Math 440, Linear Algebra II Solutions to Homework 9 Problems 11-5 ...
Math 440, Linear Algebra II Solutions to Homework 9 Problems 11-5 ...
Math 440, Linear Algebra II Solutions to Homework 9 Problems 11-5 ...
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
<strong>Math</strong> <strong>440</strong>, <strong>Linear</strong> <strong>Algebra</strong> <strong>II</strong><br />
<strong>Solutions</strong> <strong>to</strong> <strong>Homework</strong> 9 <strong>Problems</strong><br />
<strong>11</strong>-5. Let L = R 3 . In this problem you will show that L is a Lie algebra via the bracket<br />
operation given by the vec<strong>to</strong>r cross product:<br />
[v, w] = v × w<br />
for v, w ∈ L = R 3 by verifying (a)-(c) of Problem (1). Assume the properties of the<br />
vec<strong>to</strong>r cross product from linear algebra, including the fact that<br />
u × (v × w) = (u · w)v − (u · v)w.<br />
Proof. By definition, a Lie algebra is a vec<strong>to</strong>r space L <strong>to</strong>gether with a binary operation<br />
[ , ] : L × L → L that satisfies the axioms<br />
(L1) the bracket product is bilinear;<br />
(L2) the bracket product is skew-symmetric;<br />
(L3) the bracket satisfies the Jacobi identity.<br />
In this problem, we are given a vec<strong>to</strong>r space L = R 3 and a binary operation on R 3 ,<br />
namely the cross product of vec<strong>to</strong>rs as defined in Calculus 3:<br />
[v, w] := v × w.<br />
Therefore, <strong>to</strong> prove that L = R 3 with the cross product is a Lie algebra, it suffices<br />
<strong>to</strong> show that the cross product satisfies axioms (L1), (L2) and (L3) above. First,<br />
from Calculus 3, we recall the properties of the cross product: (see Stewart, Calculus:<br />
Concepts & Contexts, Section 9.4):<br />
Proposition (Properties of the Cross Product). Let u, v, w ∈ R 3 and α, β ∈ R.<br />
(a) v × w = −(w × v)<br />
(b) v × v = ⃗0<br />
(c) (αv) × (βw) = (αβ)(v × w)<br />
(d) u × (v + w) = (u × v) + (u × w) and (u + v) × w = (u × w) + (v × w).<br />
Therefore, the cross product is bilinear and skew-symmetric by this Proposition (bilinear<br />
by parts (c) and (d), skew-symmetric by parts (a) and (b)). Hence it only remains<br />
<strong>to</strong> prove that the cross product satisfies the Jacobi identity:<br />
[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0.<br />
To show this, recall that the cross product satisfies the equation<br />
a × (b × c) = (a · c)b − (a · b)c<br />
(this is a Proposition in Section 0.1 of the notes). Thus, for u, v, w ∈ R 3 , consider<br />
u × (v × w) + v × (w × u) + w × (u × v)<br />
= [(u · w)v − (u · v)w] + [(v · u)w − (v · w)u] + [(w · v)u − (w · u)v]<br />
= [−(v · w) + (w · v)]u + [(u · w) − (w · u)]v + [−(u · v) + (v · u)]w<br />
= ⃗0.<br />
Therefore the cross product satisfies the Jacobi identity, so L = R 3 with the cross<br />
product is a Lie algebra.
