Take the T basis T = ft + 4; 2t + 1g : t + 4 ! ; 2t + 1 !
4 1
1 1 1 2 1 1 1 2
Construct A = 3R1 + R2 ! R2
3 4 4 1 ! 0 1 1 7
1 1 1 2 1 0 0 9
R2 ! R2 R2 + R1 ! R1
! 0 1 1 7 ! 0 1 1 7
= [I2 jPS T ]
0 9
Then PS T =
1 7
(c) Using PS T …nd the coordinate vector [p(t)]S :
For p(t) = t + 14:
0 9 3 18
[p(t)]S = PS T [p(t)]T = =
1 7 2 17
Then we have t + 14 = 18(t + 3) 17( t 4) = t 54 + 68 = t + 14
— — — — —8 —2— — —3— 2 ——— 3 —2— —3—9– 82 3 2 3 2 39
< 1 1 2 = < 3 2 6 =
2. Let S = 4 1 5 ; 4 1 5 ; 4 1 5 and T = 4 0 5 ; 4 3 5 ; 4 2 5
: ; : ;
1 1 1 2 1 4
be two ordered bases for R3 : Find the transition matrix PS T from T basis
to S basis. 2 3
1 1 2 3 2 6
Construct A = 4 1 1 1 0 3 2 5,
21 1 1 2 1 4 3
1 0 0 1 1 1
row echelon form: 4 0 1 0 0 1 1 5 = [I3 jPS T ]
0 0 1 1 1 2

1
 2 3
1 1 1
Then PS T = 4 0 1 1 5 is the transition matrix from T to S basis.
1 1 2
——————————————————
2 3
1 0 2 1 6
6 1 1 1 0 1 7
3. Let A = 6 4 1 1 5 0
7:
3 5
2 1 7 0 4
(a) Find a basis for the row space of A:
(b) Find the rank of A:
(c) Find a basis for the null space of A:
(d) Find the nullity
2 of A: 3
1 0 2 1 6
6 1 1 1 0 1 7
Transform A = 6 4 1 1 5 0
7 to a REF or RREF.
3 5
2 2 1 3 7 0 4 2 3
1 0 2 1 6 1 0 2 1 6
6 1 1 1 0 1 7 R1 + R2 ! R2 6 0 1 3 1 5 7
A=6 4 1 1 5 0
7 R1 + R3 ! R3 6 7
3 5 4 0 1 3 1 9 5
2R1 + R4 ! R4
2 1 7 0 4 0 1 3 2 16
2 3 ! 2 3
1 0 2 1 6 1 0 2 1 6
R2 + R3 ! R3 6 6 0 1 3 1 5 7 1
7
6 0 1 3 1 5 7
R ! R3 6 7
R2 + R4 ! R4 4 0 0 0 2 14 5 2 3 !4 0 0 0 1 7 5
! 0 0 0 33 21 0 0 0 3 21
2
1 0 2 1 6
6 0 1 3 1 5 7
2R3 + R4 ! R4 6 7 = C in REF
!4 0 0 0 1 7 5
0 0 0 0 0
(a) Find a basis for the row space of A:
A basis for the row space of A is T = f 1 0 2 1 6 ; 0 1 3 1 5 ; 0 0 0 1 7 g
(the nonzero rows of C)
(b) Find the rank of A:
rank(A) = 3 (the dimension of the row space of A)
(c) Find a basis for the null space of A:
Solve Ax = 0( or equivalently Cx = 0):
Cx = 0 : x1 + 2x3 + x4 + 6x5 = 0
x2 + 3x3 + x4 + 5x5 = 0
x4 + 7x5 = 0
Here x1 ; x2 ; x4 are basic and x3 ; x5 are free variables.
Let x3 = s; x5 = r:
Then x4 = 7r
x2 = 3s ( 7r) 5r = 3s + 2r
x1 = 2s ( 7r) 6r = 2s + r
The solutions of Ax = 0 ( The null space of A)

