Rank-Nullity Theorem

The rank-nullity theorem is a conservation law for dimensions. When a linear map acts on a vector space, no dimensions disappear without a trace: they either survive in the image or get absorbed into the kernel. The theorem makes this precise and immediately explains why injectivity and surjectivity coincide for square linear maps.

Statement

Rank-nullity theorem: Let $V$ be a finite-dimensional vector space over a field $F$, and let $T: V \to W$ be a linear map. Then

\dim(V) = \text{rank}(T) + \text{nullity}(T). \tag{1}

In words: the dimension of the domain equals the dimension of the image plus the dimension of the kernel.

Proof

Let $k = \text{nullity}(T) = \dim(\ker(T))$, and choose a basis $\{u_1, \ldots, u_k\}$ of $\ker(T)$.

Step 1: Extend to a basis of $V$. Since $\ker(T)$ is a subspace of the finite-dimensional space $V$, we can extend $\{u_1, \ldots, u_k\}$ to a full basis of $V$. Call the extra vectors $v_1, \ldots, v_r$, so

$\mathcal{B} = \{u_1, \ldots, u_k, v_1, \ldots, v_r\}$

is a basis of $V$. Then $\dim(V) = k + r$.

Step 2: $\{T(v_1), \ldots, T(v_r)\}$ spans $\text{im}(T)$. Take any $w \in \text{im}(T)$; write $w = T(x)$ for some $x \in V$. Expand $x$ in the basis $\mathcal{B}$:

$x = \sum_{i=1}^k c_i\, u_i + \sum_{j=1}^r d_j\, v_j.$

Apply $T$ and use linearity. Since each $u_i \in \ker(T)$, we get $T(u_i) = \mathbf{0}$:

$w = T(x) = \sum_{i=1}^k c_i\, T(u_i) + \sum_{j=1}^r d_j\, T(v_j) = \sum_{j=1}^r d_j\, T(v_j).$

So $w$ is a linear combination of $T(v_1), \ldots, T(v_r)$, confirming they span $\text{im}(T)$.

Step 3: $\{T(v_1), \ldots, T(v_r)\}$ is linearly independent. Suppose $\sum_{j=1}^r d_j\, T(v_j) = \mathbf{0}$. By linearity, $T\!\left(\sum_{j=1}^r d_j\, v_j\right) = \mathbf{0}$, so $\sum_{j} d_j\, v_j \in \ker(T)$. Therefore this vector is a linear combination of the basis of $\ker(T)$:

$\sum_{j=1}^r d_j\, v_j = \sum_{i=1}^k e_i\, u_i$

for some scalars $e_i$. Rearranging: $\sum_{j} d_j\, v_j - \sum_{i} e_i\, u_i = \mathbf{0}$. Since $\mathcal{B}$ is a basis of $V$, all coefficients are zero — in particular $d_1 = \cdots = d_r = 0$.

Step 4: Conclude. $\{T(v_1), \ldots, T(v_r)\}$ is a basis of $\text{im}(T)$, so $\text{rank}(T) = r$. Therefore:

$\dim(V) = k + r = \text{nullity}(T) + \text{rank}(T). \qquad \square$

Interpretation

Informally, $V$ splits into two parts: the $k$-dimensional kernel (which $T$ crushes entirely to zero) and an $r$-dimensional complement (which $T$ maps isomorphically onto $\text{im}(T)$). The total dimension is preserved: $k + r = \dim(V)$. No dimensions are created or destroyed — they are simply rerouted.

Consequences for square matrices

Now suppose $T: F^n \to F^n$ is a linear map with matrix $A \in M_{n,n}(F)$ (a square matrix). Applying (1):

$n = \text{rank}(T) + \text{nullity}(T).$

The domain and codomain have the same dimension $n$, and the dimensions of the kernel and image must sum to exactly $n$. This rigid budget creates the following equivalence:

Theorem: For $T: F^n \to F^n$ (equivalently, for an $n \times n$ matrix $A$), the following statements are all equivalent:

1. $T$ is injective ($\ker(T) = \{\mathbf{0}\}$, i.e., $\text{nullity}(T) = 0$).
2. $T$ is surjective ($\text{im}(T) = F^n$, i.e., $\text{rank}(T) = n$).
3. $T$ is bijective (and hence has an inverse $T^{-1}$).
4. $\text{rank}(A) = n$ (all $n$ columns — equivalently, all $n$ rows — of $A$ are linearly independent).

The reason: if nullity is $0$, then rank must be $n$ (from (1)), so $T$ is both injective and surjective. There is no room for the rank to be $n$ without the nullity being $0$, and vice versa — the two quantities share a fixed budget of $n$.

This is a purely dimension-counting phenomenon: in infinite dimensions, injectivity and surjectivity can fail independently. In finite dimensions, they are inseparable for square maps.

A concrete example

Let $A$ be a $4 \times 6$ matrix with $\text{rank}(A) = 3$. The domain is $F^6$, so $\dim = 6$. By (1):

$\text{nullity}(A) = 6 - \text{rank}(A) = 6 - 3 = 3.$

The kernel is three-dimensional: there are three linearly independent vectors in $F^6$ that $A$ maps to zero. The image is three-dimensional inside $F^4$ — $A$ cannot be surjective (since $3 < 4$).

Summary

• Rank-nullity theorem: $\dim(V) = \text{rank}(T) + \text{nullity}(T)$ for any linear map $T: V \to W$ with $V$ finite-dimensional.
• Proof idea: extend a basis of $\ker(T)$ to a basis of $V$; the extra vectors map to a basis of $\text{im}(T)$.
• Interpretation: the domain splits between the kernel (dimensions collapsed to zero) and a complement (dimensions mapped isomorphically to the image).
• For square maps $T: F^n \to F^n$: injectivity, surjectivity, and bijectivity (invertibility) are all equivalent — they run out of the same $n$-dimensional budget simultaneously.