Null space (kernel) and column space (image) — basis, dimension: Exercises

Before the ladder, a one-screen refresher of the only five facts you need. Every solution below is just these facts applied carefully.

Recall The five load-bearing facts (cover and recite)

Pivot = a leading $1$ in RREF (reduced row echelon form). Count of pivots $= r = rank (A)$ .
$dim Col (A) = r$ ; a basis of Col = the pivot-position columns taken from the original $A$ .
$dim Null (A) = n - r$ (the nullity); a basis of Null = the special solutions, one per free column.
$Col (A) \subseteq R^{m}$ (outputs), $Null (A) \subseteq R^{n}$ (inputs). Here $A$ is $m \times n$ : $m$ rows, $n$ columns.
Rank–Nullity: $r + (n - r) = n$ = total number of columns.

Here is the map of the whole exercise flow, so you always know where you are:

L1 — Recognition

Goal: read structure off a matrix or its RREF, no heavy computation.

Recall Solution 1.1

There are $3$ pivots, so $r = 3$ . (a) $rank (A) = 3$ . (b) $dim Col (A) = r = 3$ . (c) $dim Null (A) = n - r = 5 - 3 = 2$ (the free columns are $2$ and $5$ ). (d) $Col (A) \subseteq R^{m} = R^{3}$ — outputs have as many entries as there are rows.

Recall Solution 1.2

$R_{1}$ : leading $1$ 's sit in columns $1$ and $3$ → pivot columns ${1, 3}$ , free column ${2}$ . So $r = 2$ , one free variable. $R_{2}$ : leading $1$ 's in columns $1$ and $2$ → pivot columns ${1, 2}$ , free column ${3}$ . So $r = 2$ , one free variable.

L2 — Application

Goal: fully compute both bases from a concrete matrix.

Recall Solution 2.1

Step 1 — row reduce (WHAT): eliminate below the first pivot. $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 3 R_{1}$ : $100200111 .$ Now $R_{3} \to R_{3} - R_{2}$ , then clear above the second pivot with $R_{1} \to R_{1} - R_{2}$ : $R = 100200010 .$ Why RREF: it exposes pivots and hands us the pivot-in-terms-of-free formulas directly.

Step 2 — pivots in columns $1, 3$ → $r = 2$ . Col basis = columns $1, 3$ of the original $A$ : ${(1, 2, 3)^{⊤}, (1, 3, 4)^{⊤}}$ . $dim Col = 2$ . Why original columns: row operations moved the entries inside $R$ 's columns, so $R$ 's columns no longer point in the true output directions.

Step 3 — null space. Free variable $x_{2}$ . From $R$ : $x_{1} = - 2 x_{2}$ , $x_{3} = 0$ . Set $x_{2} = 1$ : special solution $s = (- 2, 1, 0)^{⊤}$ . Null basis $= {(- 2, 1, 0)^{⊤}}$ , $dim Null = 1$ .

Step 4 — check: $r + nullity = 2 + 1 = 3 = n$ . ✓ Sanity: $A s = - 2 (1, 2, 3)^{⊤} + 1 (2, 4, 6)^{⊤} = (0, 0, 0)^{⊤}$ . ✓

Recall Solution 2.2

Row reduce: $R_{2} \to R_{2} - 2 R_{1}$ gives a zero second row: $R = 000100200 .$ The first nonzero entry of row $1$ is in column 2 — so the single pivot is in column $2$ , not column $1$ . Pivot column: ${2}$ ; free columns: ${1, 3}$ . $r = 1$ . Col basis = column $2$ of original $A$ : ${(1, 2, 0)^{⊤}}$ . $dim Col = 1$ . Null space: free $x_{1}, x_{3}$ . Row $1$ : $x_{2} + 2 x_{3} = 0 \Rightarrow x_{2} = - 2 x_{3}$ ; $x_{1}$ is unconstrained.

$x_{1} = 1, x_{3} = 0$ : $(1, 0, 0)^{⊤}$ .
$x_{1} = 0, x_{3} = 1$ : $(0, - 2, 1)^{⊤}$ . Null basis $= {(1, 0, 0)^{⊤}, (0, - 2, 1)^{⊤}}$ , $dim Null = 2$ . Check: $1 + 2 = 3 = n$ . ✓

L3 — Analysis

Goal: reason about dimension and structure, sometimes without a specific matrix.

Recall Solution 3.1

Here $m = 4$ , $n = 6$ . (a) Rank–Nullity: $r = n - dim Null = 6 - 2 = 4$ . (b) Injective $⟺ Null (A) = {0} ⟺$ nullity $0$ . Here nullity is $2 \neq = 0$ , so not injective — different inputs can give the same output. (c) Surjective onto $R^{4}$ $⟺ dim Col (A) = 4 = m$ . Here $dim Col = r = 4 = m$ , so yes, it is surjective — every output vector is reachable. Reading it: a wide matrix ( $n > m$ ) at full row rank flattens the extra input directions (nullity $> 0$ ) while still covering all of output space.

