This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MAS212 Linear Algebra I Exercises 7 (with solutions) © Francis J. Wright, 2006 Assessed submission deadline 5pm on Friday 24th November 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. Do not use echelon methods for these exercises — that comes later! 1. [20] Determine explicitly the maximum number of linearly independent rows and columns, and
hence the row rank and column rank, of each of the following matrices over the real ﬁeld R : 12 2301
(a) 34,(b) 321—1
56 —11—12 Solution
(a) [10] The columns are linearly independent since one is not a scalar multiple of the other,
therefore the column rank is 2. Similarly, any two rows are linearly independent, therefore the row rank is at least 2. Check
whether all three rows are linearly independent by solving a( l, 2) + b(3, 4) + c(5, 6) = (0, 0)
for a, b, c E R, i.e. solve a + 3b + 5c = 0 2a + 4b + 6c = 0 One solution is obviously a = b = c = 0, but two equations for three variables cannot have a
unique solution, so there are nonzero solutions. (To be speciﬁc, the general solution can be
written as a = c, b = —2c, so a nonzero solution is a = c = l, b = —2.) Therefore the rows are not all linearly independent and the row rank is also 2 (as it must be). (b) [10] By inspection, row( 1 ) — r0w(2) = row( 3 ) , but any two rows are linearly independent
since none is a scalar multiple of another, therefore the row rank is 2. Also by inspection, col ( l ) = col (2 ) —col (4) so the column rank is less than 4. Check whether the
ﬁrst three columns are linearly independent by solving 2 3 0 0
a 3 +b 2 +c 1 = 0
—l 1 l 0
whichimples
2a+3b=0
3a+2b+c=0
—a+b—c=0. Adding the last two equations implies that 2a + 3b = 0, which is identical to the ﬁrst equation, so
the equations are not independent and therefore do not have a unique solution; they are
equivalent to two equations for three unknowns. The ﬁrst equation is satisﬁed by a = 3k, b = —2k. Then the last equation implies that c = —5k. Hence, a nonzero solution of the equations is l a = 3, b = —2, c = —5 and the ﬁrst three columns are not linearly independent. Hence the column rank is less than 3. Any two columns are clearly linearly independent, therefore the column rank
is 2 (as it must be). Maple Solution
1 0 1
2. [10] Determine the rank of the matrix 1 l 0 over (a) the rational ﬁeld 0 and (b) the
0 1 l
Boolean ﬁeld F2.
Solution (a) [5] Over Q, the rows (and columns) are linearly independent since
ur0w(l) +br0w(2) +cr0w(3) =0 => a +b=0,b +c=0,a +c=0 => a=b=c=0 =>
the rank is maximal, i.e. 3. Alternatively, use columns. (b) [5] Over [F2, r0w(3) = r0w( l) + r0w(2) by inspection, but any two rows are linearly
independent since none is a scalar multiple of another, therefore the rank is 2. [Note that
l + l = 0 in [Ff] Alternatively, use columns. Maple Solution 3. [15] Use rank computations to determine the nature of the solution space of the matrix equation
1 2 3 x 2 3 4 5 y = 3 over IR, i.e. whether a solution exists and if so what the dimension of the 5672 5 solution space is. Verify this by solving (if possible) the corresponding system of simultaneous
linear equations. Solution The rank of the square matrix is 2 because r0w(1 ) + r0w(3) = 2 r0w(2) so the rank is not
maximal but any two rows are linearly independent. However, this is no longer true when the column vector is adj oined, so check that the three rows
of the augmented matrix are linearly independent by solving
a(l, 2, 3, 2) +b(3, 4, 5, 3) + c(5, 6, 7, 5) = (0, 0, 0, 0), i.e. a+3b+5c=0, 2a+4b+6c=0,
3a+5b+7c=0,
2a+3b+56=0. Eliminating a from the ﬁrst and third equations implies 4b + 8c = 0, i.e. b + 20 = 0 and
eliminating a from the second and fourth equations implies b + c = 0. These two equations
together have the unique solution b = c = 0. Then any of the original equations implies that also
a = 0. Hence a = b = c = 0 is the unique solution and the rows are linearly independent, so the
augmented matrix has rank 3. Since the ranks of the augmented and original matrices differ, there is no solution, i.e. the
equations are inconsistent. The scalar equations are x+2y+3z=2, 3 x + 4 y + 5 z = 3, 5 x + 6 y + 7 z = 5.
Eliminatingx between the ﬁrst and second equations gives 2 y + 4 z = 3 and eliminatingx between
the ﬁrst and third equations gives 4 y + 8 z = 5. These equations are inconsistent and so have no
solution; for example, multiplying the ﬁrst new equation by 2 and subtracting the second in an
attempt to eliminate either y orz leads to the false assertion that 0 = 1. Maple Solution 4. [25] Compute the determinant of each of the following matrices and if its inverse exists then
compute it and check that it satisﬁes the deﬁnition of the inverse of a matrix: 010 123 123
(a)101,(b)345,(c)203.
