112 CHAPTER 5. LINEAR TRANSFORMATIONS

Thus in terms of the basis γ ≡ {u1,u2,u3} , the matrix of this transformation is

[T ]γ≡

 cosθ −sinθ 0sinθ cosθ 0

0 0 1

 .

This is not desirable because it involves a funny basis. I want to obtain the matrix of thetransformation in terms of the usual basis β ≡ {e1,e2,e3} because it is in terms of thisbasis that we usually deal with vectors in R3. From Proposition 5.2.9, if [T ]

βis this matrix, cosθ −sinθ 0

sinθ cosθ 00 0 1

=

(u1 u2 u3

)−1[T ]

β

(u1 u2 u3

)and so you can solve for [T ]

βif you know the ui.

Recall why this is so.R3 [T ]

γ−−→R3

qγ ↓ ◦ qγ ↓R3 T−→ R3

I ↑ ◦ I ↑R3 [T ]

β−−→R3

The map qγ is accomplished by a multiplication on the left by(

u1 u2 u3

). Thus

[T ]β= qγ [T ]γ q−1

γ =(

u1 u2 u3

)[T ]

γ

(u1 u2 u3

)−1.

Suppose the unit vector u3 about which the counterclockwise rotation takes place is(a,b,c). Then I obtain vectors, u1 and u2 such that {u1,u2,u3} is a right handed or-thonormal system with u3 = (a,b,c) and then use the above result. It is of course somewhatarbitrary how this is accomplished. I will assume however, that |c| ̸= 1 since otherwise youare looking at either clockwise or counter clockwise rotation about the positive z axis andthis is a problem which is fairly easy. Indeed, the matrix of such a rotation in terms of theusual basis is just  cosθ −sinθ 0

sinθ cosθ 00 0 1

 (5.6)

Then let u3 = (a,b,c) and u2 ≡ 1√a2+b2

(b,−a,0) . This one is perpendicular to u3. If

{u1,u2,u3} is to be a right hand system it is necessary to have

u1 = u2×u3 =1√

(a2 +b2)(a2 +b2 + c2)

(−ac,−bc,a2 +b2)