M20B : Linear Algebra
University of the West Indies Department of Mathematics Solution to Problem Paper #04 Lecturer: Sam McDaniel
Solution 1 (a) If (a2 ; a; 1) is in the span of B = f(1; 2; 3); (1; 1; 1); (0; 1; 2)g then there exists constants x; y; z such that (a2 ; a; 1) = x(1; 2; 3) + y(1; 1; 1) + z(0; 1; 2) In this case, x + y = a2 2x + y + z = a 3x + y + 2z = 1 x + R2 ! R2 R3 ! R3
2R1 3R1 x +
R3 ! R3
2R2
y = a2 y + z = a 2a2 2y + 2z = 1 3a2 a2 a 2a2 3a2 2(a
y = y + z = 0 = 1
For consistency we require 1
3a2
2(a 2a2 ) a2 2a + 1 (a 1)2 a
= = = =
0 0 1 1
2a2 )
M20B Solution to Problem Paper 04
2
(b) For M; N 2 V (M + N ) 0
L(M + N ) = In general, (M + N )
1
6= M
1
1
+N
if M + N is nonsingular if M + N is singular 1
and so L is not linear.
An Example: For 1 2 3 4
M = (M + N ) and M
1
+N
1
1
1
2 2 4 5
= 1 2 1 = 2 =
1 0 1 1
;N = =
4 3
2 1
1 2
5 4
+
2 2 1 0 1 1
2 2 1 1
Solution 2 (a) Since L(0) = 0; 0 2 Im L and so Im L is non empty. Let y 1 ; y 2 2 Im L: There exists x1 ; x2 2 V such that L(x1 ) = y 1 and L(x2 ) = y 2 Now, k1 y 1 + k2 y 2 = k1 L(x1 ) + k2 L(x2 ) = L(k1 x1 ) + L(k2 x2 ) = L(k1 x1 + k2 x2 ) So k1 y 1 + k2 y 2 2 Im L and hence Im L is a subspace of W: (b) Let L : V ! W be a linear transformation. (i) Prove that dim V = dim(ker L) + dim(Im L) Let dim V = n and dim(ker L) = k: If k = n then ker L = V and so L(x) = 0 for all x 2 V: Hence Im L = f0g and dim(Im L) = 0. So the result holds when k = n: Now suppose 1 < k < n and let fx1 ; x2 ; ; xk g be basis for ker L: We may extend this basis to form a basis for V; fx1 ; x2 ; ; xk ; xk+1 ; Let y 2 Im L: There exists x 2 V such that L(x) = y and since x 2 V , x may be written as x=
n X i=1
i xi
xn g:
M20B Solution to Problem Paper 04
3
Therefore y = L(x) = L( 1 x1 + 2 x2 + + k xk + k+1 xk+1 + + n xn ) = 1 L(x1 ) + 2 L(x2 ) + + k L(xk ) + k+1 L(xk+1 ) + = k+1 L(xk+1 ) + + n L(xn ) Hence L(xk+1 ); L(xk+2 );
+
; L(xn ) span Im L:
We now show that this set is linearly independent. Suppose k+1 L(xk+1 )
+
k+2 L(xk+2 )
+
+
n L(xn )
=0
then L(
k+1 xk+1
+
k+2 xk+2
and so k+1 xk+1 + k+2 xk+2 + may be expressed as k+1 xk+1
+
k+2 xk+2
+
+
+
+ n xn
+ n xn
=
n xn )
=0
2 ker L and therefore it
1 x1
+
2 x2
+
+
k xk
and so 1 x1
+
2 x2
+
+
Since fx1 ; x2 ; 1
=
k xk
k+1 xk+1
=
=
k
n xn
=0
xn g is a basis for V;
; xk ; xk+1 ; 2
k+2 xk+2
=
k+1
=
k+2
=
n
=0
So the vectors L(xk+1 ); L(xk+2 );
; L(xn )
are linearly independent. We conclude that dim(Im L) = n
k
and so dim V = dim(ker L) + dim(Im L) (ii) If L is one-one, ker L = f0g or dim(ker L) = 0 and dim V = dim(Im L): Conversely, if dim V = dim(Im L) then from the above result, dim(ker L) = 0 or ker L = f0g implying that L is one to one.
