Math 440, Linear Algebra II Solutions to Homework 9 Problems 11-5 ...

Math 440, Linear Algebra II
Solutions to Homework 9 Problems
11-5. Let L = R 3 . In this problem you will show that L is a Lie algebra via the bracket
operation given by the vector cross product:
[v, w] = v × w
for v, w ∈ L = R 3 by verifying (a)-(c) of Problem (1). Assume the properties of the
vector cross product from linear algebra, including the fact that
u × (v × w) = (u · w)v − (u · v)w.
Proof. By definition, a Lie algebra is a vector space L together with a binary operation
[ , ] : L × L → L that satisfies the axioms
(L1) the bracket product is bilinear;
(L2) the bracket product is skew-symmetric;
(L3) the bracket satisfies the Jacobi identity.
In this problem, we are given a vector space L = R 3 and a binary operation on R 3 ,
namely the cross product of vectors as defined in Calculus 3:
[v, w] := v × w.
Therefore, to prove that L = R 3 with the cross product is a Lie algebra, it suffices
to show that the cross product satisfies axioms (L1), (L2) and (L3) above. First,
from Calculus 3, we recall the properties of the cross product: (see Stewart, Calculus:
Concepts & Contexts, Section 9.4):
Proposition (Properties of the Cross Product). Let u, v, w ∈ R 3 and α, β ∈ R.
(a) v × w = −(w × v)
(b) v × v = ⃗0
(c) (αv) × (βw) = (αβ)(v × w)
(d) u × (v + w) = (u × v) + (u × w) and (u + v) × w = (u × w) + (v × w).
Therefore, the cross product is bilinear and skew-symmetric by this Proposition (bilinear
by parts (c) and (d), skew-symmetric by parts (a) and (b)). Hence it only remains
to prove that the cross product satisfies the Jacobi identity:
[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0.
To show this, recall that the cross product satisfies the equation
a × (b × c) = (a · c)b − (a · b)c
(this is a Proposition in Section 0.1 of the notes). Thus, for u, v, w ∈ R 3 , consider
u × (v × w) + v × (w × u) + w × (u × v)
= [(u · w)v − (u · v)w] + [(v · u)w − (v · w)u] + [(w · v)u − (w · u)v]
= [−(v · w) + (w · v)]u + [(u · w) − (w · u)]v + [−(u · v) + (v · u)]w
= ⃗0.
Therefore the cross product satisfies the Jacobi identity, so L = R 3 with the cross
product is a Lie algebra.

Math 440, Linear Algebra II
Solutions to Homework 9 Problems
11-6. In this problem you will look at an example of a Lie algebra homomorphism. Let G be
the Euclidean orthogonal group O(3, R), so Lie(G) consists of all 3×3 skew-symmetric
matrices. Let L = R 3 be the Lie algebra of Problem (5). Define Φ : L → Lie(G) by
⎡
0 −x
⎤
−y
Φ(v) = Φ(x, y, z) = ⎣x 0 −z⎦
y z 0
where v = (x, y, z) ∈ L = R 3 . (You’ll need to work with coordinates.)
(a) Show that Φ is a linear transformation.
Proof. To prove that Φ is a linear transformation, we must show that Φ(v +w) =
Φ(v) + Φ(w) and that Φ(αv) = αΦ(v) for all v, w ∈ L and all α ∈ R. Hence,
suppose (x 1 , y 1 , z 1 ), (x 2 , y 2 , z 2 ) ∈ L and α ∈ R. Then (x 1 , y 1 , z 1 ) + (x 2 , y 2 , z 2 ) =
(x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) and α(x 1 , y 1 , z 1 ) = (αx 1 , αy 1 , αz 1 ), so
⎡
⎤
0 −x 1 − x 2 −y 1 − y 2
Φ(x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) = ⎣x 1 + x 2 0 −z 1 − z 2
⎦
y 1 + y 2 z 1 + z 2 0
⎡
⎤ ⎡
⎤
0 −x 1 −y 1 0 −x 2 −y 2
= ⎣x 1 0 −z 1
⎦ + ⎣x 2 0 −z 2
⎦
y 1 z 1 0 y 2 z 2 0
= Φ(x 1 , y 1 , z 1 ) + Φ(x 2 , y 2 , z 2 ),
⎡
⎤
0 −(αx 1 ) −(αy 1 )
Φ(αx 1 , αy 1 , αz 1 ) = ⎣(αx 1 ) 0 −(αz 1 ) ⎦
(αy 1 ) (αz 1 ) 0
⎡
⎤
0 −x 1 −y 1
= α ⎣x 1 0 −z 1
⎦
y 1 z 1 0
Therefore, Φ is a linear transformation.