<strong>Math</strong> <strong>440</strong>, <strong>Linear</strong> <strong>Algebra</strong> <strong>II</strong><br />
<strong>Solutions</strong> <strong>to</strong> <strong>Homework</strong> 9 <strong>Problems</strong><br />
<strong>11</strong>-6. In this problem you will look at an example of a Lie algebra homomorphism. Let G be<br />
the Euclidean orthogonal group O(3, R), so Lie(G) consists of all 3×3 skew-symmetric<br />
matrices. Let L = R 3 be the Lie algebra of Problem (5). Define Φ : L → Lie(G) by<br />
⎡<br />
0 −x<br />
⎤<br />
−y<br />
Φ(v) = Φ(x, y, z) = ⎣x 0 −z⎦<br />
y z 0<br />
where v = (x, y, z) ∈ L = R 3 . (You’ll need <strong>to</strong> work with coordinates.)<br />
(a) Show that Φ is a linear transformation.<br />
Proof. To prove that Φ is a linear transformation, we must show that Φ(v +w) =<br />
Φ(v) + Φ(w) and that Φ(αv) = αΦ(v) for all v, w ∈ L and all α ∈ R. Hence,<br />
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 ) =<br />
(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<br />
⎡<br />
⎤<br />
0 −x 1 − x 2 −y 1 − y 2<br />
Φ(x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) = ⎣x 1 + x 2 0 −z 1 − z 2<br />
⎦<br />
y 1 + y 2 z 1 + z 2 0<br />
⎡<br />
⎤ ⎡<br />
⎤<br />
0 −x 1 −y 1 0 −x 2 −y 2<br />
= ⎣x 1 0 −z 1<br />
⎦ + ⎣x 2 0 −z 2<br />
⎦<br />
y 1 z 1 0 y 2 z 2 0<br />
= Φ(x 1 , y 1 , z 1 ) + Φ(x 2 , y 2 , z 2 ),<br />
⎡<br />
⎤<br />
0 −(αx 1 ) −(αy 1 )<br />
Φ(αx 1 , αy 1 , αz 1 ) = ⎣(αx 1 ) 0 −(αz 1 ) ⎦<br />
(αy 1 ) (αz 1 ) 0<br />
⎡<br />
⎤<br />
0 −x 1 −y 1<br />
= α ⎣x 1 0 −z 1<br />
⎦<br />
y 1 z 1 0<br />
Therefore, Φ is a linear transformation.<br />
(b) Show that Φ is one-<strong>to</strong>-one and on<strong>to</strong>.<br />
= αΦ(x 1 , y 1 , z 1 ).<br />
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<br />
since matrix equality is determined entry-wise. Hence Φ is one-<strong>to</strong>-one. Since<br />
Lie(G) = {B ∈ M(3, R) : B T + B = O 3 } implies that for each B ∈ Lie(G), if<br />
B = [b ij ] then b ii = 0 and b ij = −b ji for 1 ≤ i < j ≤ 3. Hence each B ∈ Lie(G) is<br />
determined by the values b 12 , b 13 , b 23 , i.e.,<br />
⎡<br />
0 b 12<br />
⎤<br />
b 13<br />
B = ⎣−b 12 0 b 23<br />
⎦ = Φ(−b 12 , −b 13 , −b 23 )<br />
−b 13 −b 23 0<br />
is in the image of Φ. Therefore, Φ is on<strong>to</strong>.
<strong>Math</strong> <strong>440</strong>, <strong>Linear</strong> <strong>Algebra</strong> <strong>II</strong><br />
<strong>Solutions</strong> <strong>to</strong> <strong>Homework</strong> 9 <strong>Problems</strong><br />
(c) Show that Φ preserves the Lie bracket. That is, show<br />
[Φ(v), Φ(w)] = Φ([v, w])<br />
for any v, w ∈ L = R 3 . Remember that the bracket [v, w] is given by the vec<strong>to</strong>r<br />
cross product, while the bracket [Φ(v), Φ(w)] is the usual bracket operation for<br />
matrices.<br />
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 =<br />
