Image & Surjectivity

A linear map $T:\mathbb{R}^n \to \mathbb{R}^m$ eats an input vector $\mathbf{x} \in \mathbb{R}^n$ and returns an output vector $T(\mathbf{x}) \in \mathbb{R}^m$. We call $\mathbb{R}^n$ the domain (where inputs live) and $\mathbb{R}^m$ the codomain (where outputs are allowed to live).

Every such map can be written as $T(\mathbf{x}) = A\mathbf{x}$ for a unique $m\times n$ matrix $A$ (so $A$ has $m$ rows and $n$ columns).

The image (also called the range) collects every output the map can actually produce: $\operatorname{im}(T) = \{\, T(\mathbf{x}) : \mathbf{x}\in\mathbb{R}^n \,\} = \{\, A\mathbf{x} : \mathbf{x}\in\mathbb{R}^n \,\}.$

Two things to keep straight from the start:

• The image is a subset of the codomain $\mathbb{R}^m$, not of the domain. (Outputs live where outputs are allowed to live.)
• Not every vector of the codomain has to be hit. The image is the part of $\mathbb{R}^m$ that genuinely gets reached.

Intuition. Think of $T$ as a machine. The codomain is the whole wall it could in principle paint; the image is the patch it does paint as the input ranges over everything.

Write $A$ in terms of its columns $\mathbf{a}_1,\dots,\mathbf{a}_n \in \mathbb{R}^m$, and write $\mathbf{x}=(x_1,\dots,x_n)$. The key identity of matrix multiplication is $A\mathbf{x} = x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots + x_n\mathbf{a}_n.$ So $A\mathbf{x}$ is just a linear combination of the columns, with the entries of $\mathbf{x}$ as the weights. As $\mathbf{x}$ ranges over all of $\mathbb{R}^n$, the weights range over all possible values, so the outputs sweep out exactly the span of the columns: $\operatorname{im}(T) = \operatorname{span}(\mathbf{a}_1,\dots,\mathbf{a}_n).$ This set is also called the column space of $A$, written $\operatorname{Col}(A)$. Because a span is always closed under addition and scalar multiplication and contains $\mathbf{0}$, the image is a subspace of $\mathbb{R}^m$.

The rank of $T$ (equivalently of $A$) is the dimension of this subspace: $\operatorname{rank}(A) = \dim\big(\operatorname{im}(T)\big).$ Concretely, you compute it by row-reducing $A$ and counting the pivot columns — the rank is the number of pivots. (A useful fact: the column rank always equals the row rank, so row reduction gives the right number either way.)

A map $T:\mathbb{R}^n\to\mathbb{R}^m$ is surjective (or onto) if every vector of the codomain is an output — that is, for each $\mathbf{b}\in\mathbb{R}^m$ there is some $\mathbf{x}$ with $T(\mathbf{x})=\mathbf{b}$. In symbols, surjective means $\operatorname{im}(T)=\mathbb{R}^m$.

Since $\operatorname{im}(T)$ is always a subspace sitting inside $\mathbb{R}^m$, a subspace equals the whole space exactly when it has the full dimension. This gives the clean test: $T \text{ is surjective} \iff \operatorname{im}(T)=\mathbb{R}^m \iff \operatorname{rank}(A)=m,$ where $m=\dim(\text{codomain})$.

Consequences worth memorizing:

• To be surjective you need $\operatorname{rank}(A)=m$. But the rank can never exceed the number of columns $n$, so you need $n\ge m$. A map with fewer inputs than the codomain dimension ($n<m$, a 'tall' matrix) can never be surjective — there simply aren't enough columns to span $\mathbb{R}^m$.
• For a square matrix ($n=m$), surjective $\iff$ full rank $\iff$ invertible $\iff$ injective. (Surjectivity and injectivity coincide only in the square case.)

Given a specific target $\mathbf{b}\in\mathbb{R}^m$, asking '$\mathbf{b}\in\operatorname{im}(T)$?' is asking whether some input maps to it. Three equivalent phrasings: $\mathbf{b}\in\operatorname{im}(T) \iff \mathbf{b} \text{ is a linear combination of the columns of } A \iff A\mathbf{x}=\mathbf{b} \text{ has a solution}.$

Procedure. Form the augmented matrix $[\,A \mid \mathbf{b}\,]$ and row-reduce.

• If you ever get a row of the form $[\,0\ 0\ \cdots\ 0 \mid c\,]$ with $c\neq 0$ (a pivot in the last column), the system is inconsistent: $\mathbf{b}\notin\operatorname{im}(T)$.
• Otherwise the system is consistent and $\mathbf{b}\in\operatorname{im}(T)$; the solution(s) give the input(s) that map to $\mathbf{b}$.

Shortcut when the image is a hyperplane. If $\operatorname{im}(T)$ is a plane in $\mathbb{R}^3$ (or more generally a hyperplane), it can be described by a single equation $\mathbf{n}\cdot\mathbf{v}=0$, where the normal vector $\mathbf{n}$ is orthogonal to every column. Then $\mathbf{b}$ is in the image precisely when $\mathbf{n}\cdot\mathbf{b}=0$ — one dot product instead of a full row reduction.