Rank-nullity theorem — proof: Worked examples

This is the drill-ground for the rank–nullity theorem. The theorem says one thing: $rank (T) + nullity (T) = dim V,$ where $rank (T) = dim (im T)$ (dimensions that survive) and $nullity (T) = dim (ker T)$ (dimensions crushed to $0$ ). See Kernel and Image of a Linear Map if either word feels unfamiliar.

Our goal here: hit every kind of situation the theorem can appear in, so no exam problem is a surprise.

The scenario matrix

Every rank–nullity problem is really a question about how many input directions collapse. The cases below cover all the shapes that can occur.

Cell	Case class	What makes it distinct	Example
A	Injective, nullity $= 0$	nothing crushed; image as big as domain	Ex 1
B	Surjective, rank $= dim W$	image fills the codomain	Ex 2
C	General middle case ( $0 <$ nullity $< dim V$ )	some crush, some survive	Ex 3
D	Zero map (fully degenerate)	everything crushed, rank $= 0$	Ex 4
E	Wide matrix ( $m < n$ , forced nullity)	more inputs than outputs	Ex 5
F	Tall matrix ( $m > n$ , forced non-surjective)	fewer outputs than inputs	Ex 6
G	Abstract space (not $R^{n}$ )	polynomials / functions	Ex 7
H	Real-world word problem	translate a scenario into $T$	Ex 8
I	Exam-style twist (solve for unknown dim)	rank/nullity given, find one	Ex 9

Worked examples

Cell A — Injective map, nothing crushed

Worked example Example 1

$T : R^{2} \to R^{3}$ , $T (x, y) = (x, y, x + y)$ . Find rank and nullity.

Forecast: the two inputs look independent after mapping — guess nullity, then rank.

Find the kernel. Set $T (x, y) = (0, 0, 0)$ : this forces $x = 0$ , $y = 0$ . So $ker T = {(0, 0)}$ . Why this step? The kernel is always the first thing to compute — its dimension is nullity, and rank comes free from the theorem.
Nullity $= 0$ . The kernel is just the zero vector, a $0$ -dimensional space. Why this step? A single point (the origin) has no independent directions, so $dim = 0$ . See Dimension of a Vector Space.
Rank by the theorem. $rank = dim V - nullity = 2 - 0 = 2$ . Why this step? We never had to describe the image directly — rank–nullity hands it to us.

Verify: $T (1, 0) = (1, 0, 1)$ and $T (0, 1) = (0, 1, 1)$ are independent in $R^{3}$ (neither is a scalar multiple of the other), so the image is $2$ -dimensional. $2 + 0 = 2 = dim R^{2}$ . ✓ Nullity $0$ $\Leftrightarrow$ injective — see Invertible Linear Maps and Isomorphisms.

Cell B — Surjective map, image fills the codomain

Cell C — General middle case

Cell D — The zero map (fully degenerate)

Cell E — Wide matrix (forced nonzero nullity)

Cell F — Tall matrix (forced non-surjective)

Cell G — Abstract space (polynomials)

Cell H — Real-world word problem

Worked example Example 8

A factory blends $4$ raw ingredients (input vector in $R^{4}$ ) into $2$ measured outputs — total mass and total cost: $T (a, b, c, d) = (a + b + c + d, 2 a + 2 b + 2 c + 2 d) .$ How many independent "recipe adjustments" leave both outputs unchanged?

Forecast: the second output is proportional to the first — guess how constrained the outputs really are.

Model as a matrix. $A = (12121212)$ , $T : R^{4} \to R^{2}$ . Why this step? "Recipe adjustments that change nothing" is exactly the kernel.
Rank. Row $2 = 2 \times$ Row $1$ , so one pivot. Rank $= 1$ . Why this step? Mass and cost carry only one independent piece of information.
Nullity = the answer. $nullity = dim V - rank = 4 - 1 = 3$ . Why this step? Three independent ways to tweak the ingredients while keeping both outputs fixed.

Verify: $A x = 0$ is one equation $a + b + c + d = 0$ in $4$ unknowns $\Rightarrow 3$ free variables. $1 + 3 = 4 = dim R^{4}$ . ✓ Concretely, $(1, - 1, 0, 0)$ , $(1, 0, - 1, 0)$ , $(1, 0, 0, - 1)$ are $3$ independent kernel vectors.

Cell I — Exam-style twist (solve for an unknown dimension)

Worked example Example 9

A linear map $T : V \to W$ has $dim V = 9$ . You are told $rank (T) = 6$ and that the codomain has $dim W = 12$ . Find the nullity, and state whether $T$ can be injective or surjective.

Forecast: you are given rank; the theorem gives nullity in one line — but watch which dimension it uses.

Nullity. $nullity = dim V - rank = 9 - 6 = 3$ . Why this step? The theorem balances against $dim V = 9$ , not $dim W = 12$ . The $12$ is a distractor.
Injective? Injective needs nullity $= 0$ . Here nullity $= 3 \neq = 0$ , so not injective. Why this step? $3$ independent input directions collapse to $0$ .
Surjective? Surjective needs $rank = dim W = 12$ . Here rank $= 6 < 12$ , so not surjective. Why this step? The image is a $6$ -dimensional sliver of a $12$ -dimensional codomain.

Verify: $rank + nullity = 6 + 3 = 9 = dim V$ . ✓ The $dim W = 12$ never enters the sum — only affects the surjectivity verdict.

Recall Quick self-test across the matrix

Injective means nullity equals what? ::: $0$ Surjective means rank equals what? ::: $dim W$ (the codomain's dimension) The zero map $R^{4} \to R^{5}$ has rank and nullity? ::: rank $0$ , nullity $4$ A wide $2 \times 3$ rank-1 matrix has nullity? ::: $2$ If $dim V = 9$ and rank $= 6$ , nullity is? ::: $3$ Rank–nullity sums to $dim V$ or $dim W$ ? ::: $dim V$ (the domain)