f(x, y, z, w) = (y+2z-w; 2x+8y+2z-6w; 2x+7y-5w)
The kernel or null-space of a linear transformation is the set of all the vectors of the input space that are mapped under the linear transformation to the null vector of the output space. To compute the kernel, find the null space of the matrix of the linear transformation, which is the same to find the vector subspace where the implicit equations are the homogeneous equations obtained when the components of the linear transformation formula are equalled to zero.
y+2z-w = 0 2x+8y+2z-6w = 0 2x+7y-5w = 0
┌ ┐ │ 0 1 2 -1 | 0 │ │ 2 8 2 -6 | 0 │ │ 2 7 0 -5 | 0 │ └ ┘
r1 <———> r3 ┌ ┐ │ 2 7 0 -5 | 0 │ │ 2 8 2 -6 | 0 │ │ 0 1 2 -1 | 0 │ └ ┘ r2 <———> r2 - r1 ┌ ┐ │ 2 7 0 -5 | 0 │ │ 0 1 2 -1 | 0 │ │ 0 1 2 -1 | 0 │ └ ┘ r3 <———> r3 - r2 ┌ ┐ │ 2 7 0 -5 | 0 │ │ 0 1 2 -1 | 0 │ │ 0 0 0 0 | 0 │ └ ┘
2x+7y-5w = 0 y+2z-w = 0
The implicit equations of the kernel are the equations obtained in the previous step.
Ker(f) = {(x, y, z, w)∈ R4 | 2x+7y-5w = 0; y+2z-w = 0 }
Select the free variables.
{ z, w }
Back substitution.
y+2z-w = 0 y = -2z+w 2x+7y-5w = 0 x = (-7y+5w)/2 x = (-7•(-2z+w)+5w)/2 x = 7z-w
Parametric representation.
Ker(f) = {(7z-w ; -2z+w ; z ; w) | z, w ∈ R }
For each free variable, give the value 1 to that variable and value 0 to the others, obtaining a vector of the kernel. The set of vectors obtained is a basis for the kernel.
Basis(Ker(f)) = {(-1 1 0 1); (7 -2 1 0)}