Linear-Algebra-Done-Right · Unit 6 · Video 4 · Interactive Practice

Column Rank Equals Row Rank: Column-Row Factorization

IKey Formulas

Formula	Name	Setting
$\dim \operatorname{span}(A_{\cdot,1}, \dots, A_{\cdot,n}) \le \min\{m, n\}$	Column rank of $A$ : the dimension of the column span, a subspace of $\mathbf{F}^{m,1}$	$A \in \mathbf{F}^{m,n}$ ; the row rank $\dim \operatorname{span}(A_{1,\cdot}, \dots, A_{m,\cdot})$ obeys the same bound
$(A^{\mathrm{t}})_{k,j} = A_{j,k}$ , $\ (AC)^{\mathrm{t}} = C^{\mathrm{t}} A^{\mathrm{t}}$	Transpose	$A \in \mathbf{F}^{m,n}$ , $A^{\mathrm{t}} \in \mathbf{F}^{n,m}$ , $C \in \mathbf{F}^{n,p}$
$A = CR$ , $\ C \in \mathbf{F}^{m,c}$ , $R \in \mathbf{F}^{c,n}$	Column–row factorization	Column rank $c \ge 1$ ; the columns of $C$ are a basis of the column span chosen from the columns of $A$
$\operatorname{rank} A = \text{column rank of } A = \text{row rank of } A$	Rank of a matrix	Every $A \in \mathbf{F}^{m,n}$

Key Insight: Read $A = CR$ by columns and each column of $A$ combines the $c$ columns of $C$ ; read it by rows and each row of $A$ combines the $c$ rows of $R$ , so row rank $\le c$ . Applying that to $A^{\mathrm{t}}$ gives the reverse inequality.

IIBuilding C and R from Any Matrix

Keep each column of $A$ not in the span of the kept ones; the kept columns form $C$ , the coefficients form $R$ .

Rows m Columns n

IIIOne Product, Read Two Ways

Columns of $CR$ combine the columns of $C$ ; rows of $CR$ combine the $c$ rows of $R$ , which bounds the row rank by $c$ .

Example Slice of A

💡 In the $3 \times 3$ example, three rows of $A$ lie in the span of only two rows of $R$ , so the row rank is at most $2$ even though $m = 3$ .

IVThe Transpose Swaps Rows and Columns

Column $k$ of $A$ becomes row $k$ of $A^{\mathrm{t}}$ , so the column rank of $A^{\mathrm{t}}$ is the row rank of $A$ .

Matrix Columns moved 0 / 2

VQuiz Questions

Problem 1 · Compute a Column Rank

Given: $A = \begin{pmatrix} 1 & 2 & 0 & 3 \\ 2 & 4 & 1 & 7 \\ 3 & 6 & 1 & 10 \end{pmatrix} \in \mathbf{F}^{3,4}$ . Find the column rank of $A$ .

✅ Correct! Columns

1

and

3

form a basis of the column span. The rows agree: row

3

is row

1

plus row

2

, and rows

1

2

are not multiples, so the row rank is

2

as well.

❌ Not quite.

4

is the number of columns, not the dimension of their span. Column

2

2

times column

1

, so the list is not linearly independent; besides, a subspace of

\mathbf{F}^{3,1}

has dimension at most

3

❌ Not quite.

\min\{3, 4\} = 3

is only an upper bound. Column

4

equals

3 \cdot

column

1

+

column

3

, so after removing columns

2

and

4

only two columns remain in a basis.

❌ Not quite. Column

1 = (1, 2, 3)^{\mathrm{t}}

and column

3 = (0, 1, 1)^{\mathrm{t}}

are not scalar multiples of each other, so the span has dimension at least

2

Show solution

Go through the columns, keeping each one that is not in the span of the columns already kept.

Column $1 = (1, 2, 3)^{\mathrm{t}}$ is nonzero: keep it.
Column $2 = (2, 4, 6)^{\mathrm{t}} = 2 \cdot$ column $1$ : remove it.
Column $3 = (0, 1, 1)^{\mathrm{t}}$ is not a multiple of column $1$ (its first entry is $0$ but it is nonzero): keep it.
Column $4 = (3, 7, 10)^{\mathrm{t}} = 3\,(1, 2, 3)^{\mathrm{t}} + (0, 1, 1)^{\mathrm{t}}$ : remove it.

Columns $1$ and $3$ are linearly independent and span the column span, so

\text{column rank of } A = 2.

Check with rows: $(3, 6, 1, 10) = (1, 2, 0, 3) + (2, 4, 1, 7)$ , and the first two rows are not multiples of each other, so the row rank is also $2$ .

Problem 2 · Sizes in A = CR

Given: $A \in \mathbf{F}^{5,3}$ has column rank $2$ , and $A = CR$ is its column–row factorization. Find the sizes of $C$ and $R$ .

✅ Correct!

C

holds

c = 2

basis columns with

m = 5

entries each, and

R

holds one coefficient column (length

2

) for each of the

n = 3

columns of

A

❌ Not quite. The middle size is the column rank

c = 2

, not

n = 3

C

keeps only a basis of the column span, which has length

2

❌ Not quite. The columns of

C

are columns of

A

, so each has

m = 5

entries:

C

5 \times 2

, not

2 \times 5

. And

C

times your

R

would not even be defined.

