1. Show that f is the function form R3 to R3 which rotates every vector A counterclockwise about some axis L. In the book, we see if the axis L is the unit vector u = (a,b,c) then the standard matrix for the function f is

a). Prove that this transformation is linear.

By definition, a transformation is linear if and only if 
 
             (1).   f(u+v) = f(u) + f(v)             (1) 
             (2).   f(cu) = cf(u).                   (2)

In order to prove that these properties hold for the rotation we shall use geometry. If we take the sum of two vectors and then rotate the resulting vector, it will be the same as rotating the two vectors and then taking the sum. This is because the angle between the two vectors does not change when they are rotated separately, since they are rotated through the same angle, and their lengths do not change.

The second property also applies, because if you multiply the vector by a scalar the result will be parallel to the original vector. So, when you rotate the vector A it will be parallel to the same line as the vector rotated without multiplying the vector A by a scalar. The magnitude will be the same for either method. Since magnitude is not changed during rotation, it will be the same if you multiply it by a scalar before or after the rotation.

b). Show that the column-vectors in the standard matrix A of f are pairwise orthogonal and their lengths are equal to 1. We can write the standard matrix, T, for any linear transformation as [T] = [T(v1) | T(v2) | T(v3)], where v1, v2, v3 are standard basis vectors. Standard basis vectors are defined as the vectors of length 1 along the coordinate axes. The coordinate axes are always pairwise orthogonal. The column-vectors, T(v1), T(v2), T(v3), are transformed basis vectors. If we apply the linear transformation [T] to the standard basis vectors, they will remain pairwise orthogonal because each vector is rotated through the same angle around the same axis. Their lengths do not change because [T] is a simple rotation. So, the column-vectors of the standard matrix are pairwise orthogonal and their lengths are all 1.

2. Find the standard matrix of the linear transformation of R3 which reflects every vector V about the plane

ax+by+cz=0

We know that the normal vector of this plane will have components identical to the coefficients from the equation for the plane. Thus, we can define a normal vector to the plane (along which we will want to project the head of the vector v) to be k(a,b,c) for some constant k. We apply this constant k to scale the vector in such a way as to have it be the vector from the plane to the head of vector v. Thus we have a triad of vectors: v (the original vector), w (the vector normal to the plane defined as k(a,b,c) ) and p (the vector that is the projection of v on the plane, and is v-w). We also know that vectors p and w are orthogonal to each other. We are trying to find the vector q = v-2w. Let us first, though, solve for k. 
 
p=v-w 
p=(x-ka,y-kb,z-kc) 
p·(a,b,c) = 0 
(x-ka,y-kb,z-kc)·(a,b,c) = 0 
ax-ka2 + by-kb2 + cz - kc2 = 0 
(ax + by + cz) - k(a2+b2+c2) = 0 
(ax + by + cz) = k(a2+b2+c2) 
k = (ax + by + cz)/(a2+b2+c2)    
 
q =  v-2k(a,b,c) 
q =  (x-2ka, y-2kb, z-2kc)  
 
put as a function, 

f(V) = (x-2a(ax + by + cz)/(a2+b2+c2),   
            y-2b(ax + by + cz)/(a2+b2+c2),   
            z-2c(ax + by + cz)/(a2+b2+c2))

The coefficients of x, y, z in this formula form the standard matrix.

So the standard matrix is:

> A:=matrix([[1-2*a^2/(a^2+b^2+c^2),-2*b*a/(a^2+b^2+c^2),-2*c*a/(a^2+b^2+c^2)],[-2*a*b/(a^2+b^2+c^2),1-2*b^2/(a^2+b^2+c^2),-2*c*b/(a^2+b^2+c^2)],[-2*a*c/(a^2+b^2+c^2),-2*b*c/(a^2+b^2+c^2),1-2*c^2/(a^2+b^2+c^2)]]);
A := [ 1-2a2 / %1 -2ba / %1 -2ca / %1 ]
[ -2ba / %1 1-2b2 / %1 -2cb / %1 ]
[ -2ca / %1 -2cb / %1 1-2c2 / %1 ] 

%1 := 				a2 + b2 + c2