1 0 0
0 1 0
1 −1 0 3 −4 0
1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
→
1 0 0 0 1 0 0 0 0 0 0 0
1 0 0 0
0 1 0 0
−1 1 1 0
−3 4 0 1
so VT =
1 0 0 0
0 1 0 0
−1 1 1 0
−3 4 0 1
.
Hence, (U AV )T = (RV )T = VTRT =
1 0 0 0 1 0 0 0 0 0 0 0
, so U AV =
1 0 0 0 0 1 0 0 0 0 0 0
.
16. We need a sequence of elementary operations to carry [U A] to [I U−1A]. By Lemma 1 these operations can be achieved by left multiplication by elementary matrices. Observe
[I U−1A] = [U−1U U−1A] = U−1[U A]. (*) Since U−1 is invertible, it is a product of elementary matrices (Theorem 2), say U−1 = E1E2. . . Ekwhere the Eiare elementary. Hence (*) shows that [I U−1A] = E1E2. . . Ek[U A], so a sequence of k row operations carries [U A] to [I U−1A]. Clearly [I U−1A] is in re-duced row-echelon form.
17(b) A∼ A because A = IA. If Ar ∼ B, let A = UB, U invertible. Then B = Ur −1A so B ∼ A.r Finally if A∼ B and Br ∼ C, let A = UB and B = V C where U and V are invertible. Hencer A = U (V C) = (U V )C so A∼ C.r
19(b) The matrices row-equivalent to A =
0 0 0 0 0 1
are the matrices U A whre U is invertible. If U =
a b c d
then U A =
0 0 b 0 0 d
where b and d are not both zero (as U is invertible).
Every such matrix arises – use U =
a b
−b a
– it is invertible as a2+ b2 = 0 (Example 4
§2.3).
22(b) By Lemma 1, B = EA where E is elementary, obtained from I by multiplying row i by k = 0.
Hence B−1= A−1E−1 where E−1 is elementary, obtained from I by multiplying row i by 1k. But then forming the product A−1E−1 is obtained by multiplying column i of A−1 by k1.
Exercises 2.6 Matrix Transformations
1(b) Write A =
3 2
−1
, B =
2 0 5
and X =
5 6
−13
. We are given T (A) and T (B), and are asked to find T (X). Since T is linear it is enough (by Theorem 1) to express X as a linear combination of A and B. If we set X = rA + sB, equating entries gives equations 3r + 2s = 5, 2r = 6 and −r + 5s = −13. The (unique) solution is r = 3, s = −2, so X = 3A − 2B. Since T is linear we have
T (X) = 3T (A) − 2T (B) = 3
3 5
− 2
−1 2
=
11 11
.
36 Section 2.6: Matrix Transformations express X as a linear combination of A and B, and use the assumption that T is linear. If we write X = rA + sB, equate entries, and solve the linear equations, we find that r = 2 and s = −3. Hence X = 2A − 3B so, since T is linear,
Of course, T
x we can easily see directly that T has matrix
−1 0 0 −1
. However, sometimes Theorem 2 is necessary.
. If these vectors are rotated counterclockwise through π4, some simple trigonometry shows that T (e1) =
√2 R3 is reflection in the u-z-plane, we have:
T (e1) = −e1 because e1 is perpendicular to the u-z-plane; while
Section 2.6: Matrix Transformations 37
Note that T1 also fails for this transformation T, as you can verify.
8(b) We are given T transformation induced by the matrix A = √12
1 1 trans-formation induced by the matrix A = 101
−8 −6
−6 8
. Looking at Theorem 5, we see that A is the matrix of Q−3. Hence T = Q−3 is reflection in the line y = −3x.
