### Kernel and Null Space

##### 1. Kernel

The definition of the kernel of a linear transformation between euclidean spaces can be adopted without any change to the more general case. For any linear transformation T: VW,

kernelT = {uV: T(u) = 0}.

We note that the concept of kernel is not defined for nonlinear transformations.

Similar to the range of a linear transformation, we have

The kernel of a linear transformation T: VW is a subspace of V.

Proof u, vkernelTT(u) = T(v) = 0T(u + v) = T(u) + T(v) = 0 + 0 = 0u + vkernelT, where the linearity of T is used in the red equality. The proof for ukernelTcukernelT is similar.

Example The subset (see this exercise)

{(x, y, z): x + y - z = 0, 2x - 3y + z = 0} ⊂ R3

is a subspace because it is the kernel of the linear transformation

T(x, y, z) = (x + y - z, 2x - 3y + z): R3R2.

The subset (see this exercise)

{fF(R): f(1) = f(2) = 0} ⊂ F(R)

is a subspace because it is the kernel of the linear transformation

E(f) = (f(1), f(2)) F(R) → R2.

The subset (see this exercise)

{fC2(R): f'' - 4f' + 3f = 0} ⊂ C2(R)

is a subspace because it is the kernel of the linear transformation

T(f) = f'' - 4f' + 3f: C2(R) → C(R).

We also recall the fact that a linear transformation between euclidean spaces is one-to-one (injective) if and only if the kernel contains only the zero vector. The proof of the fact can be applied to the general case without any change. Therefore we also have

A linear transformation T: VW is one-to-one ⇔ kernelT = {0}.

Example Consider the derivative transformation

D(f) = f': C1(R) → C(R)

The kernel consists of those functions with zero derivative. From calculus, we know that such functions are exactly constant functions. Therefore kernelD = {c: cR}. In particular, the kernel is not the zero subspace. Thus D is not one-to-one.

Example Consider the linear transformation

M(f) = (t2+1)f: F(R) → F(R)

of multiplying the function t2+1. If (t2+1)f = 0, then by the fact that t2 + 1 is never zero, we see that f = 0. Therefore kernelM = {0} and M is one-to-one.

If we change t2 + 1 to t2 - 1, however, then M is no longer onto. In fact, the kernel of the linear transformation f → (t2-1)f consists of functions of the form

 f(t) = { 0 if t ≠ -1, 1 c if t = -1 d if t = 1

where c and d are arbitrary numbers.

On the other hand, if we restrict the source and target by considering

M(f) = (t2-1)f: C(R) → C(R),

then the continuity implies that c = d = 0. Therefore the linear transformation is again one-to-one.