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 fo
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 , whi
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 i
(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 .
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
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 t
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
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.
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
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 tak
(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 o
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
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 us
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 vec
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 o
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 ha
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) =
(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
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
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 t
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 .
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.
(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)
[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 f
(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
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
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 t
(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