Nonlinear; it fails to be homogeneous: f(2,2)=(8,2), which is not equal to 2*f(1,1)=(8,4).
Linear:
f(a*u + b*v)
=[a*u1+b*v1 + 3*(a*u2+b*v2), 3*(a*u1+b*v1) - 2*(a*u2+b*v2)] =a[u1 + 3*u2, 3*u1 - 2*u2] + a[u1 + 3*u2, 3*u1 - 2*u2]=a*f(u) + b*f(v)
The function f(X)=AX+b is affine, but not linear. We
must first find the constant vector b before we find
the matrix A. Since f(0)=A*0 + b=b, we see that
b=[1,1]. Thus, AX=f(X)-[1,1]. From here, the
kth column rule applies, so the first column of
A is
f(ê1)= 2 1 1
- =
0 1 -1
and similarly for the other columns. Thus,
A = 1 -1 -2
-1 1 0
Reflection through the x2-x3 plane means that x1 --> -x1. (See 3.2.22). Since f is linear, we only need the three columns f(ê1), f(ê2), and f(ê3). These are just
f(ê1)=-ê1 f(ê2)=ê2 f(ê3)=ê3Thus the matrix is
A=
-1 0 0
0 1 0
0 0 1