Rank-nullity theorem — proof: Question bank

Recall What the Basis Extension Theorem actually says (needed below)

Statement. If ${u_{1}, \dots, u_{k}}$ is a linearly independent set (e.g. a basis of a subspace) inside a finite-dimensional space $V$ , then there exist vectors $v_{1}, \dots, v_{n - k} \in V$ such that ${u_{1}, \dots, u_{k}, v_{1}, \dots, v_{n - k}}$ is a basis of $V$ . Hypothesis that matters: the starting set must be independent, and $V$ must be finite-dimensional. The added $v_{j}$ are chosen precisely so independence is preserved — you cannot just grab any vectors. This is the tool that lets us extend a kernel basis to a full basis of $V$ .

Recall The proof steps this page refers to (from the parent)

The proof shows $B = {T (v_{1}), \dots, T (v_{n - k})}$ is a basis of $im T$ in two moves. Spanning: any $v$ splits into a kernel part (mapped to $0$ ) plus a survivor part, so every output is a combination of the $T (v_{j})$ . Independence (the step doing real work): if $\sum c_{j} T (v_{j}) = 0$ then $\sum c_{j} v_{j} \in ker T$ ; rewriting that via the $u_{i}$ and invoking independence of the full basis of $V$ forces every $c_{j} = 0$ . Wherever this page says "the independence step," it means exactly this argument.

True or false — justify

Rank + nullity equals

dim W

, the codomain.

False. It equals

dim V = n

, the domain. The image can sit inside a huge

W

with tons of room unused — that spare room is irrelevant to the count of how

V

's directions are sorted.

T : R^{5} \to R^{2}

, its rank can be at most

2

True. The image is a subspace of

R^{2}

, so

rank \leq dim W = 2

; then nullity

= 5 - rank \geq 3

, so a lot of the domain must be crushed.

A linear map

T : R^{2} \to R^{5}

can have rank

5

False. Rank

\leq dim V = 2

as well (the image is spanned by images of a basis of

V

, which has only

2

vectors), so rank

\leq min (dim V, dim W) = 2

ker T = {0}

then

T

is injective (one-to-one).

True. Nullity

0

means only

0

maps to

0

; since

T

is linear,

T (a) = T (b) \Leftrightarrow T (a - b) = 0 \Leftrightarrow a = b

. This is exactly why the kernel measures failure of injectivity.

For

T : V \to V

with

dim V = n

, injective implies surjective.

True. Injective gives nullity

0

, so rank

= n = dim V

, meaning the image fills all of

V

. This equivalence is special to equal finite dimensions — see Invertible Linear Maps and Isomorphisms.

T : R^{3} \to R^{4}

then

T

can be surjective.

False. Rank

\leq dim V = 3 < 4 = dim W

, so the image never fills

R^{4}

. You cannot ship out more independent directions than you fed in.

Nullity is always strictly less than

dim V

False. The zero map has

ker T = V

, so nullity

= dim V

(that is

k = n

) and rank

= 0

. Every input direction is crushed.

Rank

0

forces

T

to be the zero map.

True. Rank

0

means

im T = {0}

, so

T (v) = 0

for every

v

. That is the definition of the zero map.

If two maps

T, S : V \to W

have the same rank, they have the same kernel.

False. Same rank forces same nullity (both equal

dim V - rank

), but equal dimension of kernel does not mean the same kernel subspace — different lines through the origin can share a dimension.

The independence of the survivors

{v_{1}, \dots, v_{n - k}}

V

guarantees

{T (v_{1}), \dots, T (v_{n - k})}

is independent.

False in general — a map can collapse independent vectors. It holds here only because those

v_{j}

were chosen outside the kernel via the Basis Extension Theorem; the independence step of the proof (recalled above) does real work to earn it.

Rank–nullity applies to

T = \frac{d}{d x}

on all polynomials

R [x]

(unbounded degree).

False as stated. That space is infinite-dimensional, and the counting proof needs

dim V < \infty

(see Dimension of a Vector Space). The finite-degree version

P_{n}

is fine.