(b) Show that Φ is one-to-one and onto.
= αΦ(x 1 , y 1 , z 1 ).
Proof. If Φ(x 1 , y 1 , z 1 ) = Φ(x 2 , y 2 , z 2 ), then we must have x 1 = x 2 , y 1 = y 2 , z 1 = z 2
since matrix equality is determined entry-wise. Hence Φ is one-to-one. Since
Lie(G) = {B ∈ M(3, R) : B T + B = O 3 } implies that for each B ∈ Lie(G), if
B = [b ij ] then b ii = 0 and b ij = −b ji for 1 ≤ i < j ≤ 3. Hence each B ∈ Lie(G) is
determined by the values b 12 , b 13 , b 23 , i.e.,
⎡
0 b 12
⎤
b 13
B = ⎣−b 12 0 b 23
⎦ = Φ(−b 12 , −b 13 , −b 23 )
−b 13 −b 23 0
is in the image of Φ. Therefore, Φ is onto.

Math 440, Linear Algebra II
Solutions to Homework 9 Problems
(c) Show that Φ preserves the Lie bracket. That is, show
[Φ(v), Φ(w)] = Φ([v, w])
for any v, w ∈ L = R 3 . Remember that the bracket [v, w] is given by the vector
cross product, while the bracket [Φ(v), Φ(w)] is the usual bracket operation for
matrices.
Proof. Let v = (v 1 , v 2 , v 3 ), w = (w 1 , w 2 , w 3 ) ∈ L, so their bracket in L is v × w =
(v 2 w 3 − v 3 w 2 , v 3 w 1 − v 1 w 3 , v 1 w 2 − v 2 w 1 ). We must then show that Φ(v × w) =
[Φ(v), Φ(w)], where this second bracket is the Lie bracket for matrices, [A, B] =
AB − BA. Thus consider
[Φ(v), Φ(w)] = [Φ(v 1 , v 2 , v 3 ), Φ(w 1 , w 2 , w 3 )]
= Φ(v 1 , v 2 , v 3 )Φ(w 1 , w 2 , w 3 ) − Φ(w 1 , w 2 , w 3 )Φ(v 1 , v 2 , v 3 )
⎡
⎤ ⎡
⎤
0 −v 1 −v 2 0 −w 1 −w 2
= ⎣v 1 0 −v 3
⎦ ⎣w 1 0 −w 3
⎦
v 2 v 3 0 w 2 w 3 0
⎡
⎤ ⎡
⎤
0 −w 1 −w 2 0 −v 1 −v 2
− ⎣w 1 0 −w 3
⎦ ⎣v 1 0 −v 3
⎦
w 2 w 3 0 v 2 v 3 0
⎡
⎤
−v 1 w 1 − v 2 w 2 −v 2 w 3 v 1 w 3
= ⎣ −v 3 w 2 −v 1 w 1 − v 3 w 3 −v 1 w 2
⎦
v 3 w 1 −v 2 w 1 −v 2 w 2 − v 3 w 3
⎡
⎤
−w 1 v 1 − w 2 v 2 −w 2 v 3 w 1 v 3
− ⎣ −w 3 v 2 −w 1 v 1 − w 3 v 3 −w 1 v 2
⎦
w 3 v 1 −w 2 v 1 −w 2 v 2 − w 3 v 3
⎡
⎤
0 v 3 w 2 − v 2 w 3 v 1 w 3 − v 3 w 1
= ⎣v 2 w 3 − v 3 w 2 0 v 2 w 1 − v 1 w 2
⎦
v 3 w 1 − v 1 w 3 v 1 w 2 − v 2 w 1 0
= Φ(v 2 w 3 − v 3 w 2 , v 3 w 1 − v 1 w 3 , v 1 w 2 − v 2 w 1 )
= Φ(v × w).
Therefore, as Φ is a linear transformation (by part (a)) and Φ preserves the Lie
brackets, Φ is a Lie algebra homomorphism. Furthermore, since Φ is both oneto-one
and onto (by part (b)), Φ is a Lie algebra isomorphism.
