The Range and Surjectivity

The concept of range underlies solvability analysis throughout applied mathematics: determining whether a linear system $Ax = b$ has a solution for a given $b$ is exactly asking whether $b \in \operatorname{range}(A)$, which is central to numerical linear algebra, optimization (feasibility of linear programs), and control theory (reachability of a linear system's state space). In quantum mechanics, the range of an operator determines which states are physically producible from an initial ensemble via that operator, relevant to measurement theory and the study of Hermitian and unitary operators. In statistics and data science, the range of a design matrix in linear regression determines exactly which response vectors can be fit exactly (zero residual), and the gap between the range and the full space of possible outcomes quantifies unavoidable prediction error. In coding theory and cryptography, the range of a linear encoding map determines the set of valid codewords, and surjectivity questions arise when checking whether an encoding scheme can represent every possible message digest. In computer graphics and robotics, the range of a Jacobian linear map captures which velocities or forces are achievable by a mechanism, directly informing controllability and singularity analysis.

Ask the tutor: How does the dimension of the range (the rank) combine with the dimension of the kernel to recover $\dim V$ exactly, and why is that relationship forced by linearity?

Think of $T$ as a machine: you feed in vectors from $V$, and it produces vectors in $W$. The range is the collection of all products this machine can ever make — the set of outputs, forgetting entirely which inputs produced them. Some vectors in $W$ might never come out of the machine no matter what you feed in; those live in $W \setminus \operatorname{range}(T)$. The key structural fact, which is not automatic for arbitrary functions, is that this set of outputs is always a subspace. If $T$ can produce $w_1$ and $w_2$, linearity guarantees it can also produce $w_1 + w_2$ and $c w_1$ for any scalar $c$ — you simply add or scale the corresponding inputs. So the range is never a jagged or disconnected shape; it is always a flat, linear piece of $W$: a point, a line through the origin, a plane through the origin, or a higher-dimensional subspace, and so on. Surjectivity is then the extreme case where this subspace is not a proper piece at all — it is everything. Picture a linear map from $\mathbb{R}^3$ to $\mathbb{R}^3$ that squashes space onto a plane through the origin: its range is a 2-dimensional subspace of a 3-dimensional codomain, so it misses "almost everything" and is very far from surjective. Contrast this with a map whose range genuinely fills out all of $\mathbb{R}^3$.

Example 1: A map $\mathbb{R}^3 \to \mathbb{R}^3$ that is not surjective. Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be given by the matrix

By Theorem part (c) applied to the standard basis $\{e_1, e_2, e_3\}$ of $\mathbb{R}^3$, $\operatorname{range}(T)$ is spanned by the columns of $A$:

Note $T(e_2) = 2\,T(e_1)$, so the second column is redundant; $\operatorname{range}(T) = \operatorname{span}(T(e_1), T(e_3))$. We check $T(e_1)$ and $T(e_3)$ are linearly independent: if $a\,T(e_1) + b\,T(e_3) = 0$, then

The first equation gives $a = 0$; substituting into the third gives $b = 0$. So $\{T(e_1), T(e_3)\}$ is linearly independent, and

Since $\operatorname{range}(T)$ is a 2-dimensional subspace of the 3-dimensional space $\mathbb{R}^3$, we have $\operatorname{range}(T) \ne \mathbb{R}^3$. By Theorem part (b), $T$ is not surjective. We can verify concretely: is $w = (0,0,1)^T$ in the range? We'd need $x = (x_1,x_2,x_3)^T$ with

From the first equation $x_1 = -2x_2$. Substituting into the second: $-4x_2 + 4x_2 + x_3 = 0 \Rightarrow x_3 = 0$. Substituting into the third: $2x_2 - 2x_2 + 0 = 1 \Rightarrow 0 = 1$, a contradiction. So indeed $w \notin \operatorname{range}(T)$, confirming the map misses vectors in $\mathbb{R}^3$.

Example 2: A surjective map $\mathbb{R}^3 \to \mathbb{R}^2$, verified via spanning columns. Let $S: \mathbb{R}^3 \to \mathbb{R}^2$ be given by

The range of $S$ is spanned by the columns

It suffices to show that just two of these already span all of $\mathbb{R}^2$: consider $S(e_1) = (1,-1)^T$ and $S(e_2) = (0,2)^T$. These are linearly independent, since the matrix $\begin{pmatrix} 1 & 0 \\ -1 & 2\end{pmatrix}$ has determinant $1 \cdot 2 - 0 \cdot (-1) = 2 \ne 0$. Two linearly independent vectors in the 2-dimensional space $\mathbb{R}^2$ automatically span $\mathbb{R}^2$ (any spanning-independent set of size $\dim W$ is a basis). Hence $\operatorname{span}(S(e_1), S(e_2)) = \mathbb{R}^2$, and since $\operatorname{range}(S) \supseteq \operatorname{span}(S(e_1), S(e_2)) = \mathbb{R}^2$ while also $\operatorname{range}(S) \subseteq \mathbb{R}^2$ trivially, we get $\operatorname{range}(S) = \mathbb{R}^2$. By Theorem part (b), $S$ is surjective. Concretely, given any target $w = (w_1, w_2)^T \in \mathbb{R}^2$, we can solve $Bx = w$ directly using only the first two columns (set $x_3 = 0$): we need $x_1 = w_1$ and $-x_1 + 2x_2 = w_2$, so $x_2 = (w_2 + w_1)/2$. For instance, if $w = (5, -1)^T$, then $x_1 = 5$, $x_2 = (-1+5)/2 = 2$, $x_3 = 0$, and indeed