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Unformatted text preview: MAS212 Linear Algebra I Exercises 1 (with solutions) © Francis J. Wright, 2006 Assessed submission deadline 5pm on Friday 6th October 2006 in the blue box on the ground ﬂoor.
Your submission must have your FULL NAME and STUDENT NUMBER PRINTED CLEARLY at
the top and separate sheets must be stapled together. Marks out of 100 are shown in square brackets after each question number. 1. [15] (a) Verify that the set of rational numbers, Q, forms a group under addition by using the following deﬁnitions. A rational number is a quotient 5 of integers p, q such thatq 9b 0 and the sum of two rational numbers A + L = m, where you may assume all the normal properties
q S q S
of integer arithmetic. You may assume that Q is associative under addition. (b) Does Q form a group under multiplication if the product of two rational numbers 5  f = ﬁ ? If not, give a
simple way to construct a multiplicative group from Q. Solution
Letp, q, r,s EZq $0,s $0,sothat§,§ E Q.
(a) Then? + f =%+srq—,Whereps+rq andqsare integers andqs i 030% +§ E Q and Q is closed under addition. E + L = m = m = L + E by commutativity of integer arithmetic, so Q is
q s q s s q s q
commutative under addition. % E Q and 5 + % = 5, so % E Q is an additive identity element. HE 5 chen 1 e to and E + 1 = 9, so 1 e o is the additive inverse ofﬂ e o.
q q q q 1 q q (b) % E Q does not have a multiplicative inverse so Q does not form a multiplicative group. However, Q \ {% } does form a multiplicative group, i.e. just remove the additive identity element. Maple Solution 2. [20] Let a, b, c, x be elements of a ﬁeld K Show explicitly how the ﬁeld axioms allow the
equation a + b x = c to be solved forx and state any conditions that must be satisﬁed for a solution to exist.
Solution The ﬁeld axioms assert that there exists an additive inverse of a, namely —a. Add this inverse to both sides of the equation. By additive closure the result is in K; by additive commutativity and
associativity it does not matter where in the sum we add it and we can group the terms as
(a + (—a)) + b x =c + (—a) to giveO + b x =0 + (—a). Since 0 is the additive identity we have b x = c + (—a ) , which we conventionally write as b x = c —a. 1 The ﬁeld axioms assert that provided b 9b 0 there exists a multiplicative inverse of b, namely b_1.
Multiply both sides of the equation by this inverse. By multiplicative closure the result is in K;
by multiplicative commutativity and associativity it does not matter where in the product we multiply it and we can group the terms as (b b_1) x = (c—a) b_1 to give 1 x = (c—a) b_1. Since 1
is the multiplicative identity we have x = (c—a) b_1, provided b i 0, which we conventionally
( c ‘0 ) b write as x = [20] Consider the sets of 2 x 2 real matrices consisting of (a) all matrices, (b) all nonsingular (i. e. invertible) matrices, (c) all singular (i.e. noninvertible) matrices, with standard matrix addition
and multiplication. State whether each set is a ﬁeld and if not state which ﬁeld axioms fail. Where
possible, illustrate each failure with a simple example. Solution
In all cases the set is not a ﬁeld because matrix multiplication does not generally commute. See
Q4 for an example. But even if we ignore this (and so allow the set to be a skew ﬁeld or division
algebra) then (a) No, because no singular matrix has a multiplicative inverse.
