The Null Space and Injectivity

The null space is one of the load-bearing concepts of linear algebra, and its consequences appear throughout mathematics and applied fields. In differential equations, the solution set of a homogeneous linear ODE (or system) is exactly the null space of a linear differential operator, and understanding its dimension tells you how many independent solutions exist. In statistics, the null space of a design matrix in linear regression governs identifiability: if the null space is nontrivial, distinct parameter vectors give identical predictions, and coefficients cannot be uniquely estimated. In coding theory and cryptography, the null space of a parity-check matrix defines a linear code, and the injectivity of the associated encoding map is precisely what guarantees error detection is meaningful. In computer graphics and robotics, the null space of a Jacobian matrix describes redundant degrees of freedom — motions that produce no change in the observed output, critical for controlling robotic arms with more joints than needed. In quantum mechanics, null spaces of linear operators identify degenerate or "silent" states invisible to a given measurement. Across all these settings, the same rigid fact from this article recurs: for a linear map, checking whether information is lost reduces to checking a single equation, $T(v) = 0$.

Ask the tutor: How does the dimension of the null space precisely determine how far a linear map is from being injective and from being invertible?

Think of a linear map $T$ as a machine that takes vectors in $V$ and produces vectors in $W$. The null space asks a single, sharply focused question: which inputs get flattened to the origin? Because $T$ is linear, the set of such inputs cannot be some scattered, irregular collection of points — it is forced to be a flat, linear piece of $V$ itself: a subspace. This rigidity is the first hint that linear maps behave far more predictably than arbitrary functions. The connection to injectivity is best understood through translation. If $T(u) = T(v)$, linearity lets us rewrite this as $T(u - v) = 0$, so $u - v$ lies in the null space. If the null space contains only the zero vector, then $u - v$ must be $0$, forcing $u = v$. Conversely, if some nonzero vector $n$ lies in the null space, then $n$ and $0$ are distinct vectors mapping to the same output $0$, so injectivity fails immediately. The entire global comparison "do any two vectors collide?" reduces to the question "does anything besides $0$ collapse to $0$?" The picture below shows a concrete linear map $T: \mathbb{R}^3 \to \mathbb{R}^2$ given by a matrix with a nontrivial null space — a whole line of vectors gets sent to the origin, and correspondingly the map fails to be injective, visibly collapsing parallel planes onto single points.

Example 1: A map $\mathbb{R}^3 \to \mathbb{R}^3$ with a genuinely nontrivial null space. Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be given by the matrix

We check linearity is automatic since $T(x) = Ax$ is matrix multiplication. To find $\operatorname{null}(T)$, solve $Ax = 0$ for $x = (x_1, x_2, x_3)$:

Row-reduce the augmented matrix $[A \mid 0]$. Start with rows $R_1, R_2, R_3$ as above. Replace $R_2 \to R_2 - 2R_1$: $R_2 = (2-2, 1-4, 3+2) = (0, -3, 5)$. Replace $R_3 \to R_3 - 3R_1$: $R_3 = (3-3, 3-6, 2+3) = (0, -3, 5)$. Now $R_3 - R_2 = (0,0,0)$, so $R_3$ becomes the zero row, confirming the third equation is redundant. We are left with:

From the second equation, $x_2 = \tfrac{5}{3}x_3$. Substituting into the first,

Let $x_3 = 3t$ for a free parameter $t \in \mathbb{R}$ (chosen to clear denominators). Then $x_2 = 5t$, $x_1 = -7t$, so

This is a one-dimensional subspace (a line through the origin), spanned by $(-7,5,3)$. We verify: $A(-7,5,3)^T = (-7+10-3,\ -14+5+9,\ -21+15+6) = (0,0,0)$. Correct. Since $\operatorname{null}(T) \ne \{0\}$ — for instance $(-7,5,3) \ne 0$ lies in it — the theorem tells us $T$ is not injective. Concretely, $T(-7,5,3) = T(0,0,0) = (0,0,0)$, exhibiting two distinct vectors with equal image, exactly as guaranteed by the failure of $\operatorname{null}(T) = \{0\}$.

Example 2: A map $\mathbb{R}^2 \to \mathbb{R}^3$ that is injective, checked via the null space. Define $S: \mathbb{R}^2 \to \mathbb{R}^3$ by

First confirm linearity directly from the formula: writing $S(x) = Bx$ with

$S$ is matrix multiplication, hence linear by the same argument as in the corollary above. Compute $\operatorname{null}(S)$ by solving $Bx = 0$:

From the first equation, $x_2 = 2x_1$. Substitute into the second: $x_1 + 3(2x_1) = 7x_1 = 0$, forcing $x_1 = 0$, hence $x_2 = 2(0) = 0$. Check against the third equation: $-0 + 4(0) = 0$, consistent. So the only solution is $x_1 = x_2 = 0$, meaning

By the theorem, $S$ is injective. We can sanity-check this concretely: suppose $S(a_1, a_2) = S(b_1, b_2)$. Then $(a_1 - b_1, a_2-b_2) \in \operatorname{null}(S)$ by linearity (exactly as in the proof), forcing $a_1 - b_1 = 0$ and $a_2 - b_2 = 0$, i.e. $(a_1,a_2) = (b_1,b_2)$. So indeed no two distinct input pairs can produce the same output triple, confirming injectivity directly, not just via the abstract criterion.