8.3. THE MATRIX OF A LINEAR TRANSFORMATION 225

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

[T ]γ ≡

 cos θ − sin θ 0

sin θ cos θ 0

0 0 1

 .

I want to obtain the matrix of the transformation in terms of the usual basis β ≡ {e1, e2, e3}because it is in terms of this basis that we usually deal with vectors. From Proposition 8.3.10,if [T ]β is this matrix,  cos θ − sin θ 0

sin θ cos θ 0

0 0 1

=

(u1 u2 u3

)−1

[T ]β

(u1 u2 u3

)and so you can solve for [T ]β if 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−1γ =

(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 orthonor-mal 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 has been dealt with earlier. (If c = −1, it amounts to clockwiserotation about the positive z axis while if c = 1, it is counter clockwise rotation about thepositive z axis.)

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

)Now recall that u3 is a unit vector and so the above equals

1√(a2 + b2)

(−ac,−bc, a2 + b2

)Then from the above, A is given by

−ac√(a2+b2)

b√a2+b2

a

−bc√(a2+b2)

−a√a2+b2

b√a2 + b2 0 c

 cos θ − sin θ 0

sin θ cos θ 0

0 0 1



−ac√(a2+b2)

b√a2+b2

a

−bc√(a2+b2)

−a√a2+b2

b√a2 + b2 0 c

−1

8.3. THE MATRIX OF A LINEAR TRANSFORMATION 225Thus in terms of the basis y = {u,, ug, us}, the matrix of this transformation iscos? —sin@ 0[T],= | sin? cosd 00 0 1I want to obtain the matrix of the transformation in terms of the usual basis 6 = {e1, e2, e3}because it is in terms of this basis that we usually deal with vectors. From Proposition 8.3.10,if [7], is this matrix,cos@ —sind 0sin? cos@d 0O0 0 1= (m ug U3 ) ts (m u2 us )and so you can solve for [T], if you know the uj.Recall why this is so.R? |T R?,dy { ° qy aR3 TT. R?It ° I*tR? [T], R?—sThe map q, is accomplished by a multiplication on the left by ( Uu;, U2 U3 ) . ThusT= Tl, ay' = (m1 u2 us ) [7], ( m U2 Us yoSuppose the unit vector us about which the counterclockwise rotation takes place is(a, b,c). Then I obtain vectors, u; and ug such that {uj, ug, us} is a right handed orthonor-mal system with ug = (a,b,c) and then use the above result. It is of course somewhatarbitrary how this is accomplished. I will assume however, that |c| 4 1 since otherwise youare looking at either clockwise or counter clockwise rotation about the positive z axis andthis is a problem which has been dealt with earlier. (If ¢ = —1, it amounts to clockwiserotation about the positive z axis while if c = 1, it is counter clockwise rotation about thepositive z axis.)Then let u3 = (a,b,c) and ug = Ta (b, —a,0). This one is perpendicular to ug. If{uj, Ug, us} is to be a right hand system it is necessary to have1J (a? + b?) (a? + 6? + c?)Now recall that ug is a unit vector and so the above equals1(+R)Then from the above, A is given byU, = Ug X UZ = (—ac, —be, a? + b*)(—ae, —be, a? + b*)—ac b —ac bV(a2-+b?) erp © cos? —sin@ O V(ar+e2) Va? +0? a—be —a . —be —aWarsi ae b sinO cos@d 0 Waasty Vath bava? + b? 0 c 0 0 va? + b2 0 c