and you can now easily graph it.
6.5 Which, if either, of the following matrices corresponds to an area preserving transformation? A= 2 1 3 3 B= 2 1 3 3 .
SOLUTION We compute the magnication factor for each of these matrices, and see ab whether it is 1 or
which is the equation of a line in the u, v plane. (c) As in part (a), we use () to express the equation for S , here x2 + y 2 = 1, in terms of u and v . The result is: (3u 2v )/3)2 + (v/3)2 = 1 , which simplies to 9u2 + 5v 2 6uv = 9 , which is the equati
6.1 Let A be the matrix A =
12 . Find the equation describing the image of S under the linear 03 transformation corresponding to A, and graph this image of S , where: (a) S is the line y axis. (b) S is the line x + y = 3. (c) S the unit circle.
SOLUTION T
(b) Find the vector af such that f (x) = af x, which we know exists by Problem 5.23, and is unique by Problem 5.24.
SOLUTION Computing the integral,
1
a + bx + cx2
0
a 1 dx = a + b/2 + c/3 = b 1/2 . c 1/3
Hence, if we dene 1 af = 1/2 , 1/3 we have f (x)
SOLUTION Since f is a linear transformation from IRn to IR, Theorem 2 says that f (x) = Af x where Af is the 1 n matrix Af = [f (e1 ), f (e2 ), . . . , f (en )] . The single row of this matrix written in vertical form, as is our convention with vectors is
5.21 Determine whether the following assertions are true or false. If true, explain why. If false, give a counterexample. (a) If A is any p m matrix, and let B be a m n, and the columns of B are all the same, then the columns of AB are all the same. (b) I
and then take B = [C v1 , C v2 ]. Using the given information, we then have CA = B with A= 1 2 2 1 and B= 2 1 . 1 1
Then CA = B , we deduce C = BA1 . Using the formula for 2 2 matrix inverses, we then nd 1 4 5 . C= 11 3
5.15 Are there any m n isometries w
5.9 Let A be an m n matrix, and B be an n p matrix. If B has at least one zero column, can it ever be the case that AB has no zero columns? Explain why not, or give an example.
SOLUTION No. By V.I.F. 3, which is the denition of matrixmatrix multiplication
By the denition of the transpose, and what we have just seen, we have: (row k of AB ) = (column k of (AB )t ) = (column k of B t A) = = That is, (row i of AB ) = (row i of B ) which tells us what the rows of AB are. SECOND SOLUTION FOR (a) We can also sta
column of AB is the linear combination of the columns of A with coecients coming from B 1 ,1 B2,1 , the rst column of B . Therefore, B 3 ,1 v2 + v3 = B1,1 v1 + B2,1 v2 + B3,1 v3 , so we can always take B1,1 = 0, B2,1 = 1 and B3,1 = 1. In the same way we n
(c) Write the second row of AB as a linear combination of the rows of B .
SOLUTION (a) By the rst part of Theorem 13, or what is the same, V.I.F. 4, the entry in question is given by the dot product of the second row of A with the thirst column of B . Thu
1 4.17 Let u be the unit vector u = 6 given by u.
2 1 1
and x =
1 1 . Compute x and x , where the direction is 1
SOLUTION We use the formula x = (x u)u to compute x . Since xu= this gives 2 1 1 . x= 3 1 Now, since x = x + x , we have x = x x , which gives
SECOND SOLUTION One can also use summation notation to do this, but one has k to be careful with the indices: We have x = j =1 xj . Therefore,
k
xi
k
xj .
|x|2 = x x =
i=1
j =1
Note that we have used two dierent summation indices i and j for the two di
x y in IR3 that are orthogonal to the xed vector z Find an equation specifying this plane. 4.11 The set of all vectors
1 2 1
is a plane in IR3 .
x 1 SOLUTION Computing the dot product between the vectors x = y and 2 you z 1 will obtain the equation x + 2
A nicer way to express this: Note that since d = (d/a)d when a = 0, d bd/a = d a Thus, with d arbitrary, the vectors of the form of the vector b . a Notice that the prescription all of the multiples of the vector a b b a b a b a bd/a d .
are precisely all
Therefore, the angle n between these vectors is 1 (rs)n n = 1 rs = 1 r 2n 1 r2
1/2
1 s2n 1 s2
1/2
(1 r2 )1/2 (1 s2 )1/2 1 rs
1 (rs)n (1 r2n )1/2 (1 s2n )1/2
.