2
 82 3 9
>
> 2s + r >
>
>
> >
<66 3s + 2r 7
7
>
=
Null space of A is 6 6 s 7 j s; r 2 R
> 7 >
>
> 4 7r 5 >
>
>
: >
;
2 3 2 3 r 2 3
2s + r 1 2
6 3s + 2r 7 6 2 7 6 3 7
6 7 6 7 6 7
6 s 7 = r6 0 7 + s6 1 7
6 7 6 7 6 7
4 7r 5 4 7 5 4 0 5
r 1 802 3 2 39
>
> 1 2 >>
>
> 7>
<6 7 6
6 2 7 6 3 7=
>
A basis for the Null space of A is 6 6 0 7;6 1 7
7 6 7
>4 7 5 4 0 5>
> >
>
> >
>
: ;
1 0
(d) Find the nullity of A:
N ulluty(A) = 2 (Dimension of the Null space of A)
————————————-
4. (a) Is the expression (x; y) = 2x1 y1 x1 y2 x2 y1 + 5x2 y2
an inner product on R2 ?
Check the conditions:
(1): Is (x; x) 0:
(x; x) = 2x1 x1 x1 x2 x2 x1 + 5x2 x2 = 2x21 2x1 x2 + 5x22
= x21 2x1 x2 + x22 + x21 + 4x22
= (x1 x2 )2 + x21 + 4x22 0
0
If (x; x) = 0 is x =
0
If (x; x) = 0 then (x1 x2 )2 + x21 + 4x22 = 0 : x1 x2 = 0; x1 = 0; x2 = 0
0
So x =
0
(2) Is (x; y) = (y; x)?
(x; y) = 2x1 y1 x1 y2 x2 y1 + 5x2 y2
(y; x) = 2y1 x1 y1 x2 y2 x1 + 5y2 x1
Then (x; y) = (y; x)
(3) Is (x + y; z) = (x; z) + (y; z)?
(x + y; z) = 2(x1 + y1 )z1 (x1 + y1 )z2 (x2 + y2 )z1 + 5(x2 + y2 )z2
= (2x1 z1 x1 z2 x2 z1 + 5x2 z2 ) + (2y1 z1 y1 z2 y2 z1 + 5y2 z2 )
(x; z) + (y; z)
(4) Is (cx; y) = c(x; y)?
(cx; y) = 2cx1 y1 cx1 y2 cx2 y1 + 5cx2 y2
= c(2x1 y1 x1 y2 x2 y1 + 5x2 y2 ) = c(x; y)
All the conditions hold so (x; y) = 2x1 y1 x1 y2 x2 y1 + 5x2 y2
is an inner product on R2 :
(b) Is the expression (x; y) = x1 y1 + x1 y2 + x1 y3
an inner product on R3 ?

3
Check the conditions:
(1) Is (x; x) 0?
(x; x) = x12x1 + x31 x2 + x1 x3 is not always 0:
1
Take x = 4 1 5 then (x; x) = 1:1 + 1( 1) + 1( 2) = 2 < 0
2
Then the expression (x; y) = x1 y1 + x1 y2 + x1 y3 is not an inner product.
(c) (b) Is the expression (u; v) = u1 v1 u2 v1 + u1 v2
an inner product on R2 ?
Exercise.
—————— 8— 2—— 3 —2— —3— 2 ——— 39————————
>
> 1 2 0 > >
<6 7=
0 7 6 7 6
7 ; 6 1 7 ; 6 0 7 be a basis for the subspace W
5. Let B = 6 4 1 5 4 2 5 4 1 5>
>
> >
: ;
0 0 1
of the inner product space R4 with the standard inner product:
(a) Find an orthogonal 8 basis
2 for 3 W:2 3 2 39
> 1
> 2 0 >>
<6 6 1 7 6 0 7=
6 0 7
7 6 7 6 7
Let B = fu1 ; u2 ; u3 g = 4 ; ;
> 1 5 4 2 5 4 1 5>
> >
: ;
0 0 1
Use Gram-Schmidt 2 method3 :
1
6 0 7
Take v1 = u1 = 6 4 1 5
7

0
v2 = u2 (v (v
1 ;u2 )
v 1
21 ;v1 )3 2 3
1 2
6 0 7 6 1 7
(v1 ; u2 ) = (6 7 6 7
4 1 5 ; 4 2 5) = 2 + 0 + 2 + 0 = 4
0 0
2 3 2 3
1 1
6 0 7 6 0 7
(v1 ; v1 ) = (6 7 6 7
4 1 5 ; 4 1 5) = 1 + 0 + 1 + 0 = 2
0 02
2 3 3 2 3 2 3
2 1 2 2 0
6 1 7 46 0 7 6 1 0 7 6 1 7
Then v2 = 6 7 6
4 2 5 24 1 5=4 2 2 5=4 0 5
7 6 7 6 7

