Kernel

65. Kernel

For a linear transformation $𝑇 : ℝ^{𝑛} \to ℝ^{𝑚}$ , the kernel (or null space of $𝑇$ ) is the set of vectors that map to the zero vector:

ker (𝑇) = {\vec{𝑥} \in ℝ^{𝑛} | 𝑇 (\vec{𝑥}) = 𝟎}

The kernel is the preimage of $𝟎$ .

65.1. Why the kernel is a subspace

For any $\vec{𝑥}, \vec{𝑦} \in ker (𝑇)$ and scalar $𝑐$ :

$𝑇 (\vec{𝑥} + \vec{𝑦}) = 𝑇 (\vec{𝑥}) + 𝑇 (\vec{𝑦}) = 𝟎 + 𝟎 = 𝟎$ → closed under addition
$𝑇 (𝑐 \vec{𝑥}) = 𝑐 𝑇 (\vec{𝑥}) = 𝑐 𝟎 = 𝟎$ → closed under scalar multiplication
$𝟎 \in ker (𝑇)$ since $𝑇 (𝟎) = 𝟎$

Hence $ker (𝑇)$ is a subspace of $ℝ^{𝑛}$ .

65.2. Matrix form: kernel ↔ null space

If $𝑇$ is represented by a matrix $𝐴$ (so $𝑇 (\vec{𝑥}) = 𝐴 \vec{𝑥}$ ), the kernel of $𝑇$ equals the null space of $𝐴$ :

ker (𝑇) = null (𝐴) = {\vec{𝑥} | 𝐴 \vec{𝑥} = 𝟎}

Example

Let $𝑇 : ℝ^{2} \to ℝ^{2}$ with $𝐴 = [\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}]$ . Solve $𝐴 \vec{𝑥} = 𝟎$ :

\begin{matrix} 2 𝑥_{1} = 0 \Rightarrow 𝑥_{1} = 0 \\ 3 𝑥_{2} = 0 \Rightarrow 𝑥_{2} = 0 \end{matrix}

Only the trivial solution: $ker (𝑇) = {𝟎}$ . The transformation is injective.

65.3. Injectivity

A linear transformation $𝑇$ is injective (one-to-one) if and only if $ker (𝑇) = {𝟎}$ .

Proof sketch: if $𝑇 ({\vec{𝑥}}_{1}) = 𝑇 ({\vec{𝑥}}_{2})$ , then $𝑇 ({\vec{𝑥}}_{1} - {\vec{𝑥}}_{2}) = 𝟎$ , so ${\vec{𝑥}}_{1} - {\vec{𝑥}}_{2} \in ker (𝑇)$ . The kernel being trivial forces ${\vec{𝑥}}_{1} = {\vec{𝑥}}_{2}$ .

65.4. Connections

Null Space — kernel of the matrix representation
Rank–Nullity: $dim (ker (𝑇)) + dim (im (𝑇)) = 𝑛$
Image — the “output side” counterpart
Preimage — generalization to non-zero target sets
Zero Vector — what kernel elements map to