(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) =<br />
[Φ(v), Φ(w)], where this second bracket is the Lie bracket for matrices, [A, B] =<br />
AB − BA. Thus consider<br />
[Φ(v), Φ(w)] = [Φ(v 1 , v 2 , v 3 ), Φ(w 1 , w 2 , w 3 )]<br />
= 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 )<br />
⎡<br />
⎤ ⎡<br />
⎤<br />
0 −v 1 −v 2 0 −w 1 −w 2<br />
= ⎣v 1 0 −v 3<br />
⎦ ⎣w 1 0 −w 3<br />
⎦<br />
v 2 v 3 0 w 2 w 3 0<br />
⎡<br />
⎤ ⎡<br />
⎤<br />
0 −w 1 −w 2 0 −v 1 −v 2<br />
− ⎣w 1 0 −w 3<br />
⎦ ⎣v 1 0 −v 3<br />
⎦<br />
w 2 w 3 0 v 2 v 3 0<br />
⎡<br />
⎤<br />
−v 1 w 1 − v 2 w 2 −v 2 w 3 v 1 w 3<br />
= ⎣ −v 3 w 2 −v 1 w 1 − v 3 w 3 −v 1 w 2<br />
⎦<br />
v 3 w 1 −v 2 w 1 −v 2 w 2 − v 3 w 3<br />
⎡<br />
⎤<br />
−w 1 v 1 − w 2 v 2 −w 2 v 3 w 1 v 3<br />
− ⎣ −w 3 v 2 −w 1 v 1 − w 3 v 3 −w 1 v 2<br />
⎦<br />
w 3 v 1 −w 2 v 1 −w 2 v 2 − w 3 v 3<br />
⎡<br />
⎤<br />
0 v 3 w 2 − v 2 w 3 v 1 w 3 − v 3 w 1<br />
= ⎣v 2 w 3 − v 3 w 2 0 v 2 w 1 − v 1 w 2<br />
⎦<br />
v 3 w 1 − v 1 w 3 v 1 w 2 − v 2 w 1 0<br />
= 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 )<br />
= Φ(v × w).<br />
Therefore, as Φ is a linear transformation (by part (a)) and Φ preserves the Lie<br />
brackets, Φ is a Lie algebra homomorphism. Furthermore, since Φ is both one<strong>to</strong>-one<br />
and on<strong>to</strong> (by part (b)), Φ is a Lie algebra isomorphism.<br />
<strong>11</strong>-9. This is a different approach <strong>to</strong> Proposition 4.1.3(c). (If we define ad as the differential<br />
of Ad, then we already know ad is a linear transformation.) Assume G is a Lie group<br />
and Lie(G) is its tangent space. Assume a bracket operation has been defined on Lie(G)<br />
(so Lie(G) is closed under the bracket operation) satisfying the properties (a)-(c) in<br />
Problem (1). For a fixed A ∈ Lie(G), define ad(A) : Lie(G) → Lie(G) by<br />
ad(A)(B) = [A, B].<br />
Use this definition in terms of the bracket and the properties in (1) <strong>to</strong> show ad(A) is<br />
a linear transformation.
<strong>Math</strong> <strong>440</strong>, <strong>Linear</strong> <strong>Algebra</strong> <strong>II</strong><br />
<strong>Solutions</strong> <strong>to</strong> <strong>Homework</strong> 9 <strong>Problems</strong><br />
Proof. Let A ∈ Lie(G), so ad(A) : Lie(G) → Lie(G) defined by ad(A)(B) = [A, B]<br />
is at least a function from Lie(G) <strong>to</strong> itself. To prove that this mapping is a linear<br />
transformation, we only need <strong>to</strong> prove that for all B, C ∈ Lie(G) and all scalars r ∈ R,<br />
we have ad(A)(B + C) = ad(A)(B) + ad(A)(C) and ad(A)(rB) = r[ad(A)(B)]. Thus<br />
consider, for B, C ∈ Lie(G) and r ∈ R, using Properties of the bilinearity of the<br />
bracket:<br />
ad(A)(B + C) = [A, B + C] = [A, B] + [A, C] = ad(A)(B) + ad(A)(C),<br />
adA(rB) = [A, rB] = r[A, B] = r[ad(A)(B)].<br />
Therefore, ad(A) is a linear transformation.<br />