0 0 0 0 0
(v1 ;u3 ) (v2 ;u3 )
v3 = u3 (v1 ;v1 ) v1 (v2 ;v2 ) v2
2 3 2 3
1 0
6 0 7 6 0 7
(v1 ; u3 ) = (6 7 6 7
4 1 5 ; 4 1 5) = 0 + 0 + 1 + 0 = 1
0 1
(v1 ; v1 ) = 2

4
 2 3 2 3
0 0
6 1 7 6 0 7
(v2 ; u3 ) = (6
4
7;6
5 4 1
7) = 0 + 0 + 0 + 0 = 0
5
0
0 1
2 3 2 3
0 0
6 17 6 1 7
(v2 ; v2 ) = (6
4
7;6
5 4 0
7) = +1 + 0 + 0 = 1
5
0
0 0
2 3 2 3 2 3 2 1 3
0 1 0 2
6 0 7 6
1 6 0 7
7 6 7 6 0 7
0 6 1 7
v3 = 6 7
4 1 5 2 4 1 5
6
1 4 0 5=4 1 5
7
2
1 0 0 1
2 1
3 2 3
2 1
6 0 7 6 0 7
6
Take v3 = 2 4 1 7 6
5=4 1 5
7
2
1 2
2 3 2 3 2 3
1 0 1
6 0 7 6 1 7 6 0 7
Then the set fv1 ; v2 ; v3 g = f6 7 6 7 6
4 1 5;4 0 5;4 1
7gis orthogonal.
5
0 0 2
Verify: 2 3 2 3
1 0
6 0 7 6 1 7
(v1 ; v2 ) = (6 7 6 7
4 1 5 ; 4 0 5) = 0
0 0
2 3 2 3
1 1
6 0 7 6 0 7
(v1 ; v3 ) = (6 7 6 7
4 1 5 ; 4 1 5) = 1 + 0 + 1 + 0 = 0
2 0 3 2 2 3
0 1
6 1 7 6 0 7
(v2 ; v3 ) = (6 7 6 7
4 0 5 ; 4 1 5) = 0 + 0 + 0 + 0 = 0
0 2
(b) Find an orthonormal basis for W:
2 3 2 p1 3
1 2
6
1 6 0 7
7 6 0 7
1
De…ne w1 = jjv1 jj v1 = 2 4 p = 6 7
1 5 4 p12 5
0 0
2 3 2 3
0 0
6 1 7 6 1 7
w2 = jjv12 jj v2 = 11 6 7 6 7
4 0 5=4 0 5
0 0

5
 2 3 2 p1 3
1 6
6 0 7 6 6 0 77
1 1 6
w3 = jjv3 jj v3 = p1+0+1+4 4 7 =6 7
1 5 4 p16 5
2 p2
6
Then the set fw1 ; w2 ; w3 g is an orthonormal basis for W:
—————————————–
6. Consider the subspace W = f a + c a + b c b + c ; j a; b; c 2 Rg
of the inner product space R4 with the standard inner product.
(a) Find a basis for W:
a + c a + b c b + c = a 1 1 0 0 +b 0 1 0 1 +c 1 0 1 1
Then W = spanf 1 1 0 0 ; 0 1 0 1 ; 1 0 1 1 g
To …nd a2basis for W3: 2 3
1 0 1 1 0 0
6 1 1 0 7 6 0 1 0 7
Let A = 6 7 6
4 0 0 1 5, row echelon form: 4 0 0 1 5
7

0 1 1 0 0 0
All columns contains leading 1s. Then the vectors
f 1 1 0 0 ; 0 1 0 1 ; 1 0 1 1 g
are linearly independent and a basis for W is
T = fu1 ; u2 ; u3 g = f 1 1 0 0 ; 0 1 0 1 ; 1 0 1 1 g
(b) Find an orthogonal basis for W:
Exercise.
(c) Find an orthonormal basis for W:
Exercise.
————————————————-