Recall Solution 3.2

Here $n = 3$ columns, so the rank satisfies $0 \leq r \leq min (m, n) = min (5, 3) = 3$ . Nullity $= n - r = 3 - r$ . The largest nullity occurs at the smallest rank, $r = 0$ (the zero matrix), giving nullity $3$ . So nullity can reach $3$ , not just $2$ . The statement is FALSE. The tightest correct bound is $0 \leq dim Null (A) \leq 3$ .

L4 — Synthesis

Goal: combine null space, column space and solvability of $A x = b$ .

Recall Solution 4.1

Recall $A = 123246134$ , columns $c_{1} = (1, 2, 3)^{⊤}$ , $c_{2} = (2, 4, 6)^{⊤}$ , $c_{3} = (1, 3, 4)^{⊤}$ , and $Col (A) = span {c_{1}, c_{3}}$ (from Ex 2.1). (a) Is $b$ reachable? (solvable iff $b\in\mathrm{Col}(A)$.) Try $b = α c_{1} + γ c_{3}$ : $α + γ = 2, 2 α + 3 γ = 5, 3 α + 4 γ = 7.$ From the first: $α = 2 - γ$ . Sub into the second: $2 (2 - γ) + 3 γ = 5 \Rightarrow 4 + γ = 5 \Rightarrow γ = 1$ , so $α = 1$ . Check third: $3 (1) + 4 (1) = 7$ . ✓ So yes, $b \in Col (A)$ . (b) Particular solution. We wrote $b = 1 \cdot c_{1} + 0 \cdot c_{2} + 1 \cdot c_{3}$ , i.e. $x_{1} = 1, x_{2} = 0, x_{3} = 1$ . So $x_{p} = (1, 0, 1)^{⊤}$ . (Verify: $A x_{p} = c_{1} + c_{3} = (1, 2, 3)^{⊤} + (1, 3, 4)^{⊤} = (2, 5, 7)^{⊤} = b$ . ✓) (c) Full solution set = particular + null space: $x = x_{p} + t s = 101 + t - 2 10, t \in R,$ using the null-space basis $s = (- 2, 1, 0)^{⊤}$ from Ex 2.1. It is a line in $R^{3}$ : infinitely many solutions because the kernel is nontrivial.

The picture below shows the two flavours of solution set — a single point when the kernel is trivial, a whole line when it is not.

Figure — Null space (kernel) and column space (image) — basis, dimension

Recall Solution 4.2

The null-space basis has $1$ vector, so $dim Null (A) = 1$ . Rank–Nullity: $r = n - 1 = 3 - 1 = 2$ . Invertible $⟺$ full rank $r = n = 3$ $⟺$ nullity $0$ . Here nullity is $1 \neq = 0$ , so $A$ is NOT invertible — it collapses the direction $(1, 1, 1)^{⊤}$ to zero, so it cannot be undone.

L5 — Mastery

Goal: prove a general statement and design a matrix to specification.

Recall Solution 5.1

Read the requirements as column relations. $A (1, - 1, 0, 0)^{⊤} = 0$ means $c_{1} - c_{2} = 0$ , i.e. $c_{1} = c_{2}$ . $A (0, 0, 1, - 1)^{⊤} = 0$ means $c_{3} - c_{4} = 0$ , i.e. $c_{3} = c_{4}$ . So columns come in two equal pairs. For $rank = 2$ we need $c_{1}$ and $c_{3}$ independent. Pick $c_{1} = (1, 0, 0)^{⊤}$ , $c_{3} = (0, 1, 0)^{⊤}$ . Then: $A = 100100010010 .$ Verify rank: RREF is $A$ itself; pivots in columns $1, 3$ → $r = 2$ . ✓ Verify kernel vectors: $A (1, - 1, 0, 0)^{⊤} = (0, 0, 0)^{⊤}$ ✓; $A (0, 0, 1, - 1)^{⊤} = (0, 0, 0)^{⊤}$ ✓. Consistency with Rank–Nullity: nullity $= n - r = 4 - 2 = 2$ , exactly the two independent kernel vectors we prescribed. ✓

Recall Solution 5.2

Setup. Rank counts pivots, and each pivot occupies its own row and its own column, so $r \leq m$ and $r \leq n$ ; in particular $r \leq m$ . Core step. By Rank–Nullity, $dim Null (A) = n - r$ . Since $r \leq m$ , $dim Null (A) = n - r \geq n - m .$ Use the hypothesis. Given $m < n$ , we have $n - m > 0$ , hence $dim Null (A) \geq n - m > 0$ . A subspace of positive dimension contains a nonzero vector, so $Null (A) \neq = {0}$ . $■$ Meaning: more inputs than outputs forces the machine to squash at least $n - m$ independent directions to zero — a wide map can never be injective.

Connections

Rank of a matrix — every "find $r$ " step above.
Rank–Nullity Theorem — the identity powering L3–L5.
Row reduction & RREF — the engine behind Ex 2.1, 2.2, 5.1.
Linear independence and basis — why pivot columns form a basis.
Injective and surjective linear maps — Ex 3.1, 4.2, 5.2.
Solving Ax=b — Ex 4.1's particular + kernel structure.
Four fundamental subspaces — the bigger picture these bases sit in.