Partial Solution Set, Leon §4.1 Tuesday 16th October, 2012 at time
First, recall that to show L is a linear transformation, one must show that either
(1) L(x + y) = L(x) + L(y) and L(αx) = αL(x), or
(2) L(αx + βy) = αL(x) + βL(y), or
(3) L(αx + y) = αL(x) + L(y).
Either one of the three would suffice.
4.1.1 For each of five transformations, we are to verify linearity and describe geometrically
the effect of the transformation. Verification is straightforward for all; a sketch might
make it easier to see what’s going on geometrically.
(a) L(x) = (−x1 , x2 )T . Verifying linearity is simple: we simply show that
L(αx + y) =
=
=
=
(−(αx1 + y1 ), (αx2 + y2 ))T
(−αx1 , αx2 )T + (−y1 , y2 )T
α(−x1 , x2 )T + (−y1 , y2 )T
αL(x) + L(y).
The geometric effect is reflection across the x2 -axis.
(c) Reflection across the identity line x1 = x2 .
(e) Projection onto the x2 -axis.
4.1.2 Let L be the linear transformation mapping R2 into itself defined by
L(x) = (x1 cos α − x2 sin α, x1 sin α + x2 cos α)T .
Express x1 , x2 , and L(x) in terms of polar coordinates. Describe geometrically the effect
of the linear transformation.
Solution: p
Rewriting in terms of polar coordinates, we have x1 = r cos θ and x2 = r sin θ,
where r = x21 + x22 and θ = arctan(x2 /x1 ). But now we have
L(x) = (x1 cos α − x2 sin α, x1 sin α + x2 cos α)T
= (r cos θ cos α − r sin θ sin α, r cos θ sin α + r sin θ cos α)T
= (r cos(θ + α), r sin(θ + α))T ,
from which we see that the original vector has been rotated counterclockwise through an
angle α.
4.1.3 Let a be a fixed nonzero vector in R2 . A mapping of the form L(x) = x + a is called a
translation. Show that a translation is not a linear transformation. Illustrate geometrically the effect of a translation.
Solution: The verification that L is nonlinear is easy:
L(αx + y) = αx + y + a
6= α(x + a) + (y + a)
= αL(x) + L(y).
The geometric effect is to shift the line containing x away from the origin, the distance
and direction of the shift determined by the length and direction of a. In the event that a
and x are collinear, the shifted line still passes through the origin, but the transformation
is nonetheless nonlinear, as shown by the preceding step.
4.1.6 Determine whether the following are linear transformations from R2 into R3 .
(a) L(x) = (x1 , x2 , 1)T . This is nonlinear, since
L(αx) = (αx1 , αx2 , 1)T 6= (αx1 , αx2 , α)T = αL(x).
(c) L(x) = (x1 , 0, 0)T . This is linear, since
L(αx + y) = (αx1 + y1 , 0, 0)T = α(x1 , 0, 0)T + (y1 , 0, 0)T = αL(x) + L(y).
4.1.14 Let L be a linear operator on R1 and let a = L(1). Show that L(x) = ax for all x ∈ R1 .
Proof: Assume that L is as described. Let x ∈ R1 . Then
L(x) = L(x1) = xL(1) = xa = ax,
2
and we’re done.
4.1.15 Let L be a linear operator mapping a vector space V into itself. Recursively define Ln ,
by L1 = L and Ln+1 (v) = L(Ln (v)) for all n ≥ 1 and all v ∈ V . Show that Ln is a linear
operator on V for each n ≥ 1.
Proof: The proof is by induction on n. By definition of Ln , the result holds for n = 1.
Let u, v ∈ R, let α be a scalar, and let k ≥ 1. Then
Lk+1 (αu + v) =
=
=
=
L(Lk ((αu + v)
L(αLk (u) + Lk (v)
αL(Lk (u)) + L(Lk (v)
αLk+1 (v) + Lk+1 (v), and we’re done.
2
2
4.1.16 Let L1 : U → V and L2 : V → W be linear transformations and let L = L2 ◦ L1
be the mapping defined by L(u) = L2 (L1 (u)) for each u ∈ U . Show that L is a linear
transformation mapping U into W .
Solution: That L is a mapping from U into W follows from elementary properties of
functions. We must show that L is linear. So let u1 , u2 ∈ U , and let α be a scalar. Then
L(αu1 + u2 ) =
=
=
=
L2 (L1 (αu1 + u2 ))
L2 (αL1 (u1 ) + L1 (u2 ))
αL2 (L1 (u1 )) + L2 (L1 (u2 ))
αL(u1 ) + L(u2 )
(Definition of L)
(linearity of L1 )
(linearity of L2 )
(Definition of L.)
2
4.1.17 Determine the kernel and range of each of the following linear transformations from
R3 into itself. Solution:
(a) L(x) = (x3 , x2 , x1 )T . The kernel is the zero vector from R3 , since L(x) = 0 if and
only if x1 = x2 = x3 = 0. The range is all of R3 : let y = (y1 , y2 , y3 )T ∈ R3 . Then
y = L (y3 , y2 , y1 )T , so L is an onto mapping.
(c) L(x) = (x1 , x1 , x1 )T . The kernel contains all vectors of the form (0, x2 , x3 ), and is
therefore a two-dimensional subspace of R3 . The range is Span (1, 1, 1)T .
4.1.21 A linear transformation L : V → W is one-to-one if L(v1 ) = L(v2 ) implies that
v1 = v2 . Show that L is one-to-one iff Ker(L) = {0V }.
Proof: (⇒) Method 1: Let x ∈ Ker(L) ⇒ L(x) = 0W . Also L(0v ) = 0W since L
is a linear transformation. Thus L(x) = L(0V ) and since L is one-to-one it follows that
x = 0V . Thus Ker(L) = {0V }.
(⇒) Method 2 (Contrapositive): Suppose that Ker(L) 6= {0V }. Then {0V , x} ⊆ Ker(L)
for some nonzero x. But L(x) = 0W = L(0V ), so L is not one-to-one.
(⇐) Method 1: Assume that Ker(L) = {0V }. To prove L is one-to-one, let L(x) = L(y)
and prove that x = y:
L(x)
L(x) − L(y)
L(x) − y)
x−y
x−y
x
3
=
=
=
∈
=
=
L(y)
0W
0W
Ker(L)
0V
y
(⇐) Method 2 (Contrapositive): Suppose L is not one-to-one. Let L(v1 ) = L(v2 ) for
some v1 6= v2 . Then v1 − v2 6= 0V , but L(v1 − v2 ) = 0W , so {0V , v1 − v2 } ⊂ ker(L), the
negation of Ker(L) = {0V }.
2
4