❌ Not quite.

R

has one column per column of

A

, so it has

n = 3

columns. With

R \in \mathbf{F}^{2,5}

the product

CR

would be

5 \times 5

, not

5 \times 3

Show solution

The theorem: if $A \in \mathbf{F}^{m,n}$ has column rank $c \ge 1$ , then $A = CR$ with $C \in \mathbf{F}^{m,c}$ and $R \in \mathbf{F}^{c,n}$ .

$C$ : its $c$ columns are a basis of the span of the columns of $A$ , each an element of $\mathbf{F}^{m,1}$ . So $C$ is $m \times c = 5 \times 2$ .
$R$ : column $k$ of $R$ lists the $c$ coefficients that express column $k$ of $A$ in the columns of $C$ , for $k = 1, \dots, n$ . So $R$ is $c \times n = 2 \times 3$ .

Size check: $(5 \times 2)(2 \times 3) = 5 \times 3$ , the size of $A$ .

Problem 3 · Find R

Given: $A = \begin{pmatrix} 3 & 1 & 5 & -1 \\ 2 & 1 & 4 & 0 \end{pmatrix}$ , and $C = \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}$ is formed from its first two columns. Find the matrix $R$ with $A = CR$ .

✅ Correct! Column

3

1\,(3, 2)^{\mathrm{t}} + 2\,(1, 1)^{\mathrm{t}} = (5, 4)^{\mathrm{t}}

. Column

4

-1\,(3, 2)^{\mathrm{t}} + 2\,(1, 1)^{\mathrm{t}} = (-1, 0)^{\mathrm{t}}

❌ Not quite. Row

1

R

holds the coefficients of column

1

C

. With your column

3

2\,(3, 2)^{\mathrm{t}} + 1\,(1, 1)^{\mathrm{t}} = (7, 5)^{\mathrm{t}} \ne (5, 4)^{\mathrm{t}}

❌ Not quite. Check column

4

1\,(3, 2)^{\mathrm{t}} + 2\,(1, 1)^{\mathrm{t}} = (5, 4)^{\mathrm{t}}

, not

(-1, 0)^{\mathrm{t}}

. The coefficient of column

1

C

must be

-1

❌ Not quite. These are the right numbers in the wrong shape: this is

R^{\mathrm{t}}

R

must be

c \times n = 2 \times 4

, with the coefficients for column

k

A

written as column

k

R

Show solution

The columns $(3, 2)^{\mathrm{t}}$ and $(1, 1)^{\mathrm{t}}$ are not multiples of each other, so they are linearly independent; they span $\mathbf{F}^{2,1}$ , and the column rank of $A$ is $2$ . Column $k$ of $R$ holds the coefficients $a, b$ with $a\,(3, 2)^{\mathrm{t}} + b\,(1, 1)^{\mathrm{t}} = A_{\cdot,k}$ .

Columns $1$ and $2$ of $A$ are the columns of $C$ : coefficients $(1, 0)$ and $(0, 1)$ .
Column $3$ : $3a + b = 5$ , $2a + b = 4$ . Subtracting gives $a = 1$ , then $b = 2$ .
Column $4$ : $3a + b = -1$ , $2a + b = 0$ . Subtracting gives $a = -1$ , then $b = 2$ .

A = \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 1 & -1 \\ 0 & 1 & 2 & 2 \end{pmatrix}

Row check: row $1$ of $A$ is $3\,(1, 0, 1, -1) + 1\,(0, 1, 2, 2) = (3, 1, 5, -1)$ ✓.

Problem 4 · What Must Be True?

Given: $A \in \mathbf{F}^{4,6}$ is any $4 \times 6$ matrix. Which statement must be true?

✅ Correct! The column rank is at most

\min\{4, 6\} = 4

. If the

6

columns were linearly independent they would be a basis of their span, forcing the column rank to be

6

❌ Not quite. Nothing forces that: the zero matrix in

\mathbf{F}^{4,6}

has row rank

0

. The row rank can be anything from

0

4

❌ Not quite. The row span is spanned by

4

rows, so its dimension is at most

4

. Rows living in

\mathbf{F}^{1,6}

only gives the weaker bound

6

❌ Not quite.

A^{\mathrm{t}}

has only

4

columns, so its column rank is at most

4

. In fact the column rank of

A^{\mathrm{t}}

equals the row rank of

A

, which is at most

4

Show solution

Here $m = 4$ and $n = 6$ . Both ranks are at most $\min\{m, n\} = 4$ :

the column span is a subspace of $\mathbf{F}^{4,1}$ , which has dimension $4$ ;
the row span is spanned by the $4$ rows, and a spanning list is at least as long as the dimension.

(d) If the $6$ columns were linearly independent, they would form a basis of their span, and the column rank would be $6 > 4$ . So they are linearly dependent, for every $A \in \mathbf{F}^{4,6}$ .

(a) fails for $A = 0$ . (b) fails because row rank $\le 4$ . (c) fails because the columns of $A^{\mathrm{t}}$ are the rows of $A$ written as columns, so the column rank of $A^{\mathrm{t}}$ is the row rank of $A$ , at most $4$ .

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