(c) Since both |r| < 1 and |s| < 1, we have that limn r2n = 0, limn s2n = 0 and also, limn (rs)n
As you can check, this is the same result that one gets from the formula for inverting 2 2 matrices. Also in exercise 3.30, we found that with q (x) = x3 5x2 + 7x 2, q (B ) = 0. In this case, q (0) = 2, and 1 q (0) q (x) = (x2 5x + 7) . xq (0) 2 Therefore
(b) Let A and B be the matrices from problem 3.30. Find the inverses of A and B using part (a) of this problem.
SOLUTION (a) Write q (x) as q (x) = a0 + a1 x + a2 x2 + + ak xk . Notice that q (0) = a0 , so that q (0) q (x) = a1 x a2 x2 ak xk , and hence p
(b) Things are easy now because we have already computed A2 and B 2 . We nd: A2 2I = and 0 3 2 3 B 2I = 4 1 5 1 1 20 5 10 0 0 1 0 0 2 3 1 0 = 3 2 5 . 1 1 5 8 1 4 8 1 2 7 0 0 3 = 1 4 8 5 ,
3.31 Let A be an n n matrix such that Ak+1 = 0 for some k. Show I A
and so
n
tr(BC ) =
i,j =1
Bi,j Cj,i .
()
In the same way, we see that
n
tr(CB ) =
i,j =1
Ci,j Bj,i .
( )
Since the names of the variables that we are summing over the dummy variables do not matter, we can exchange i and j in (), with the result that
n
tr(
SOLUTION To see what is going on here, we square the matrix 1 2a . Cubing it we next nd 01 1 0
1a . The result is 01
3a . The pattern is clear. We claim that 1
n
1a 01
=
1 na 01
.
()
We have checked this for n = 2 and n = 3, and it is a tautology for n =
3.19 Given an example of noninvertible matrices A and B such that A + B is invertible.
SOLUTION
1 0
0 10 0 = + 1 00 0
0 . 1
4 5 5 , 6
3.21 Let A B and C be three invertible 2 2 matrices. Let D = ABC . Suppose that D1 = A= 5 7 7 10 and C 1 = 1 1 1 . Comput
Looking at the oof diagonal entries we see that either x = 0 or y = 0. But if y = 0, then the diagonal entries would be x2 , which cannot equal 4. Hence it must be that x = 0 and y 2 = 4, or y = 2.
3.15 Find all 2 2 matrices A such that A2 = A.
ab a2 + bc
(b) For any numbers x and y , form the matrix xI + yJ . Show that the product of any two such matrices is another matrix of the same form. More specically, (xI + yJ )(uI + vJ ) = (xu yv )I + (xv + yu)J . (c) Find all values of x and y so that (xI + yJ )2
[B, [C, A] = BCA + ACB BAC CAB [C, [A, B ] = CAB + BAC CBA ABC Notice that each of the six products ABC CBA BCA ACB CAB BAC
shows up once with a plus sign and once with a minus sign, so that when we form the sum [A, [B, C ] + [B, [C, A] + [C, [A, B ], eve
(b) Find all possible values of a b and c so that A2 = B where B =
1 0
3 . How many dierent sets of 4
values for a b and c are there for which A2 = B ? (The matrices A that you compute here are square roots of the matrix B .)
SOLUTION (a)
a2 0
ab + bc . (
3.5 Let A =
0 0 0
1 0 0
0 1 . Compute A2 and A3 . 0
SOLUTION One can just do the computations, using the denition of matrixmatrix multiplication, and the result is: 0 2 0 A= 0 0 0 0 1 0 0 0 3 0 A= 0 0 0 0 0 0 . 0
SECOND SOLUTION Although the computations
and recall that multiplication by a matrix is always a linear transformation. (Here we are using the converse part of Theorem 2.) Hence the matrix representing f is one in ().
Section 3
3.1 Consider the four matrices A, B , C and D from Exercise 2.7. Ther
(The polynomial transformation is multiply by x2 , then dierentiate twice). Is f linear? If so, nd the corresponding matrix. If not, explain why not.
SOLUTION As we saw in exercise 1.15, since d2 2 (x (a + bx + cx2 ) = 2a + 6bx + 12cx2 , dx2 the formula f