10(b) Since T is linear, we have T
is on the y axis. Now observe that T is rotation of the x-z-plane through the angle θ from the x axis to the z axis. By Theorem 4 the effect of T on the x-z-plane is given by
x
Hence the matrix of T is
12(b) Let Q0 denote reflection in the x axis, and let Rπ denote rotation through π. Then Q0 has matrix A =
38 Section 2.6: Matrix Transformations
transformation Q0◦ Rπ, and this has matrix AB =
−1 0 0 1
by Theorem 3. This is the matrix of reflection in the y axis.
(d) Let Q0 denote reflection in the x axis, and let Rπ2 denote rotation through π2. Then Q0 has matrix A =
1 0 0 −1
, and Rπ
2 has matrix B =
0 −1
1 0
. Then Q0 followed by Rπ
2 is the transformation Rπ
2 ◦ Q0, and this has matrix BA =
0 1 1 0
by Theorem 3. This is the matrix of reflection Q1 in the line with equation y = x.
(f) Let Q0 denote reflection in the x axis, and let Q1 denote reflection in the line y = x. Then Q0 has matrix A =
1 0 0 −1
, and Q1 has matrix B =
0 1 1 0
. Then Q0 followed by Q1 is the transformation Q1◦ Q0, and this has matrix BA =
0 −1 1 0
by Theorem 3. This is the matrix of rotation Rπ2 about the origin through the angle π2.
13(b) Since R has matrix A, we have R(x) = Ax for all x in Rn. By the definition of T we have T (x) = a R(x) = a(Ax) = (aA)x
for all x in Rn. This shows that the matrix of T is aA.
14(b) We use Axiom T2: T (−x) = T [(−1)x] = (−1) T (x) = −T (x).
17(b) The matrix of T is B, so T (x) = Bx for all x in Rn. Let B2 = I. Then
T2(x) = T [T (x)] = B[Bx] = B2x= Ix = x = 1R2(x) for all x in Rn. Hence T2 = 1Rn since they have the same effect on every column x.
Conversely, if T2 = 1Rn then
B2x= B(Bx) = T (T (x)) = T2(x) = 1R2(x) = x = Ix for all x in Rn. This implies that B2= I by Theorem 5 §2.2.
18 The matrices of Q0, Q1, Q−1 and Rπ2 are
1 0 0 −1
,
0 1 1 0
,
0 −1
−1 0
and
0 −1 1 0
, respectively. We use Theorem 3 repeatedly: If S has matrix A and T has matrix B then S ◦T has matrix AB.
(b) The matrix of Q1◦ Q0 is
0 1 1 0
1 0 0 −1
=
0 −1 1 0
, which is the matrix of Rπ2. (d) The matrix of Q0◦ Rπ2 is
1 0 0 −1
0 −1 1 0
=
0 −1
−1 0
which is the matrix of Q−1.
19(b) We have Pm[Qm(x)] = Pm(x) for all x in R2 because Qm(x) lies on the line y = mx. This means Pm◦ Qm= Pm.
Section 2.6: Matrix Transformations 39 20 To see that T is linear, write x = [x1 x2 · · · xn]T and y = [y1 y2 · · · yn]T. Then:
T (x + y) = T
[x1+ y1 x2+ y2 · · · xn+ yn]T
= (x1+ y1) + (x2+ y2) + · · · + (xn+ yn)
= (x1+ x2+ · · · + xn) + (y1+ y2+ · · · + yn)
= T (x) + T (y),
T (ax) = T ([ax1 ax2 · · · axn]T)
= ax1+ ax2+ · · · + axn
= a(x1+ x2+ · · · + xn)
= a T (x).
Hence T is linear, so its matrix is A = [T (e1) T (e2) · · · T (en)] = [1 1 · · · 1] by Theorem 2.
Note that this can be seen directly because
T
x1
x2
... xn
= x1+ · · · + xn= [1 1 · · · 1]
x1
x2
... xn
so we see immediately that T is the matrix transformation induced by [1 1 · · · 1]. Note that this also shows that T is linear, and so avoids the tedious verification above.
