1. 1. The wedge product of two vectors u, v ∈ Rn
, u ∧ v, is given by the matrix u ∧ v = 1/2(uT
v − vT
u).
Show that u ∧ v is skew symmetric.
Solution: (u ∧ v)
T
= 1/2(uT
v − vT
u)
T
= 1/2((uT
v)T
− (vT
u)T
) = 1/2(vT
(uT
)T
− uT
(vT
)T
) =
1/2(vT
u − uT
v).
2. The α−rotation matrix, Rα, is given by:
Rα =
cos α − sin α
sin α cos α
.
2.1. Show that the determinant of Rα is independent of α.
Solution: |Rα| = cos2
α + sin2
α = 1 and the conclusion follows.
2.2. Verify that indeed Rα rotates a vector in R2
by an angle of amplitude α.
Solution: Let u = (x, y) ∈ R2
be such vector. Setting x = r cos θ and y = r sin θ, one can write
u in the form u = (r cos θ, r sin θ), where r = u = x2 + y2 and θ = arctan(y/x) and thus such
coordinates r and θ are uniquely determined. Hence, using those coordinates, we have:
Rα =
cos α − sin α
sin α cos α
r cos θ
r sin θ
=
r cos α cos θ − r sin α sin θ
r sin α cos θ + r cos α sin θ
=
r cos(α + θ)
r sin(α + θ)
.
This shows that Rα rotates a vector in R2
by an angle of amplitude α.
2.3. Show that Rα is orthogonal.
Solution: There is a quicker way to answer this exercise: noticing that the columns of Rα form an
orthonormal basis of R2
. Nevertheless, we can use the definition to show that R−1
α = RT
α . Clearly,
by exercise 2.1., Rα is invertible. Thus:
cos α sin α
− sin α cos α
cos α − sin α
sin α cos α
= I2.
where I2 denotes the identity matrix of order 2.
2.4. Let O(2) be the set of orthogonal matrices of order 2 (clearly, Rα ∈ O(2)). Show that O(2) is a
group under matrix multiplication.
Solution: The existence of an identity, inverse element is obvious (just take I2). Associativity is also
straightforward. Finally, we just need to show that O(2) is closed under matrix multiplication. Take
A, B ∈ O(2). Then (AB)(AB)T
= ABBT
AT
= AAT
= I2 and so (AB)T
= (AB)−1
is an element of
O(2).
2.5. Show that the correspondence T : α ∈ [0, 2π[ → O(2); α → Rα is well defined and is one–to–one. Is
it onto?
Solution: Let α = β. Then it is clear that Rα = Rβ, for all α, β ∈ [0, 2π[ and thus T is well
defined. Moreover, since cosine and sine functions are injective in [0, 2π[ and alternate in sign in the
four quadrants it follows that T is also one–to–one. Although it is not onto since O(2) also includes
matrices whose determinant equals -1.
3. Consider the transformation T : R2
→ R between vector spaces defined by T(u, v) = u + v.
3.1. Verify that T is linear.
Solution: Trivial.
3.2. Find ker(T). Conclude that T is not injective.
Solution: ker(T) = {(1, −1)}, i.e., the subspace of R2
generated by the vector (1, −1). Thus ker(T)
has many infinitely elements and hence T is not injective.
3.3. Write T as a linear combination of the elements of a basis of R2∗
, the dual of R2
.
Solution: Take the canonical basis of R2
. (φ1, φ2) is the dual basis, where φi : R2
→ R, for i = 1, 2,
are defined by φ1(u, v) = u and φ2(u, v) = v. Thus, clearly T = 1 ∗ φ1 + 1 ∗ φ2.