100 567 031 Solution
(a) [10] By diagonal expansion, det(A) = (0 + l + 0) — (0 + 0 + 0) = 1. Since this is nonzero,
the inverse exists and is the adj oint divided by the determinant. First, the matrix of minors is
0—0 0—1 0—0 0 —l 0
0—0 0—0 0—1 = 0 0 —1
1—0 0—0 0—1 1 0 —1 From this, changing signs as appropriate and transposing gives the adj oint and dividing by the
determinant give the inverse as 001
i 100.
1 01—1 Check:
001010 100
1 T100101=010.
01—1100 001 (b) [5] From the previous question, the rank of this matrix is not maximal so the determinant is
zero and it has no inverse. (c) [10] By diagonal expansion, det(A) = (0 + 0 + 18) — (0 + 4 + 9) = 5. Since this is nonzero,
the inverse exists and is the adj oint divided by the determinant. First, the matrix of minors is
0—9 2 —0 6 —0 —9 2 6
2—9 l—0 3—0 = —7 1 3
6—0 3—6 0—4 6 —3 —4 From this, changing signs as appropriate and transposing gives the adj oint and dividing by the
determinant give the inverse as — —2 1 3
5
6 —3 4
Check:
9 7 6 1 2 3 1 0 0
% —2 1 3 2 o 3 = 010
6 —3 4 0 3 l 0 0 1
Maple Solution
. a b e f ..
5. [30] (a) For two general2 x 2 matricesA = d andB= h show exphcrtly that
c g
AB =AB
Solution[10]
a b e ae+b a +bh
(a) LetA:= ,B:= f .ThenAB= g f so AB =
c d g h ce+dg cf+dh (ae+bg) (cf+dh)—(af+bh) (ce+dg)=aedh+bgcf—afdg—bhce,andalso
A B =(ad—bc) (eh—fg) =aedh+bgcf—afdg—bhce. Maple Solution
1 0 0 0 0
2 2 0 0 0
(b) Evaluate the determinant 3 3 3 0 0 efﬁciently.
4 4 4 4 0
5 5 5 5 5
Solution [10] The value of a triangular determinant is the product of its diagonal elements, so this determinant
is 1 x2x3 x4x5=5!=120. Maple Solution
“0 “1 “2 art—l
“rt—1 “0 “1 “rt—2
(c) An n x n integer matrixA of the formA = an_2 an_1 a0 an_3 , in which the entries of
“1 a2 “3 “0 successive rows are shifted to the right one position (modulo n), is called a circulant matrix and its n — l
determinant A is called a circulant or a cyclic determinant. Prove that A contains 2 at. as a i = 0
factor. [Hint Consider the sum of all the rows or all the columns. Use the properties of
determinants to rewrite A so that every element in one of the rows or columns is the same and then 4 factor it out. Remark: In fact, the elements of A could be elements of any unique factorization
domain, of which the integers provides the simplest example, but they cannot be ﬁeld elements] Solution [10] Add all columns but the ﬁrst to the ﬁrst column. This does not change the value of the
n — 1
determinant but every element in the ﬁrst column becomes equal to 2 ai. A determinant is
1': 0
linear in each of its rows and columns, so a common factor of every element of a column can be
factored out as a scalar multiple. Optional Extras (no marks) 0 0
6. (a) If A is an n x n diagonal matrix of the formA = A2 with Al. at 0, 1 S i S n, 00x n prove that Ar = for any integer r (and in particular for r = —l).
0 0 )f
71
Solution
Suppose the result is true for some integer r. Then
it: 0 0 x1 0 0 A?“ 0 0 and the result is r+1r_0;\;...0 02.2u0 0A; ...0 0 0 Ar 0 0 )‘n 0 0 Ar+1
n n true for r + l. The result is true by deﬁnition for r = 1, therefore by induction it is true for all 1nteger r 2 1. The result 1s true for r = 0 prov1ded we Interpret A as the identity matrix for any
square matrix A and the result is true for r = —1 because 7% 0 0
X1 0 0 1
1 1 0 " 0
0 0 0 — 0 0 1 0
)‘2 I 9‘2 = : : '. 5 0 0 1 0 0 A" 1 0 0 _ it n Finally, suppose the result is true for some (negative) integer r. Then L 0 0 _1
xjo 0 7‘1 7L: 0 0
1 —1
r ,_ 0’0 0—o 0 r...0
A 1=AA1= A? *2 = A? andthe
r : rI—l
0 0 an 0 0 i o 0 an
a result is true for r— 1. Therefore by induction the result is true for all negative integer r and hence
all integer r. l 0 0 0
0 co 0 0 (b) LetQ be ann Xndiagonal matrix of the form 9= 0 0 (02 0 wheremis an nth root
0 o o of” of unity, i.e. one of the n distinct complex solutions of the equation x" = 1. If A is the circulant matrix . . . 1 . . . . deﬁned 1n Questlon 5(c) above, venfy that Q A S2 IS a Circulant matrix w1th elements at. ml and that n — l
1 ' . . . I Q A 52' = A . Hence, show that 2 ai o)1 is also a factor of A and that the Circulant A 1s
i= 0 precisely equal to the product of the n such factors corresponding to the n distinct nth roots of unity to}, l S j S :1. Verify this explicitly for the circulants with n = 2 and n = 3. Solution
. . . . 'l .
