Kernel & Injectivity

A linear map $T:V\to W$ is a rule that takes a vector $v$ from the domain $V$ and returns a vector $T(v)$ in the codomain $W$, respecting addition and scalar multiplication. Think of $T$ as a machine that transforms vectors.

Some vectors survive the transformation as something nonzero. Others get crushed to the zero vector $0_W$. The kernel is the collection of all the inputs that get crushed:

$\ker(T)=\{\,v\in V : T(v)=0_W\,\}.$

• $v$ is a vector in the domain $V$.
• $0_W$ is the zero vector of the codomain $W$.
• $\ker(T)$ is read "kernel of $T$."

Why care? Because the kernel is exactly the information $T$ throws away. If two different inputs $u$ and $v$ land on the same output ($T(u)=T(v)$), then their difference is crushed: $T(u-v)=T(u)-T(v)=0$, so $u-v\in\ker(T)$. A big kernel means lots of inputs collide; a kernel containing only $0$ means no two distinct inputs ever collide. That single idea is the whole story of injectivity, which we make precise below.

The zero vector is always in the kernel: because $T$ is linear, $T(0_V)=0_W$. So $\ker(T)$ is never empty — the only question is whether it contains anything besides $0$.

When the map is given by a matrix, the kernel has a very concrete description. Suppose $T:\mathbb{R}^n\to\mathbb{R}^m$ is defined by

$T(x)=Ax,$

where $A$ is an $m\times n$ matrix (it has $m$ rows and $n$ columns) and $x\in\mathbb{R}^n$. Then

$\ker(T)=\{\,x\in\mathbb{R}^n : Ax=0\,\}.$

This set has its own name in matrix language: the null space of $A$, written $\operatorname{Null}(A)$. So for a matrix map,

$\boxed{\;\ker(T)=\operatorname{Null}(A)=\{x : Ax=0\}.\;}$

Finding the kernel is therefore nothing more than solving the homogeneous linear system $Ax=0$. Every solution vector is a kernel vector, and vice versa.

Nullity. Because the kernel is a subspace (next section), it has a dimension. We call it the nullity:

$\operatorname{nullity}(T)=\dim\big(\ker(T)\big).$

Intuitively, the nullity counts how many independent directions get crushed to zero. When you row-reduce $A$, the nullity equals the number of free variables (equivalently, the number of columns without a pivot).

A nonempty subset $S$ of a vector space is a subspace if it contains the zero vector and is closed under addition and scalar multiplication. The kernel passes all three tests, and linearity of $T$ is the reason.

Let $u,v\in\ker(T)$ (so $T(u)=0$ and $T(v)=0$) and let $c$ be any scalar.

• Contains zero: $T(0_V)=0_W$, so $0_V\in\ker(T)$.
• Closed under addition: $T(u+v)=T(u)+T(v)=0+0=0$, so $u+v\in\ker(T)$.
• Closed under scalar multiplication: $T(cu)=c\,T(u)=c\cdot 0=0$, so $cu\in\ker(T)$.

Therefore $\ker(T)$ is a subspace of the domain $V$. Two useful consequences:

1. It makes sense to talk about a basis of the kernel and its dimension (the nullity).
2. The kernel is never just a stray collection of points — it is always a line, a plane, a higher-dimensional flat through the origin, or the single point $\{0\}$.

A map $T$ is injective (one-to-one) if distinct inputs always give distinct outputs: $T(u)=T(v)\implies u=v$. The kernel detects this instantly.

Injectivity criterion. For a linear map $T$,

$T\text{ is injective} \iff \ker(T)=\{0\} \iff \operatorname{nullity}(T)=0.$

Why: if $T(u)=T(v)$ then $T(u-v)=0$, so $u-v\in\ker(T)$. If the only kernel vector is $0$, then $u-v=0$, i.e. $u=v$ — injective. Conversely, if some $w\neq0$ lies in the kernel, then $T(w)=0=T(0)$ with $w\neq0$, so $T$ is not injective.

How to compute the kernel and decide injectivity for $T(x)=Ax$:

1. Form the homogeneous system $Ax=0$.
2. Row-reduce $A$ to reduced row echelon form.
3. Identify pivot columns and free variables (non-pivot columns).
4. Solve for each pivot variable in terms of the free variables, then write the general solution as a linear combination; the vectors multiplying the free parameters form a basis of the kernel.
5. The nullity is the number of free variables. If it is $0$ (a pivot in every column), $T$ is injective; otherwise it is not.

A quick shortcut from Rank–Nullity. Since $\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim(V)=n$, an injective map needs $\operatorname{rank}(T)=n$. This forces $n\le m$: a map from a bigger space to a smaller one (like $\mathbb{R}^3\to\mathbb{R}^2$) can never be injective, because the nullity is at least $n-m>0$.