<strong>11</strong>-10. In this problem you will take a careful look at the Lie algebra of G = SL(2, R), using<br />
the following basis for Lie(G):<br />
[ ] [ ] [ ]<br />
0 1 1 0 0 0<br />
E = , H = , F = .<br />
0 0 0 −1 1 0<br />
(a) Find [E, H], [F, H] and [E, F ] (where the bracket operation is the one defined in<br />
(1)).<br />
Solution: For E, H, F as above, we have<br />
[ ] [ ] [ ] [ ] [ ]<br />
0 1 1 0 1 0 0 1 0 −2<br />
[E, H] = EH − HE =<br />
−<br />
= = −2E,<br />
0 0 0 −1 0 −1 0 0 0 0<br />
[ ] [ ] [ ] [ ] [ ]<br />
0 0 1 0 1 0 0 0 0 0<br />
[F, H] = F H − HF =<br />
−<br />
= = 2F,<br />
[E, F ] = EF − F E =<br />
1 0<br />
[ 0 1<br />
0 0<br />
0 −1<br />
] [ ] 0 0<br />
−<br />
1 0<br />
0 −1 1 0<br />
] [ ] 0 1<br />
=<br />
0 0<br />
[ 0 0<br />
1 0<br />
2 0<br />
]<br />
= H.<br />
[ 1 0<br />
0 −1<br />
(b) Using the basis {E, H, F } (in that order), find the matrix E = ad(E), the adjoint<br />
of E defined in (9). (The matrix will be 3 × 3.)<br />
Solution: Recall, the adjoint of E is the linear transformation ad(E) : Lie(G) →<br />
Lie(G) given by [ ad(E)(A) ] = [E, A] for A ∈ Lie(G). Now, if A ∈ Lie(G), then we<br />
a b<br />
know that A = for some a, b, c ∈ R. With respect <strong>to</strong> the ordered basis<br />
c −a<br />
B = {E, H, F },<br />
⎡ ⎤<br />
[ ]<br />
b<br />
a b<br />
A = = bE + aH + cF =⇒ [A]<br />
c −a<br />
B = ⎣a⎦<br />
c<br />
(see Lay, Section 4.4, for a discussion of coordinate systems relative <strong>to</strong> a basis).<br />
Then the matrix E of ad(E) with respect <strong>to</strong> the ordered basis will be<br />
E = [ ] [ ]<br />
[ad(E)(E)] B [ad(E)(H)] B [ad(E)(F )] B = [E, E]B [E, H] B [E, F ] B<br />
⎡ ⎤<br />
= [ ]<br />
0 −2 0<br />
[O 2 ] B [−2E] B [H] B = ⎣0 0 1⎦ .<br />
0 0 0
<strong>Math</strong> <strong>440</strong>, <strong>Linear</strong> <strong>Algebra</strong> <strong>II</strong><br />
<strong>Solutions</strong> <strong>to</strong> <strong>Homework</strong> 9 <strong>Problems</strong><br />
(c) Find the matrix of H = ad(H) and the matrix of F = ad(F ), as in (b).<br />
Solution: To find the matrix H of ad(H) with respect <strong>to</strong> the ordered basis<br />
B = {E, H, F }, note that [H, E] = −[E, H] = −(−2E) = 2E, [H, H] = O 2 and<br />
[H, F ] = −[F, H] = −2F . Hence<br />
H = [ ] [ ]<br />
[ad(H)(E)] B [ad(H)(H)] B [ad(H)(F )] B = [H, E]B [H, H] B [H, F ] B<br />
⎡ ⎤<br />
= [ ]<br />
2 0 0<br />
[2E] B [O 2 ] B [−2F ] B = ⎣0 0 0 ⎦ .<br />
0 0 −2<br />
The matrix F of ad(F ) with respect <strong>to</strong> B requires we compute [F, E] = −[E, F ] =<br />
−H, [F, H] = 2F and [F, F ] = O 2 , so<br />
F = [ ] [ ]<br />
[ad(F )(E)] B [ad(F )(H)] B [ad(F )(F )] B = [F, E]B [F, H] B [F, F ] B<br />
⎡ ⎤<br />
= [ ]<br />
0 0 0<br />
[−H] B [2F ] B [O 2 ] B = ⎣−1 0 0⎦ .<br />
0 2 0<br />
(d) Find exp(E).<br />
Solution: By definition,<br />