11-9. This is a different approach to Proposition 4.1.3(c). (If we define ad as the differential
of Ad, then we already know ad is a linear transformation.) Assume G is a Lie group
and Lie(G) is its tangent space. Assume a bracket operation has been defined on Lie(G)
(so Lie(G) is closed under the bracket operation) satisfying the properties (a)-(c) in
Problem (1). For a fixed A ∈ Lie(G), define ad(A) : Lie(G) → Lie(G) by
ad(A)(B) = [A, B].
Use this definition in terms of the bracket and the properties in (1) to show ad(A) is
a linear transformation.

Math 440, Linear Algebra II
Solutions to Homework 9 Problems
Proof. Let A ∈ Lie(G), so ad(A) : Lie(G) → Lie(G) defined by ad(A)(B) = [A, B]
is at least a function from Lie(G) to itself. To prove that this mapping is a linear
transformation, we only need to prove that for all B, C ∈ Lie(G) and all scalars r ∈ R,
we have ad(A)(B + C) = ad(A)(B) + ad(A)(C) and ad(A)(rB) = r[ad(A)(B)]. Thus
consider, for B, C ∈ Lie(G) and r ∈ R, using Properties of the bilinearity of the
bracket:
ad(A)(B + C) = [A, B + C] = [A, B] + [A, C] = ad(A)(B) + ad(A)(C),
adA(rB) = [A, rB] = r[A, B] = r[ad(A)(B)].
Therefore, ad(A) is a linear transformation.
11-10. In this problem you will take a careful look at the Lie algebra of G = SL(2, R), using
the following basis for Lie(G):
[ ] [ ] [ ]
0 1 1 0 0 0
E = , H = , F = .
0 0 0 −1 1 0
(a) Find [E, H], [F, H] and [E, F ] (where the bracket operation is the one defined in
(1)).
Solution: For E, H, F as above, we have
[ ] [ ] [ ] [ ] [ ]
0 1 1 0 1 0 0 1 0 −2
[E, H] = EH − HE =
−
= = −2E,
0 0 0 −1 0 −1 0 0 0 0
[ ] [ ] [ ] [ ] [ ]
0 0 1 0 1 0 0 0 0 0
[F, H] = F H − HF =
−
= = 2F,
[E, F ] = EF − F E =
1 0
[ 0 1
0 0
0 −1
] [ ] 0 0
−
1 0
0 −1 1 0
] [ ] 0 1
=
0 0
[ 0 0
1 0
2 0
]
= H.
[ 1 0
0 −1
(b) Using the basis {E, H, F } (in that order), find the matrix E = ad(E), the adjoint
of E defined in (9). (The matrix will be 3 × 3.)
Solution: Recall, the adjoint of E is the linear transformation ad(E) : Lie(G) →
Lie(G) given by [ ad(E)(A) ] = [E, A] for A ∈ Lie(G). Now, if A ∈ Lie(G), then we
a b
know that A = for some a, b, c ∈ R. With respect to the ordered basis
c −a
B = {E, H, F },
⎡ ⎤
[ ]
b
a b
A = = bE + aH + cF =⇒ [A]
c −a
B = ⎣a⎦
c
(see Lay, Section 4.4, for a discussion of coordinate systems relative to a basis).
Then the matrix E of ad(E) with respect to the ordered basis will be
E = [ ] [ ]
[ad(E)(E)] B [ad(E)(H)] B [ad(E)(F )] B = [E, E]B [E, H] B [E, F ] B
⎡ ⎤
= [ ]
0 −2 0
[O 2 ] B [−2E] B [H] B = ⎣0 0 1⎦ .
0 0 0

Math 440, Linear Algebra II
Solutions to Homework 9 Problems
(c) Find the matrix of H = ad(H) and the matrix of F = ad(F ), as in (b).
Solution: To find the matrix H of ad(H) with respect to the ordered basis
B = {E, H, F }, note that [H, E] = −[E, H] = −(−2E) = 2E, [H, H] = O 2 and
[H, F ] = −[F, H] = −2F . Hence
H = [ ] [ ]
[ad(H)(E)] B [ad(H)(H)] B [ad(H)(F )] B = [H, E]B [H, H] B [H, F ] B
⎡ ⎤
= [ ]
2 0 0
[2E] B [O 2 ] B [−2F ] B = ⎣0 0 0 ⎦ .