(b) No, because
(i) the additive identity (zero) matrix is not in the set (since it is singular);
(ii) hence the additive inverse is not deﬁned;
(iii) addition of two nonsingular matrices may produce a singular matrix, so the set is not
closed under addition. Example:
1 1
1 0
(c) No, because (i) the multiplicative identity (unit) matrix is not in the set (since it is nonsingular); (ii) hence the multiplicative inverse is not deﬁned; (iii) addition of two singular matrices may produce a nonsingular matrix, so the set is not
closed under addition. Example: 11
—10 22
00 1 0 0 0 1 0
+ =
0 0 0 1 0 1
u I 0 0 0 1
4. [20] The matr1cesA andB over the Boolean ﬁeld F2 are g1ven byA= 1 1 ,B= 0 1 . ComputeA + B, A B and B A. Does multiplication of A and B commute? Compute A2 and B2 and deduce the values of A" and B" for all integer n 2 0. [Remember that all your results must be over the Boolean ﬁeld F2 represented by the set of elements {0, 1}, which you can achieve by computing over Z and then replacing all even integers by 0 and all odd integers by 1.] Solution
0 0 0 1
A:= :B:= :
1 1 0 1
0 0 0 1 0 1 0 1 0 1 0 1
A+B= + = and mod2= ,henceA+B= over
1 1 0 1 1 2 1 2 1 0 1 0
IF.
2 0 0 0 1 0 0 0 0 0 0 0 0 AB= = and mod2= ,henceAB= overlF .
l 1 0 1 0 2 0 2 0 0 0 0 2
0 1 0 0 1 1 1 1 1 l l 1 BA= = and mod2= ,henceBA= overlF .
0 1 l 1 1 1 1 1 1 l l 1 2 Therefore, multiplication of A and B does not commute, sinceA B at B A. 2 0 0 0 0 0 0 n I A = = =A, henceA =A for all integer n 2 0.
l l l l l 1
0 1 0 1 0 1 B2 = = =B, henceB" =B for all integer n 2 0.
0 l 0 1 0 1 Such matrices are called idempotent. Maple Solution 5. [5] In any vector space Vover a ﬁeld K show, by using the vector space axioms explicitly, that
(a+b) (x+y) =ax+ay+bx+byforanya,b E Kandx,y E V. Solution Starting from the expression (a + b) (x + y) , apply distributivity over scalar addition to give
a (x + y) + b (x + y) , then apply distributivity over vector addition to each term to give ax+ay+bx+by(orviceversa). 6. [20] Let V={(a1, az)  a1, a2 E R} be a set of vectors over the real ﬁeld [Rwith addition of
elements of Vdeﬁned coordinatewise (exactly as in the vector space M) and, for (al, a2) E Vand
c 6 IR, scalar multiplication deﬁned by the (unusual) rule (0,0) c=0
C(avazr [a 2] clo
c ’ 0 Explain brieﬂy why all the additive vector space axioms must be satisﬁed for V. However, V is not a
vector space because one of the scalar multiplication axioms is not satisﬁed; by checking each of
them determine which one. Solution
The vector space addition axioms are all satisﬁed because the deﬁnition of addition is the same as
that for R2. Now consider scalar multiplication. Let a = (a1, a2) , b = (b1, b2) E V. Closure is satisﬁed since c a = c (a1, a2) E V.
Distributivity over vector addition is satisﬁed since c( a + b) = c (a1 + b1, a2 + b2) =
(0, 0) c = 0
a1 + b1 a2 + b2
[ c ’ 0
ca +cb=c (a1, a2) +c (b1,b2) and J c¢0 (0,0)+(0,0)=(0,0) c=0 = e we WW1 c c c c c c
areequal.
Distributivity over scalar addition is not satisﬁed since, for c, d at 0, (c+d)a=(c+d) a,a = a a and
<12) [1, 2) c+d c+d
d d [‘11 “2]
ca+ a—c(a1,a2)+ (a1,a2)— c, c
“1 “2 _ i L L L
+[ d’ (ii—“c +dial’ic +01%]
are not equal.
(0,0) dc=0
Associativity is satisﬁed sinced(c(a)) =d c a ,a = a a and
(<1 2)) [—1,—2) dc dc
(0,0) dc=0
(dc)a=(dc) a,a = a a areequal.
<1 2) [—1,—2) dc dc Identity is satisﬁed since 1 (a1, a2) = (a1, a2).
Hence Vis not a vector space. ...
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 Spring '08
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