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤<br />
∞∑ 1 0 0 0 −2 0<br />
1<br />
exp(E) = I 3 +<br />
k! E k = ⎣0 1 0⎦+<br />
⎣0 0 1⎦+ 1 0 0 −2 1 −2 −1<br />
⎣0 0 0 ⎦ = ⎣0 1 1 ⎦ .<br />
2<br />
k=1 0 0 1 0 0 0 0 0 0 0 0 1<br />
(e) Find exp(H).<br />
Solution: Since H is diagonal, applying the result of Problem 9-2 (see <strong>Homework</strong><br />
7), we have<br />
⎡ ⎤<br />
e 2 0 0<br />
exp(H) = ⎣ 0 e 0 0<br />
0 0 e −2<br />
⎡ ⎤<br />
e 2 0 0<br />
⎦ = ⎣ 0 1 0 ⎦ .<br />
0 0 e −2<br />
(f) Find exp(F).<br />
Solution: We’ll use the definition for this one:<br />
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤<br />
∞∑ 1 0 0 0 0 0<br />
1<br />
exp(F) = I 3 +<br />
k! F k = ⎣0 1 0⎦+<br />
⎣−1 0 0⎦+ 1 0 0 0 1 0 0<br />
⎣ 0 0 0⎦ = ⎣−1 1 0⎦ .<br />
2<br />
k=1 0 0 1 0 2 0 −2 0 0 −1 2 1<br />
(g) Find M E = exp(E), M H = exp(H) and M F = exp(F ).<br />
Solution: We have<br />
[ ] 1 1<br />
M E = exp(E) = I 2 + E = ,<br />
0 1<br />
[ ] [ ]<br />
e<br />
1<br />
0 e 0<br />
M H = exp(H) =<br />
0 e −1 =<br />
0 e −1 ,<br />
[ ] 1 0<br />
M F = exp(F ) = I 2 + F = .<br />
1 1
<strong>Math</strong> <strong>440</strong>, <strong>Linear</strong> <strong>Algebra</strong> <strong>II</strong><br />
<strong>Solutions</strong> <strong>to</strong> <strong>Homework</strong> 9 <strong>Problems</strong><br />
(h) Because trace(MAM −1 ) = trace(A), Ad(M) : Lie(G) → Lie(G). Find the matrix<br />
of Ad(M E ) with respect <strong>to</strong> the basis {E, H, F } of Lie(G).<br />
Solution: The matrix of Ad(M E ) with respect <strong>to</strong> the ordered basis B = {E, H, F }<br />
will be<br />
[Ad(M E )] B = [ ]<br />
[Ad(M E )(E)] B [Ad(M E )(H)] B [Ad(M E )(F )] B<br />
= [ [M E EM −1<br />
E ] B [M E HM −1<br />
E ] B [M E F M −1<br />
E<br />
= [ ]<br />
[E] B [−2E + H] B [−E + H + F ] B<br />
⎡ ⎤<br />
1 −2 −1<br />
= ⎣0 1 1 ⎦ = exp(E).<br />
0 0 1<br />
] B<br />
]<br />
(i) Find the matrix of Ad(M H ) with respect <strong>to</strong> the basis {E, H, F } of Lie(G).<br />
Solution: The matrix of Ad(M H ) with respect <strong>to</strong> the ordered basis B = {E, H, F }<br />
will be<br />
[Ad(M H )] B = [ ]<br />
[Ad(M H )(E)] B [Ad(M H )(H)] B [Ad(M H )(F )] B<br />
[M H HM −1<br />
H ] B<br />
= [ [M H EM −1<br />
H ] B<br />
= [ ]<br />
[e 2 E] B [H] B [e −2 F ] B<br />
=<br />
⎡ ⎤<br />
e 2 0 0<br />
⎣ 0 1 0<br />
0 0 e −2<br />
⎦ = exp(H).<br />
[M H F M −1<br />
H ] B<br />
]<br />
(j) Find the matrix of Ad(M F ) with respect <strong>to</strong> the basis {E, H, F } of Lie(G).<br />
Solution: The matrix of Ad(M F ) with respect <strong>to</strong> the ordered basis B = {E, H, F }<br />
will be<br />
[Ad(M F )] B = [ ]<br />
[Ad(M F )(E)] B [Ad(M F )(H)] B [Ad(M F )(F )] B<br />
= [ [M F EM −1<br />
F ] B [M F HM −1<br />
F ] B [M F F M −1<br />
F<br />
= [ ]<br />
[E − H − F ] B [H + 2F ] B [F ] B<br />
⎡ ⎤<br />
1 0 0<br />
= ⎣−1 1 0⎦ = exp(F).<br />
−1 2 1<br />
] B<br />
]