Linear Algebra, Theory And Applications, 2012a

64 MATRICES AND LINEAR TRANSFORMATIONS
showing that Q (t) is a linear transformation. Also, Q (t) preserves all distances because,
since the vectors, i (t) , j (t) , k (t) form an orthonormal set,
|Q (t) u| =
( 3∑
i=1
(
u
i ) ) 1/2
2
= |u| .
Lemma 2.6.3 Suppose Q (t) is a real, differentiable n×n matrix which preserves distances.
Then Q (t) Q (t) T = Q (t) T Q (t) =I. Also, if u (t) ≡ Q (t) u, then there exists a vector, Ω (t)
such that
u ′ (t) =Ω (t) × u (t) .
The symbol × refers to the cross product.
(
Proof: Recall that (z · w) = 1 4
|z + w| 2 −|z − w| 2) . Therefore,
(Q (t) u·Q (t) w) = 1 (|Q (t)(u + w)| 2 −|Q (t)(u − w)| 2)
4
= 1 (|u + w| 2 −|u − w| 2)
4
= (u · w) .
This implies (
)
Q (t) T Q (t) u · w =(u · w)
for all u, w. Therefore, Q (t) T Q (t) u = u and so Q (t) T Q (t) =Q (t) Q (t) T = I. This proves
the first part of the lemma.
It follows from the product rule, Lemma 2.6.2 that
Q ′ (t) Q (t) T + Q (t) Q ′ (t) T =0
and so
(
)
Q ′ (t) Q (t) T = − Q ′ (t) Q (t) T T
. (2.26)
From the definition, Q (t) u = u (t) ,
=u
{ }} {
u ′ (t) =Q ′ (t) u =Q ′ (t) Q (t) T u (t).
Then writing the matrix of Q ′ (t) Q (t) T with respect to fixed in space orthonormal basis
vectors, i ∗ , j ∗ , k ∗ , where these are the usual basis vectors for R 3 , it follows from (2.26) that
the matrix of Q ′ (t) Q (t) T is of the form
⎛
0 −ω 3 (t) ω 2 (t)
⎞
⎝ ω 3 (t) 0 −ω 1 (t) ⎠
−ω 2 (t) ω 1 (t) 0
for some time dependent scalars ω i . Therefore,
⎛ ⎞
⎝ u1
′ ⎛
0 −ω 3 (t) ω 2 (t)
u 2 ⎠ (t)= ⎝ ω 3 (t) 0 −ω 1 (t)
u 3
−ω 2 (t) ω 1 (t) 0
⎞ ⎛
⎠
⎝ u1
u 2
u 3
⎞
⎠ (t)