Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 87
Three.II.1.20 Each of these projections is a homomorphism. Projection to the xz-plane and to the
yz-plane are these maps.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎝ x y⎠ ↦→ ⎝ x 0⎠
z 0
⎝ x y⎠ ↦→ ⎝ x 0⎠
z z
⎝ x y⎠ ↦→ ⎝ 0 y⎠
z 0
⎝ x y⎠ ↦→ ⎝ 0 y⎠
z z
Projection to the x-axis, to the y-axis, and to the z-axis are these maps.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
And projection to the origin is this map.
⎛
⎞
⎝ x y⎠ ↦→
z
⎛
⎝ 0 0
0
Verification that each is a homomorphism is straightforward.
transformation on R 3 .)
⎞
⎠
⎝ x y⎠ ↦→ ⎝ 0 0⎠
z z
(The last one, of course, is the zero
Three.II.1.21 The first is not onto; for instance, there is no polynomial that is sent the constant
polynomial p(x) = 1. The second is not one-to-one; both of these members of the domain
( ) ( )
1 0
0 0
and
0 0
0 1
are mapped to the same member of the codomain, 1 ∈ R.
Three.II.1.22
Yes; in any space id(c · ⃗v + d · ⃗w) = c · ⃗v + d · ⃗w = c · id(⃗v) + d · id( ⃗w).
Three.II.1.23 (a) This map does not preserve structure since f(1 + 1) = 3, while f(1) + f(1) = 2.
(b) The check is routine.
( ) ( ) ( )
x1 x2 r1 x
f(r 1 · + r
y 2 · ) = f( 1 + r 2 x 2
)
1 y 2 r 1 y 1 + r 2 y 2
= (r 1 x 1 + r 2 x 2 ) + 2(r 1 y 1 + r 2 y 2 )
= r 1 · (x 1 + 2y 1 ) + r 2 · (x 2 + 2y 2 )
( ) ( )
x1
x2
= r 1 · f( ) + r
y 2 · f( )
1 y 2
Three.II.1.24 Yes. Where h: V → W is linear, h(⃗u − ⃗v) = h(⃗u + (−1) · ⃗v) = h(⃗u) + (−1) · h(⃗v) =
h(⃗u) − h(⃗v).
Three.II.1.25 (a) Let ⃗v ∈ V be represented with respect to the basis as ⃗v = c 1β1 ⃗ + · · · + c nβn ⃗ . Then
h(⃗v) = h(c 1β1 ⃗ + · · · + c nβn ⃗ ) = c 1 h( β ⃗ 1 ) + · · · + c n h( β ⃗ n ) = c 1 · ⃗0 + · · · + c n · ⃗0 = ⃗0.
(b) This argument is similar to the prior one. Let ⃗v ∈ V be represented with respect to the basis as
⃗v = c 1β1 ⃗ + · · · + c nβn ⃗ . Then h(c 1β1 ⃗ + · · · + c nβn ⃗ ) = c 1 h( β ⃗ 1 ) + · · · + c n h( β ⃗ n ) = c 1β1 ⃗ + · · · + c nβn ⃗ = ⃗v.
(c) As above, only c 1 h( β ⃗ 1 ) + · · · + c n h( β ⃗ n ) = c 1 rβ ⃗ 1 + · · · + c n rβ ⃗ n = r(c 1β1 ⃗ + · · · + c nβn ⃗ ) = r⃗v.
Three.II.1.26 That it is a homomorphism follows from the familiar rules that the logarithm of a
product is the sum of the logarithms ln(ab) = ln(a) + ln(b) and that the logarithm of a power is the
multiple of the logarithm ln(a r ) = r ln(a). This map is an isomorphism because it has an inverse,
namely, the exponential map, so it is a correspondence, and therefore it is an isomorphism.
Three.II.1.27
Where ˆx = x/2 and ŷ = y/3, the image set is
) ) (ˆx ∣∣ (2ˆx)
(ˆx
{
2
∣∣
ŷ 4
+ (3ŷ)2
9
= 1} = { ˆx
ŷ
2 + ŷ 2 = 1}
the unit circle in the ˆxŷ-plane.
Three.II.1.28 The circumference function r ↦→ 2πr is linear. Thus we have 2π · (r earth + 6) − 2π ·
(r earth ) = 12π. Observe that it takes the same amount of extra rope to raise the circle from tightly
wound around a basketball to six feet above that basketball as it does to raise it from tightly wound
around the earth to six feet above the earth.