5.4. EXERCISES 113

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

 cosθ −sinθ 0

sinθ cosθ 00 0 1



−ac√(a2+b2)

b√a2+b2

a

−bc√(a2+b2)

−a√a2+b2

b√

a2 +b2 0 c

−1

It is easy to take the inverse of this matrix on the left. You can check right away that its in-verse is nothing but its transpose. Then doing the computation and then some simplificationyields

=

 a2 +(1−a2

)cosθ ab(1− cosθ)− csinθ ac(1− cosθ)+bsinθ

ab(1− cosθ)+ csinθ b2 +(1−b2

)cosθ bc(1− cosθ)−asinθ

ac(1− cosθ)−bsinθ bc(1− cosθ)+asinθ c2 +(1− c2

)cosθ

 .

(5.7)With this, it is clear how to rotate clockwise about the unit vector, (a,b,c) . Just rotate

counter clockwise through an angle of −θ . Thus the matrix for this clockwise rotation isjust

=

 a2 +(1−a2

)cosθ ab(1− cosθ)+ csinθ ac(1− cosθ)−bsinθ

ab(1− cosθ)− csinθ b2 +(1−b2

)cosθ bc(1− cosθ)+asinθ

ac(1− cosθ)+bsinθ bc(1− cosθ)−asinθ c2 +(1− c2

)cosθ

 .

In deriving 5.7 it was assumed that c ̸=±1 but even in this case, it gives the correct an-swer. Suppose for example that c = 1 so you are rotating in the counter clockwise directionabout the positive z axis. Then a,b are both equal to zero and 5.7 reduces to 5.6.

5.4 Exercises1. If A,B, and C are each n× n matrices and ABC is invertible, why are each of A,B,

and C invertible?

2. Give an example of a 3× 2 matrix with the property that the linear transformationdetermined by this matrix is one to one but not onto.

3. Explain why Ax= 0 always has a solution whenever A is a linear transformation.

4. Recall that a line inRn is of the form x+tv where t ∈R. Recall that v is a “directionvector”. Show that if T : Rn→ Rm is linear, then the image of T is either a line or apoint.

5. In the following examples, a linear transformation, T is given by specifying its actionon a basis β . Find its matrix with respect to this basis.