0 0 −2
The matrix F of ad(F ) with respect to B requires we compute [F, E] = −[E, F ] =
−H, [F, H] = 2F and [F, F ] = O 2 , so
F = [ ] [ ]
[ad(F )(E)] B [ad(F )(H)] B [ad(F )(F )] B = [F, E]B [F, H] B [F, F ] B
⎡ ⎤
= [ ]
0 0 0
[−H] B [2F ] B [O 2 ] B = ⎣−1 0 0⎦ .
0 2 0
(d) Find exp(E).
Solution: By definition,
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
∞∑ 1 0 0 0 −2 0
1
exp(E) = I 3 +
k! E k = ⎣0 1 0⎦+
⎣0 0 1⎦+ 1 0 0 −2 1 −2 −1
⎣0 0 0 ⎦ = ⎣0 1 1 ⎦ .
2
k=1 0 0 1 0 0 0 0 0 0 0 0 1
(e) Find exp(H).
Solution: Since H is diagonal, applying the result of Problem 9-2 (see Homework
7), we have
⎡ ⎤
e 2 0 0
exp(H) = ⎣ 0 e 0 0
0 0 e −2
⎡ ⎤
e 2 0 0
⎦ = ⎣ 0 1 0 ⎦ .
0 0 e −2
(f) Find exp(F).
Solution: We’ll use the definition for this one:
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
∞∑ 1 0 0 0 0 0
1
exp(F) = I 3 +
k! F k = ⎣0 1 0⎦+
⎣−1 0 0⎦+ 1 0 0 0 1 0 0
⎣ 0 0 0⎦ = ⎣−1 1 0⎦ .
2
k=1 0 0 1 0 2 0 −2 0 0 −1 2 1
(g) Find M E = exp(E), M H = exp(H) and M F = exp(F ).
Solution: We have
[ ] 1 1
M E = exp(E) = I 2 + E = ,
0 1
[ ] [ ]
e
1
0 e 0
M H = exp(H) =
0 e −1 =
0 e −1 ,
[ ] 1 0
M F = exp(F ) = I 2 + F = .
1 1

Math 440, Linear Algebra II
Solutions to Homework 9 Problems
(h) Because trace(MAM −1 ) = trace(A), Ad(M) : Lie(G) → Lie(G). Find the matrix
of Ad(M E ) with respect to the basis {E, H, F } of Lie(G).
Solution: The matrix of Ad(M E ) with respect to the ordered basis B = {E, H, F }
will be
[Ad(M E )] B = [ ]
[Ad(M E )(E)] B [Ad(M E )(H)] B [Ad(M E )(F )] B
= [ [M E EM −1
E ] B [M E HM −1
E ] B [M E F M −1
E
= [ ]
[E] B [−2E + H] B [−E + H + F ] B
⎡ ⎤
1 −2 −1
= ⎣0 1 1 ⎦ = exp(E).
0 0 1
] B
]
(i) Find the matrix of Ad(M H ) with respect to the basis {E, H, F } of Lie(G).
Solution: The matrix of Ad(M H ) with respect to the ordered basis B = {E, H, F }
will be
[Ad(M H )] B = [ ]
[Ad(M H )(E)] B [Ad(M H )(H)] B [Ad(M H )(F )] B
[M H HM −1
H ] B
= [ [M H EM −1
H ] B
= [ ]
[e 2 E] B [H] B [e −2 F ] B
=
⎡ ⎤
e 2 0 0
⎣ 0 1 0
0 0 e −2
⎦ = exp(H).
[M H F M −1
H ] B
]
(j) Find the matrix of Ad(M F ) with respect to the basis {E, H, F } of Lie(G).
Solution: The matrix of Ad(M F ) with respect to the ordered basis B = {E, H, F }
will be
[Ad(M F )] B = [ ]
[Ad(M F )(E)] B [Ad(M F )(H)] B [Ad(M F )(F )] B
= [ [M F EM −1
F ] B [M F HM −1
F ] B [M F F M −1
F
= [ ]
[E − H − F ] B [H + 2F ] B [F ] B
⎡ ⎤
1 0 0
= ⎣−1 1 0⎦ = exp(F).
−1 2 1
] B
]