PostmultiplyingA by Q multiplies the fh column by (0] to give
2 n — l
2 n — l
an_1 H. 2 n—l ' ' '
a _ a _ m a 0) m a _ c0 . Us1ng the prev1ous result on powers of a diagonal
n 2 n 1 0 n 3
2 n — l 1 matrix shows that premultiplyingA by 9—1 multiplies the 11h row by 0) _i to give —1 n—2
an_1(D 0) an_20)
n
—2 —1 n—3 ' ' =
a _ 0) a _ w a a _ w . F1na11y, usmg the fact that a) 1 to remove
n 2 n 1 0 n 3
l—n 2—11 3—11
negative powers (or rationalize the denominators) in the lower triangular part of the matrix gives
2 n — 1
a0 C0 (D ' I I an_ 1 0)
n1 n—2
an_1 (D 0) an_2 0)
—1
n—2 n — 1 n—3 ' ' ' '
a _ 0) a _ w a a _ w . Hence!) A Q is a c1rcu1ant matr1x With
It 2 n l 0 n 3
2 3 elements at. uni. By the general product rule for determinants,
—1 —1 —1
IQ AQI = IQ A Q = '99 A = A. Thenthe same additiontechnique asbefore
n — 1 shows that 2 at. ml is also a factor of A  . There are n such factors corresponding to the n
i= 0 distinct nth roots of unity (of, l S j S n, and each such factor involves do with coefﬁcient 1, so their product involves as with coefficient 1. The general properties of determinants, or equivalently consideration of the evaluation of the determinant by recursive Laplace expansion,
shows that the circulant also involves as with coefﬁcient 1. Therefore, the circulant A is precisely equal to the product of the n factors corresponding to the n distinct nth roots of unity
(of l S j S n. a b
Verifying this explicitly for the circulant with n = 2 gives b = aZ—b2 and also
a
(a + b (o) (a + b (D) = aZ—bz; the circulant with n = 3 gives
w=1 0)=—1
a b c
_ 3 3 3
c a b —a —3abc+c +b andalso
b c a
(a+b0)+c0)2) (a+b0)+c0)2) (a+b0)+c(o2) =a3—3abc
(o=1 311: in:
co=e3 03=e 3 + c3 + b3. 7. (a) Let 0c : U—> Vand B : V—> Wbe linear maps. Prove that (i) rank(B 0c) S rank( 0L); (ii)
rank( B at) = rank( at) if B is onetoone; (iii) rank( B at) = rank( B) if on is onto. Solution (i) By deﬁnition, rank( 06) = dim ( 0c( U) ) and rank( B at) = dim ( B 0c( U) ). When B is applied
after at it can act only on 0L( U), the image of 01., so we can restrict the map B to the domain oc( U) ; Vand considerB : oc( U)—> W. Then im(B) = B oc( U) and dim(B oc( U)) =
dim(im() S dim(im(B)) + dim(ker( ) = dim(oc( U)) by the Dimension Theorem for the
map B. Hence rank( B at) S rank( cc). (ii) If B is onetoone then dim (ker( ) = 0, so rank( B on) = rank( on). (iii) Ifoc is onto then oc( U) = Vand B OL( )= B( V) = im(B), so rank( B CL) = rank( (b) LetA be an n x m matrix, P an invertible m x m matrix and Q an invertible n x 11 matrix, all over
the same ﬁeld. Prove that (i) rank(A P) = rank(A ), (ii) rank( QA) = rank(A ), (iii)
rank( QA P) = rank(A ), (iv) an intertible matrix has maximal rank. Solution Regard the matrices A, P, Q as linear maps. Since P and Q are invertible they are onto and oneto
one. (i) Applying result (iii) above with B =A and 0c = P shows that rank(A P) = rank(A ). (ii) Applying result (ii) above with B = Q and CL =A shows that rank( QA) = rank(A ). (iii) rank( QA P) = rank( QA) = rank(A). (iv) Take A to be the identity matrix (which necessarily has maximal rank) in either of (i) or (ii). ...
View
Full
Document
This note was uploaded on 02/11/2012 for the course MTH 141, 142, taught by Professor Mcallister during the Spring '08 term at SUNY Empire State.
 Spring '08
 McAllister
 Math

Click to edit the document details