The Zero Route

At a glance

• 1. ker⁡(𝑇𝐴) ={(0,0)}, while ker⁡(𝑇𝐵) ={(𝑡, −𝑡) :𝑡 ∈ℝ}.
• 2. Map A’s kernel is the origin. Map B’s is the line 𝑦 = −𝑥; for example, (1, −1) is a nonzero kernel vector.
• 3. One collision is 𝑇𝐵(0,0) =𝑇𝐵(1, −1) =(0,0).
• 4. 𝑇(𝑝) =𝑇(𝑞) ⟺ 𝑇(𝑝 −𝑞) =0 ⟺ 𝑝 −𝑞 ∈ker⁡(𝑇). If 𝑇(𝑎) =𝑠, then the fiber over 𝑠 is 𝑎 +ker⁡(𝑇).
• 5. A linear transformation is one-to-one exactly when its kernel is {0}. Map A passes the one-to-one test.

Key idea: subtract the inputs

The concrete systems and the general proof are connected by one move. If two inputs produce the same output, linearity turns their difference into a kernel vector:

Conversely, a difference in the kernel creates equal outputs. That is why studying the single fiber over zero controls every possible collision.

1. Solve the two homogeneous systems

For Map A, setting both output coordinates equal to zero gives

Adding the equations gives 2𝑥 =0, so 𝑥 =0. Then 𝑦 =0. Thus

Its kernel is only the origin, also called the trivial subspace.

For Map B, the homogeneous system is

The second equation is twice the first, so there is only one independent condition. Let 𝑥 =𝑡. Then 𝑦 = −𝑡, giving

This is the one-dimensional subspace consisting of the line 𝑦 = −𝑥.

2. Draw and verify the kernels

On Map A’s plane, mark only the origin with a visible solid dot. On Map B’s plane, draw the full line 𝑦 = −𝑥 through the origin and across the grid.

For a nonzero example, choose 𝑣 =(1, −1). Then

Therefore (1, −1) ∈ker⁡(𝑇𝐵). Every scalar multiple (𝑡, −𝑡) also disappears because linear transformations send scalar multiples of a kernel vector to scalar multiples of zero.

3. Exhibit a collision

Take 𝑝 =(0,0) and 𝑞 =(1, −1). They are different inputs, but

and

Hence

Any nonzero kernel vector creates this kind of collision with the zero vector.

4. Prove the collision lemma and describe the fibers

Suppose first that 𝑇(𝑝) =𝑇(𝑞). By linearity,

Therefore 𝑝 −𝑞 ∈ker⁡(𝑇).

Conversely, suppose 𝑝 −𝑞 ∈ker⁡(𝑇). Then 𝑇(𝑝 −𝑞) =0, so

which implies 𝑇(𝑝) =𝑇(𝑞). We have proved both directions:

Now let 𝑠 be an output with a nonempty fiber, and choose 𝑎 in that fiber, so 𝑇(𝑎) =𝑠. An input 𝑣 lies in the same fiber exactly when 𝑇(𝑣) =𝑇(𝑎). By the lemma, this happens exactly when 𝑣 −𝑎 ∈ker⁡(𝑇), or equivalently when 𝑣 =𝑎 +𝑘 for some 𝑘 ∈ker⁡(𝑇). Therefore

So the kernel does not merely find collisions with zero. It gives the common shape of every nonempty fiber: every collection of indistinguishable inputs is a shifted copy of the kernel.

5. Prove the one-to-one theorem

First suppose ker⁡(𝑇) ={0}. If 𝑇(𝑝) =𝑇(𝑞), the collision lemma gives 𝑝 −𝑞 ∈ker⁡(𝑇). The only possibility is 𝑝 −𝑞 =0, so 𝑝 =𝑞. Thus 𝑇 is one-to-one.

For the other direction, suppose 𝑇 is one-to-one. Let 𝑣 ∈ker⁡(𝑇). Then

One-to-one-ness forces 𝑣 =0. Therefore the kernel contains no vector other than zero:

This proves

Map A passes the one-to-one test because its kernel is trivial. Map B does not because it has nonzero kernel vectors.

A precise final sentence could be: The kernel records exactly which input vectors a linear transformation sends to the zero output.

Also acceptable: The kernel records the input differences that the transformation cannot distinguish. A response such as “the kernel measures how much is erased” is too vague unless it identifies the vectors or input differences sent to zero.

Check

Substitute the complete kernel descriptions back into their transformations. Map A gives only 𝑇𝐴(0,0) =(0,0). For Map B,

for every real 𝑡. The collision lemma also checks the final decision: Map A permits 𝑇𝐴(𝑝) =𝑇𝐴(𝑞) only when 𝑝 −𝑞 =0, whereas Map B permits any difference (𝑡, −𝑡).

A farther facet: zero, dimension, and lost information

For a linear transformation from a finite-dimensional space, the rank-nullity theorem says

The number dim⁡(ker⁡𝑇), called the nullity, measures how many independent input directions are lost. Map A has nullity 0; Map B has nullity 1. This is why zero is enough: linearity converts every equality of outputs into a statement about one difference vector mapping to zero.