Kernel, Image, and Rank–Nullity

What a Map Collapses and What It Reaches

Two subspaces describe any linear transformation $T : V \to W$.

The kernel (null space) is everything $T$ sends to zero: $$\ker T = \{v \in V : T(v) = \mathbf{0}\}.$$ It measures what the map collapses. The kernel is a subspace of the domain. A crucial fact: $T$ is injective (one-to-one) exactly when its kernel is just $\{\mathbf{0}\}$ — if only the zero vector maps to zero, no two distinct inputs can collide.

The image (range) is everything $T$ can produce: $$\operatorname{im} T = \{T(v) : v \in V\}.$$ It measures what the map reaches. The image is a subspace of the codomain, and $T$ is surjective (onto) exactly when its image is all of $W$. The dimension of the image is the rank of $T$; the dimension of the kernel is the nullity.

These are tied by the Rank–Nullity Theorem, one of the central results of linear algebra: $$\dim(\ker T) + \dim(\operatorname{im} T) = \dim V,$$ i.e. nullity + rank = dimension of the domain. Every dimension of the input is accounted for: some directions get collapsed (the kernel) and the rest survive into the image. If a map from $\mathbb{R}^5$ has a 2-dimensional kernel, its image is exactly 3-dimensional — the numbers must add up.

This theorem instantly settles many questions. A linear map between spaces of the same finite dimension is injective if and only if it is surjective (if and only if it is bijective) — because rank + nullity is fixed, one forces the other. This is why a square matrix is invertible exactly when its kernel is trivial.

Common pitfall: confusing the kernel (in the domain, what maps to zero) with the image (in the codomain, what is reached), and forgetting that rank and nullity must sum to the domain's dimension. If a map from an $n$-dimensional space has a $k$-dimensional kernel, its image is $(n - k)$-dimensional — not free to be anything. Rank–Nullity is a hard accounting constraint, not a rough guide.

An input space split into an accent kernel band (collapsed to the zero point) and the complementary directions surviving into the image, their dimensions summing to the domain dimension.

$$\dim(\ker T) + \dim(\operatorname{im} T) = \dim V$$