TEST BANK FOR Manifolds, Tensor and Forms An Introduction for Mathematics and Physicists By Renteln

1 Linear algebra page 1
2 Multilinear algebra 20
3 Differentiation on manifolds 33
4 Homotopy and de Rham cohomology 65
5 Elementary homology theory 77
6 Integration on manifolds 84
7 Vector bundles 90
8 Geometric manifolds 97
9 The degree of a smooth map 151
Appendix D Riemann normal coordinates 154
Appendix F Frobenius’ theorem 156
Appendix G The topology of electrical circuits 157
Appendix H Intrinsic and extrinsic curvature 158
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Linear algebra
1.1 We have
0 = c1(1, 1) + c2(2, 1) = (c1 + 2c2, c1 + c2)
⇒ c2 = −c1 ⇒ c1 − 2c1 = 0 ⇒ c1 = 0 ⇒ c2 = 0,
so (1, 1) and (2, 1) are linearly independent. On the other hand,
0 = c1(1, 1) + c2(2, 2) = (c1 + 2c2, c1 + 2c2)
can be solved by choosing c1 = 2 and c1 = −1, so (1, 1) and (2, 2) are
linearly dependent (because c1 and c2 are not necessarily zero).
1.2 Subtracting gives
0 =

i
vi ei −

i
v

i ei =

i
(vi − v

i )ei .
But the ei ’s are a basis for V, so they are linearly independent, which implies
vi − v

i
= 0.
1.3 Let V = U ⊕ W, and let E := {ei }n
i=1 be a basis for U and F := { f j }mj
=1 a
basis for W. Define a collection of vectors G := {gk}n+m
k=1 where gi = ei for
1 ≤ i ≤ n and gn+i = fi for 1 ≤ i ≤ m. Then the claim follows if we can
show G is a basis for V. To that end, assume
0 =
n+m
i=1
ci gi =
n
i=1
ci ei +
m
i=1
ci fi .
The first sum in the rightmost expression lives in U and the second sum lives
in W, so by the uniqueness property of direct sums, each sum must vanish
by itself. But then by the linear independence of E and F, all the constants
ci must vanish. Therefore G is linearly independent. Moreover, every vector
v ∈ V is of the form v = u + w for some u ∈ U and w ∈ W, each of which
1
2 Linear algebra
can be written as a linear combination of the gi ’s. Hence the gi ’s form a basis
for V.
1.4 Let S be any linearly independent set of vectors with |S| < n. The claim is
that we can always find a vector v ∈ V so that S ∪{v} is linearly independent.
If not, consider the sum
cv +
|S|
i=1
ci si = 0,
where si ∈ S. Then some of the ci ’s are nonzero. We cannot have c = 0,
because S is linearly independent. Therefore v lies in the span of S, which
says that dim V = |S| < n, a contradiction.
1.5 Let S, T : V → W be two linear maps, and let {ei } be a basis for V.
Assume Sei = Tei for all i, and that v =
i ai ei. Then Sv =
i ai Sei =

i aiTei = T v.
1.6 Let v1, v2 ∈ ker T. Then T (av1 + bv2) = aT v1 + bT v2 = 0, so ker T is
closed under linear combinations. Moreover ker T contains the zero vector of
V. All the other vector space properties are easily seen to follow, so ker T is a
subspace of V. Similarly, let w1,w2 ∈ im T and consider aw1 + bw2. There
exist v1, v2 ∈ V such that T v1 = w1 and T v2 = w2, so T (av1 + bv2) =
aT v1 + bT v2 = aw1 + bw2, which shows that imT is closed under linear
combinations.Moreover, im T contains the zero vector, so imT is a subspace
of W.
1.7 For any two vectors v1 and v2 we have
T v1 = T v2 ⇒ T (v1 − v2) = 0 ⇒ v1 − v2 = 0 ⇒ v1 = v2.
Assume the kernel of T consists only of the zero vector. Then for any two
vectors v1 and v2, T (v1 − v2) = 0 implies v1 − v2 = 0, which is equivalent
to saying that T v1 = T v2 implies v1 = v2, namely that T is injective. The
converse follows similarly.
1.8 Let V and W be two vector spaces of the same dimension, and choose a basis
{ei } for V and a basis { fi } for W. Let T : V → W be the map that sends ei to
fi , extended by linearity. Then the claim is that T is an isomorphism. Let v =

i ai ei be a vector in V. If v ∈ ker T , then 0 = T v =
i aiTei =
i ai fi .
By linear independence, all the ai ’s vanish, which means that the kernel of
T consists only of the zero vector, and hence by Exercise 1.7, T is injective.
Also, if w =
i ai fi, then w =
i aiTei = T

i ai ei , which shows that T
is also surjective.
1.9 a. Let v ∈ V and define w := π(v) and u := (1 − π)(v). Then π(u) =
(π − π2)(v) = 0, so v = w + u with w ∈ im π and u ∈ ker π. Now
Linear