Consequences of Rank-Nullity for Finite Dimensions

The equivalence of injectivity and surjectivity for same-dimension linear maps is the computational shortcut behind testing whether a square matrix is invertible: rather than verifying both "no nontrivial kernel" and "columns span the whole space," one check suffices, directly motivating the many equivalent conditions collected under Invertibility of Linear Maps. In numerical linear algebra, this underlies why solving square systems $Ax=b$ via LU decomposition or Gaussian elimination can be certified by a single rank check (no zero pivots) rather than separately verifying existence and uniqueness. In cryptography, block ciphers and hash constructions rely on linear (or affine) maps over finite fields $\mathbb{F}_2^n \to \mathbb{F}_2^n$; showing such a map is a bijection (needed for invertible encryption) reduces to checking that it has trivial kernel, a much easier computation than exhaustively verifying every output is hit. In differential equations and control theory, dimension counting on the coefficient matrix of a linear system determines whether a steady-state or boundary value problem has a solution, is unique, or has an entire family of solutions parameterized by $\ker T$, connecting directly to the structure of Systems of Linear Equations. In statistics, the rank of a design matrix in linear regression determines both whether the normal equations have a unique solution and whether every response vector can, in principle, be fit exactly — a direct application of dimension-counting for solvability. Finally, this material is a stepping stone to the general theory of Linear Independence and basis-extension arguments that make the "same dimension forces equality" auxiliary fact rigorous in every proof of this kind.

Ask the tutor: If $T:V\to W$ is linear with $\dim V=\dim W$ but I only know $T$ is neither injective nor surjective, what exactly can I say about $\dim(\ker T)$?

Think of a linear map $T: V \to W$ with $\dim V = \dim W = n < \infty$ as a machine that "spends" the $n$ dimensions of its domain in two possible ways: some dimensions get collapsed to zero (measured by nullity), and the rest survive to become the image (measured by rank). Rank-Nullity says these two uses of dimension must add up exactly to $n$, with nothing left over and no double-counting. Now bring in the codomain's dimension, which is also $n$. The image is a subspace of $W$, so its dimension can be at most $n$. If $T$ is injective, nullity is $0$, forcing rank to be exactly $n$ — but a subspace of $W$ with dimension equal to $\dim W$ must be all of $W$ (a fact about subspaces, not about maps). Hence the image is everything, i.e., $T$ is surjective. Run the logic the other way: if $T$ is surjective, rank is $n$, forcing nullity to be $0$, so $T$ is injective. Neither direction needed anything beyond arithmetic once the dimensions were fixed and finite. The picture is a balance scale with a fixed total weight $n$ on one side (the domain) and a container of fixed capacity $n$ on the other (the codomain). Injectivity says "no weight is thrown away" (nullity $0$); surjectivity says "the container is completely filled" (rank $n$). Because the total weight equals the container capacity, throwing nothing away is the same event as filling the container completely.

Example 1: a $4\times 4$ system, checking invertibility via nullity alone. Let $T : \mathbb{R}^4 \to \mathbb{R}^4$ be given by the matrix

We want to know whether $Ax = b$ has a unique solution for every $b \in \mathbb{R}^4$. Rather than compute a determinant or check surjectivity directly, we compute $\ker A$ by solving $Ax = 0$ via row reduction. Row reduce $A$ (each step is an elementary row operation, which does not change the solution set of $Ax=0$): Start:

$R_3 \to R_3 - 2R_1$, $R_4 \to R_4 - R_1$:

$R_3 \to R_3 + R_2$, $R_4 \to R_4 + 2R_2$:

$R_4 \to R_4 - R_3$:

This is upper triangular with four nonzero pivots ($1, 1, -1, -1$ on the diagonal), so it has rank $4$. Back-substitution from $Ax=0$: the last row gives $-x_4 = 0 \Rightarrow x_4 = 0$; row 3 gives $-x_3 + 2x_4 = 0 \Rightarrow x_3 = 0$; row 2 gives $x_2 + x_3 - x_4 = 0 \Rightarrow x_2 = 0$; row 1 gives $x_1 + 2x_2 + x_4 = 0 \Rightarrow x_1 = 0$. So $\ker A = \{0\}$, meaning $T$ is injective. By the Theorem (valid since domain and codomain are both $\mathbb{R}^4$, dimension $4$), injective implies surjective. So without any further computation, $T$ is also surjective: $Ax = b$ has at least one solution for every $b \in \mathbb{R}^4$. Combining with injectivity (part (c) of the corollary), that solution is unique for every $b$. We got both existence and uniqueness for free from a single nullspace computation.

Example 2: a rectangular system, where dimension-counting shows solvability must fail for some $b$. Let $A$ be the $3 \times 2$ matrix

Here $\dim(\text{domain}) = 2 \ne 3 = \dim(\text{codomain})$, so the Theorem's equivalence does not apply (this is exactly the hypothesis being used correctly, not violated: we are in the general Rank-Nullity setting, not the equal-dimension case). Compute the rank of $A$ directly. The columns are $(1,2,4)^T$ and $(3,5,9)^T$; they are linearly independent (neither is a scalar multiple of the other, e.g. $3/1 = 3 \ne 5/2$), so $\operatorname{rank}(A) = 2$. By Rank-Nullity applied to $T_A: \mathbb{R}^2 \to \mathbb{R}^3$,

So $T_A$ is injective: whenever $Ax=b$ has a solution, it is unique. But $\operatorname{rank}(A) = 2 < 3 = \dim(\mathbb{R}^3)$, so by part (a) of the corollary, $T_A$ is not surjective — there exist vectors $b \in \mathbb{R}^3$ for which $Ax = b$ has no solution. Concretely, take $b = (0,0,1)^T$. Solving $Ax = b$: from row 1, $x_1 + 3x_2 = 0 \Rightarrow x_1 = -3x_2$. Substituting into row 2: $2(-3x_2) + 5x_2 = 0 \Rightarrow -x_2 = 0 \Rightarrow x_2 = 0$, hence $x_1 = 0$. Row 3 then requires $4(0) + 9(0) = 1$, i.e. $0 = 1$, a contradiction. So indeed no solution exists for this $b$, confirming the dimension count: $\operatorname{range}(T_A)$ is a genuine $2$-dimensional subspace of $\mathbb{R}^3$, not all of it, and $(0,0,1)^T$ happens to lie outside it. This illustrates exactly why the equal-dimension hypothesis in the Theorem is essential — with unequal dimensions, injectivity and surjectivity split apart, and here we get injective-but-not-surjective.