3.6 Row Space and Column Space · Linear Algebra Refresher

This is the payoff section of Chapter 3 and the densest source of true/false questions on any exam. One number — the rank — controls consistency, uniqueness, and the dimension of every subspace attached to

A

Definition · Row space, column space, rank

For an

m \times n

matrix

A

: the row space is the span of the rows (a subspace of

R^{1 \times n}

), the column space

R (A)

is the span of the columns (a subspace of

R^{m}

), and

rank (A)

is the dimension of the row space.

Theorem 3.6.1 & 3.6.6

Row equivalent matrices have the same row space. And the dimension of the row space equals the dimension of the column space — so 'rank' is unambiguous.

Equal dimensions, but not equal spaces: for a $3 \times 5$ matrix the row space sits in $R^{5}$ and the column space in $R^{3}$ .

Theorem 3.6.2 · Consistency, upgraded

A x = b

is consistent if and only if

b

is in the column space of

A

Theorem 3.6.5 · Rank–Nullity

For any

m \times n

matrix

A

rank (A) + dim N (A) = n .

$n$ is the number of columns — the dimension of the domain. Subtracting from $m$ is the single most common error on this theorem.

The rank decision table

Let $A$ be $m \times n$ with $rank (A) = r$ . Every consistency/uniqueness question reduces to comparing $r$ against $m$ and $n$ :

Case	Consistency	Solution count when consistent
$r = n = m$	Consistent for every $b$ — columns span $R^{m}$	Exactly one ( $A$ nonsingular)
$r = n < m$ (full column rank, tall)	Only for $b$ in an $r$ -dimensional subspace of $R^{m}$	Exactly one — this is the least-squares setting of §5.3
$r = m < n$ (full row rank, wide)	Consistent for every $b$	Infinitely many, $n - r$ free parameters
$r < m$ and $r < n$	Only for $b$ in an $r$ -dimensional subspace	Infinitely many, $n - r$ free parameters

Read it as two independent questions: rows vs rank controls existence, columns vs rank controls uniqueness.

The table is easier to trust once you have seen it. Below, the rank is pinned at $r = 2$ and only the shape moves — all four rows of the table appear from that one number.

Same rank, four shapes

rank(A) = 2 held fixed

Fix rank(A) = 2 and vary only the shape. The row you are in decides whether a solution exists; the column decides whether it is unique.

n = 2

2 unknowns

n = 3

3 unknowns

m = 2

2 eqns

m = 3

3 eqns

Counts include the inconsistent case. Hover or tap a square to see what that shape looks like geometrically.

2×2Square and nonsingular — the two columns tile all of

R^{2}

A = (21 - 1 2)

. The columns are independent, so the set of all

x_{1} a_{1} + x_{2} a_{2}

is the faint lattice — it covers

R^{2}

and covers it once. Every

b

sits on exactly one crossing; here

b = 2 a_{1} + a_{2}

. Inconsistency is impossible because

R (A) = R^{2}

, and non-uniqueness is impossible because

N (A) = {0}

★The classic exam question

A

has 6 rows and rank 5 — how many solutions can

A x = b

have?' You cannot answer without the number of columns. If

A

6 \times 5

: at most one solution (and possibly none). If

A

6 \times 8

: none, or infinitely many with 3 parameters — never exactly one. Always ask for

n

Finding bases for all three subspaces of $A$

1
Row reduce $A$ to reduced row echelon form $U$
Keep track of which columns contain the pivots.
2
Row space basis: the nonzero rows of $U$
Legitimate because row operations preserve the row space (Thm 3.6.1). Dimension $= r$ .
3
Column space basis: the pivot columns of the original $A$
Row operations change the column space but preserve which columns are dependent on which. Dimension $= r$ .
4
Null space basis: set each free variable to 1 in turn
Dimension $= n - r$ , confirming Rank–Nullity.
✓
Sanity check
$dim (row) = dim (col) = r$ and $r + dim N (A) = n$ . If these don't line up, you miscounted pivots.

One row reduction, three answers. This routine appears on virtually every exam.

Worked example

All three subspaces from one reduction

0/6 steps

Find the rank and a basis for the row space, column space, and null space.

A = 121242 - 1 - 3 1 105

!Exam trap

Never take the pivot columns of

U

as a basis for

R (A)

. Row operations mix the entries within each column, so

R (U) \neq = R (A)

in general — e.g.

(1 1)

reduces to

(0 1)

, and those span different lines.

Corollary 3.6.4 · Extending the equivalence list

For an

n \times n

matrix

A

, all of the following are equivalent:

A

is nonsingular;

det (A) \neq = 0

;

A x = 0

has only the trivial solution;

A

is row equivalent to

I

; the columns of

A

are linearly independent; the columns form a basis of

R^{n}

;

rank (A) = n

;

N (A) = {0}

;

A x = b

has a unique solution for every

b

Check your work

Row reduction / rank / null space

Exact fraction arithmetic — type entries like 3, -2, or 1/2. One row per line.

Matrix

1	2	-1	1
2	4	-3	0
1	2	1	5

→

1	2	0	3
0	0	1	2
0	0	0	0

rank

nullity

cols (n)

pivot cols

1, 3

Basis for N(A)

-2

-3

-2

Basis for the column space — original columns 1, 3

-1

-3

Feed in a matrix to get rank, nullity, and bases for the null space and column space at once.

Check yourself

A

is a

6 \times 8

matrix with

rank (A) = 5

. If

A x = b

is consistent, how many solutions does it have?

A

6 \times 8

with

rank (A) = 5

. For how many

b \in R^{6}

A x = b

consistent?

T / F

U

is a row echelon form of

A

, then

A

and

U

have the same column space.

T / F

For any

m \times n

matrix,

rank (A) \leq min (m, n)

3.6Row Space and Column Space

What you must be able to do

The rank decision table

Same rank, four shapes

Finding bases for all three subspaces of $A$

All three subspaces from one reduction

Row reduction / rank / null space

Check yourself

3.6Row Space and Column Space

What you must be able to do

The rank decision table

Same rank, four shapes

Finding bases for all three subspaces of A

All three subspaces from one reduction

Row reduction / rank / null space

Check yourself

Finding bases for all three subspaces of $A$