The kernel and the image

Two subspaces tell the whole story of a linear map: what it crushes (the kernel) and what it reaches (the image).

The kernel: what gets crushed to zero

Ker f = { v in E  such that  f(v) = 0 }

It is the set of vectors sent to the zero vector. In matrix terms, it is exactly the solution set of the homogeneous system A X = 0 — an object we have already met.

The kernel is a vector subspace of the source space, and it always contains 0.

The image: what gets reached

Im f = { f(v)  as v runs over E }

It is the set of vectors actually reached by f. Since f(v) = A X is a combination of the columns of A:

Im f = Vect( columns of A )

The image is a vector subspace of the target space.

The picture to remember

        E (source)                       F (target)
   ___________________              ___________________
  |                   |            |                   |
  |    Ker f          |    f       |     Im f          |
  |  (crushed to 0)   |  ------->  |    (reached)      |
  |___________________|            |___________________|

  lives in the SOURCE               lives in the TARGET

Do not confuse them: they do not even live in the same space.

Injectivity and surjectivity, linear version

This is where these notions become easy to test:

f injective   <=>   Ker f = {0}          (nothing crushed but 0)
f surjective  <=>   Im f = F             (everything reached)
f bijective   <=>   both

The first criterion deserves justification: if f(u) = f(v), then by linearity f(u - v) = 0, so u - v lies in the kernel. If the kernel reduces to {0}, this forces u = v. A linear map is injective if and only if it crushes nothing but the zero vector — a single subspace to compute instead of comparing all pairs of vectors.

A worked example

Let f : R³ -> R² be defined by f(x ; y ; z) = (x + y ; y + z), with matrix

A = [ 1  1  0 ]
    [ 0  1  1 ]

Kernel: solve x + y = 0 and y + z = 0, giving y = -x and z = -y = x:

Ker f = { (x ; -x ; x) } = Vect( (1 ; -1 ; 1) )      dimension 1

So f is not injective: the vector (1 ; -1 ; 1) is crushed to zero.

Image: the columns are (1;0), (1;1), (0;1). The first two already span R²:

Im f = R²        dimension 2       ->  f is surjective

Summary

• Ker f = the vectors sent to 0; it lives in the source space.
• Im f = the vectors reached; it lives in the target space and is spanned by the columns.
• Both are vector subspaces.
• f injective ⟺ Ker f = {0}.
• f surjective ⟺ Im f is the whole target space.
• Computing the kernel amounts to solving the homogeneous system A X = 0.