Spot the error

"Rank

=

number of columns of the matrix, so an

m \times n

matrix has rank

n

Wrong: rank

=

number of independent columns (the dimension of the column space), which can be far less than

n

if columns repeat or depend on each other.

"Since

ker T \subseteq V

and

im T \subseteq V

, and they only meet at

0

, we get

V = ker T \oplus im T

Wrong:

im T

lives in the codomain

W

, not in

V

, so they need not even be in the same space. The theorem adds dimensions, it does not split

V

into these two subspaces.

T (v_{j})

spans the image because there are

n - k

of them and

dim (im T) = n - k

Circular: you cannot use the conclusion (

dim = n - k

) to prove spanning. The proof spans first (any

w = T (v)

collapses onto the

T (v_{j})

since kernel parts vanish), then deduces the dimension.

"If

\sum c_{j} T (v_{j}) = 0

then each

c_{j} T (v_{j}) = 0

, so

c_{j} = 0

Wrong: a single sum of vectors equalling

0

does not force each term to be

0

unless the vectors are already known independent — which is exactly what we are trying to prove. The correct route pulls

\sum c_{j} v_{j}

into

ker T

and uses full-basis independence of

V

"Extend the kernel basis by any

n - k

vectors of

V

; they'll form a basis."

Wrong: arbitrary added vectors may be dependent on the kernel basis. The Basis Extension Theorem specifically supplies vectors that keep the set independent.

"Row rank and column rank can differ, so 'rank' in the theorem is ambiguous."

Wrong: row rank

=

column rank always (see Rank of a Matrix (row vs column space)). Rank in rank–nullity is

dim (im T) =

column rank, which equals row rank.

Why questions

Why does the sum land on

dim V

and never on

dim W

Because we start by counting a basis of

V

(

n

vectors) and sort those

n

input directions into "crushed" (

k

) and "shipped independently" (

n - k

). The codomain never enters the count.

Why must we choose the kernel basis first, then extend?

To cleanly isolate the crushed directions (

u_{1}, \dots, u_{k}

) so that the remaining

v_{j}

are guaranteed to lie outside the kernel; that separation is what makes their images independent.

Why does each

T (u_{i}) = 0

matter in the spanning step?

Writing any input

v

in the full basis, the kernel-coefficient part maps to

0

, so every output is built solely from the survivors

T (v_{j})

— that is precisely why they span the image.

Why is finite-dimensionality of

V

essential?

The proof extends a basis and counts a finite number of vectors (

k + (n - k) = n

). With infinitely many basis vectors that arithmetic of dimensions breaks down and needs cardinality care.

Why does more nullity force less rank (for fixed domain)?

They are tied by

rank = dim V - nullity = n - k

: every extra crushed direction is one fewer surviving output direction — free variables trade off against pivots.

Why is rank

=

number of pivots when we row-reduce?

Each pivot marks an independent surviving direction of the column space (the image), so counting pivots counts the rank; the non-pivot (free) columns correspond to the nullity.

Edge cases

The zero map

T : V \to W

: what are its rank and nullity?

Rank

= 0

(image is just

{0}

) and nullity

= dim V

(everything is crushed,

k = n

). Their sum is

dim V

. ✓

The identity map

T : V \to V

: what are its rank and nullity?

Rank

= dim V

(nothing is lost) and nullity

= 0

(only

0

maps to

0

k = 0

). Sum

= dim V

. ✓

A map from the zero space

T : {0} \to W

: does the theorem hold?

Yes trivially:

dim V = 0

, and both rank and nullity are

0

, so

0 + 0 = 0

. The empty basis case still obeys the law.

An isomorphism

T : V \to W

with

dim V = dim W

: what does rank–nullity say?

Nullity

= 0

and rank

= dim V = dim W

, so

T

is injective and surjective — exactly the isomorphism condition.

Can nullity and rank both be zero at once?

Only if

dim V = 0

, i.e.

V = {0}

. Otherwise their sum is the positive number

dim V

, so at least one must be nonzero.

T

is injective but

dim V < dim W

, is it surjective?

No. Injective gives rank

= dim V < dim W

, so the image is a proper subspace — plenty of

W

is never reached. Injective ≠ surjective when dimensions differ.