68 CHAPTER 2. 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

(ui)2)1/2

= |u| .

Lemma 2.8.3 Suppose Q (t) is a real, differentiable n×n matrix which preserves distances.

Then Q (t)Q (t)T= Q (t)

TQ (t) = I. Also, if u (t) ≡ Q (t)u, then there exists a vector, Ω (t)

such thatu′ (t) = Ω (t)× u (t) .

The symbol × refers to the cross product.

Proof: Recall that (z ·w) = 14

(|z+w|2 − |z−w|2

). Therefore,

(Q (t)u·Q (t)w) =1

4

(|Q (t) (u+w)|2 − |Q (t) (u−w)|2

)=

1

4

(|u+w|2 − |u−w|2

)= (u ·w) .

This implies (Q (t)

TQ (t)u ·w

)= (u ·w)

for all u,w. Therefore, Q (t)TQ (t)u = u and so Q (t)

TQ (t) = Q (t)Q (t)

T= I. This proves

the first part of the lemma.It follows from the product rule, Lemma 2.8.2 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

. (2.28)

From the definition, Q (t)u = u (t) ,

u′ (t) = Q′ (t)u =Q′ (t)

=u︷ ︸︸ ︷Q (t)

Tu (t).

Then writing the matrix of Q′ (t)Q (t)T

with respect to fixed in space orthonormal basisvectors, i∗, j∗,k∗, where these are the usual basis vectors for R3, it follows from 2.28 thatthe matrix of Q′ (t)Q (t)

Tis of the form 0 −ω3 (t) ω2 (t)

ω3 (t) 0 −ω1 (t)

−ω2 (t) ω1 (t) 0

for some time dependent scalars ωi. Therefore, u1

u2

u3

′

(t)=

 0 −ω3 (t) ω2 (t)

ω3 (t) 0 −ω1 (t)

−ω2 (t) ω1 (t) 0

 u1

u2

u3

 (t)