n L(xn )
M20B Solution to Problem Paper 04
(c) To …nd ker L we solve 0 @
4 2 2 3 1 1
R1 $ R3
0 @
0
R2 ! R2 + 2R1 @ R3 ! R3 + 4R1 0
R3 ! 5R3
6R2
@
4
10 1 0 1 2 x 0 1 A@ y A = @ 0 A 2 z 0 1 1 2 3 4 2 1 1 0 5 0 6 1 1 0 5 0 0
10 2 1 A@ 2 10 2 5 A@ 6 10 2 5 A@ 0
1 0 1 x 0 y A = @ 0 A z 0 1 0 1 x 0 y A = @ 0 A z 0 1 0 1 x 0 A @ y 0 A = z 0
Put z = t to get y = t and x = t 80 9 1 t < = @ A t ker L = :t2R : ; t Since dim(ker L) = 1, L is not 1 : 1:
Since dim V = dim(ker L) + dim(Im L); dim(Im L) = 2
Solution 3 (a) Since L
L
and L
x y 1 0 0 1
0
= @
0
= @
0
= @
1 x + 3y x y A 2x + 5y 1 0 1 0 1 1 0 A @ A @ 1 =1 0 +1 1 2 0 0 1 0 1 0 3 1 A @ 1 = 3 0 A 1@ 5 0
Therefore the matrix of transformation is A=
1 3
1 2 1 5
1
0
0 A + 2@ 0 1 1 0 0 1 A + 5@ 0
1 A
1 0 0 A 1
M20B Solution to Problem Paper 04
(b) Since L[p(t)] = tp(t)
L(t) = t t = t2 = 1(t2 ) + 0(t) + 0(1) L(1) = t 1 = t = 0(t2 ) + 1(t) + 0(1) Therefore the matrix of transformation is A=
1 0 0 0 1 0
Solution 4 (a) We choose y 1 = (1; 1; 0; 1): We next choose y 2 such that y 1 y 2 = 0 and y 2 = (1; 1; 0; 1) + (2; 0; 0; 1) y 2 y 1 = 0 = (1; 1; 0; 1) (1; 1; 0; 1) + (2; 0; 0; 1) (1; 1; 0; 1) 0 = 3 +1 1 = 3 1 (1; 1; 0; 1) + (2; 0; 0; 1) y2 = 3 5 1 0 4 = ( ; ; ; ) 3 3 3 3 or take y 2 = (5; 1; 0; 4) We choose y 3 = y 1 + y 2 + x3 such that y 3 is orthogonal to both y 1 and y 2 y 3 y 1 = 0 = (1; 1; 0; 1) (1; 1; 0; 1) + (5; 1; 0; 4) (1; 1; 0; 1) +(0; 0; 1; 0) (1; 1; 0; 1) So 0 = 3 + 0 + 0 = 0 y 3 y 2 = 0 = (1; 1; 0; 1) (5; 1; 0; 4) + (5; 1; 0; 4) (5; 1; 0; 4) +(0; 0; 1; 0) (5; 1; 0; 4) 0 = 0 + 42 + 0 = 0 Therefore y 3 = 0(1; 1; 0; 1) + 0(5; 1; 0; 4) + (0; 0; 1; 0) = (0; 0; 1; 0)
5
M20B Solution to Problem Paper 04
6
An orthogonal basis for this space is B = f(1; 1; 0; 1); (5; 1; 0; 4); (0; 0:1; 0)g An orthonormal basis is 1 1 5 1 4 1 T = f( p ; p ; 0; p ); ( p ; p ; 0; p ); (0; 0; 1; 0)g 3 3 3 42 42 42 (b) Since x and y are orthogonal x y = 0 1 1 p ; 0; p 2 2
1 a a; p ; b = p 2 2 and so a = b:
b p 2
Since y is a unit vector a2 +
1 + a2 = 1 2
Solving 2a2 = a =
1 2 1 2
Solution pairs are 1 1 1 ( ; ) and ( ; 2 2 2
Please report all typographical errors to:
1 ): 2
[email protected] Thanks, S.A.M