22(b) Suppose that T : Rn→ R is linear. Let e1, e2, · · · , en be the standard basis of Rn, and write T (ej) = wj for each j = 1, 2, · · · , n. Note that each wj is in R. As T is linear, Theorem 2 asserts that T has matrix A = [T (e1) T (e2) · · · T (en)] = [w1 w2 · · · wn]. Hence, given
x=
x1
x2
... xn
in Rn, we have
T (x) = Ax = [w1 w2 · · · wn]
x1
x2
...
xn
= w1x1+ w2x2+ · · · + wnxn= w • x = Tw(x) for
all X in Rn where w = [w1 w2 · · · wn]T. This means that T = Tw. This can also be seen without Theorem 2: We have x = x1e1+ x2e2+ · · · + xnen so, since T is linear,
T (X) = T (x1e1+ x2e2+ · · · + xnen)
= x1T (e1) + x2T (e2) + · · · + xnT (en)
= x1w1+ x2w2+ · · · + xnwn
= w • x
= Tw(x).
for all x in Rn. Thus T = Tw.
40 Section 2.7: LU-factorization 24(b) Given linear transformations Rn T→ Rm S→ Rk we are to show that (S ◦ T )(ax) = a (S ◦ T )(x)
for all x in Rn and all scalars a. The proof is a straight forward computation:
(S ◦ T )(ax) = S[T (ax)]
= S[a T (x)]
= a [S[T (x)]]
= a [(S ◦ T )(x)].
Definition of S ◦ T T is linear S is linear Definition of S ◦ T
Exercises 2.7 LU-factorization
1(b)
2(b) The reduction to row-echelon form requires two row interchanges:
The elementary matrices corresponding (in order) to the interchanges are P1 =
Section 2.7: LU-factorization 41 We apply the LU -algorithm to P A:
P A =
(d) The reduction to row-echelon form requires two row interchanges:
The elementary matrices corresponding (in order) to the interchanges are P1 = We apply the LU -algorithm to P A:
P A =
and we solve this by forward
substitu-tion: y1 = −1, y2 = 13(−1 − y1) = 0, y3 = 1 + y1 − 2y2 = 0. The system U X = Y
42 Section 2.8: An Application to Input-Output Economic Models
is x1 + x2 − x4 = −1 x2 + x4 = 0
0 = 0
and we solve this by back substitution: x4 = t, x3 = 5,
x2 = −x4 = −t, x1 = −1 + x4− x2= −1 + 2t.
(d) Analogous to (b). The solution is: y =
2 8
−1 0
, x =
8 − 2t 6 − t
−1 − t t
, t arbitrary
5. If the rows in question are R1 and R2, they can be interchanged thus:
R1
R2
→
R1+ R2
R2
→
R1+ R2
−R1
→
R1
−R1
→
R2
R1
.
6(b) Let A = LU = L1U1 be LU -factorizations of the invertible matrix A. Then U and U1 have no row of zeros so (being row-echelon) are upper triangular with 1’s on the main diagonal.
Thus L−11 L = U1U−1is both lower triangular (L−11 L) and upper triangular (U1U−1) and so is diagonal. But it has 1’s on the diagonal (U1 and U do) so it is I. Hence L1 = L and U1= U . 7. We proceed by induction on n where A and B are n × n. It is clear if n = 1. In general,
write A =
a 0 X A1
and B =
b 0 Y B1
where A1 and B1 are lower triangular. Then
AB =
ab 0
Xb + A1Y A1B1
by Theorem 4 §2.2, and A1B1 is upper triangular by induction.
Hence AB is upper triangular.
9(b) Let A = LU = L1U1 be two such factorizations. Then U U1−1 = L−1L1; write this matrix as D = U U1−1 = L−1L1. Then D is lower triangular (apply Lemma 1 to D = L−1L1), and D is also upper triangular (consider U U1−1). Hence D is diagonal, and so D = I because L−1 and L1are unit triangular. Since A = LU, this completes the proof.