Find the kernel and range (Image) of the following transformation: phi: R^3 rightarrow R^2 phi (x, y, z) = (2x - y + 3z, 3x + 2y - z) Solution Let, (x,y,z) be in kernel So, phi(x,y,z)=0=(2x-y+3z,3x+2y-z) 2x-y+3z=0, 3x+2y-z=0 So, y=2x+3z 3x+2(2x+3z)-z=0 7x+5z=0 x=-5z/7 y=2x+3z=-10z/7+3z=11z/7 So, (x,y,z)=(-5z/7,11z/7,z)=z/7(-5,11,7) Hence, kernel of phi ={(-5,11,7)t: t is a real number} phi is a map from R3 to R2 So, by rank nullity theorem dim kernel +dim range =3 kernel is spanned by 1 non zero vector . Hence, dim kernel =1 So, dim range =2 Now range is a subset of R2 and R2 has dimension 2 And range has dimension 2. So Range is R2 itself..

1. Find the kernel and range (Image) of the following transformation: phi: R^3 rightarrow R^2
phi (x, y, z) = (2x - y + 3z, 3x + 2y - z)
Solution
Let, (x,y,z) be in kernel
So, phi(x,y,z)=0=(2x-y+3z,3x+2y-z)
2x-y+3z=0,
3x+2y-z=0
So, y=2x+3z
3x+2(2x+3z)-z=0
7x+5z=0
x=-5z/7
y=2x+3z=-10z/7+3z=11z/7
So, (x,y,z)=(-5z/7,11z/7,z)=z/7(-5,11,7)
Hence, kernel of phi ={(-5,11,7)t: t is a real number}
phi is a map from R3 to R2
So, by rank nullity theorem
dim kernel +dim range =3
kernel is spanned by 1 non zero vector . Hence, dim kernel =1
So, dim range =2
Now range is a subset of R2 and R2 has dimension 2
And range has dimension 2. So Range is R2 itself.