Rank of a matrix — definition, row rank = column rank theorem: Worked examples

Before any computation, one reminder in plain words.

Cell	Case class	What makes it tricky	Example
A	Square, full rank ( $r = n$ )	every direction survives	Ex 1
B	Square, rank-deficient ( $r < n$ )	a hidden dependency	Ex 2
C	Wide rectangle $m < n$	more columns than the space allows	Ex 3
D	Tall rectangle $m > n$	rank capped by columns	Ex 4
E	Zero rows / zero columns present	nonzero-row trap	Ex 5
F	All-zero (degenerate) matrix	the limiting case $r = 0$	Ex 6
G	Rank-1 witness (outer product)	column rank seen instantly	Ex 7
H	Real-world word problem (data table)	translate then rank	Ex 8
I	Exam twist — rank of a product $A B$	the bound $rank (A B) \leq min$	Ex 9
J	Parameter case — rank depends on a symbol $k$	limiting/degenerate values	Ex 10

$3 \times 3$ matrix where nothing collapses $A = 102210031$ Forecast: guess the rank before reading on. Three rows, none is an obvious copy of another — so a good guess is the maximum, $3$ . Let's verify.

Step 1 — clear column 1 below the pivot. $R_{3} \to R_{3} - 2 R_{1}$ . Why this step? The entry $2$ under the first pivot must become $0$ so the staircase can descend cleanly. $\to 100 21 - 4 031$ Step 2 — clear column 2 below its pivot. $R_{3} \to R_{3} + 4 R_{2}$ . Why? Same reason: expose whether row 3 is independent or secretly a mix of the first two. $\to 1002100313$ Step 3 — count pivots. Positions $(1, 1), (2, 2), (3, 3)$ : three pivots. $rank (A) = 3$ . Why this finishes? Three non-zero staircase rows ⇒ the row space is $3$ -dimensional ⇒ full rank.

Verify: full rank on a square matrix means $det A \neq = 0$ . Here $det A = 13 \neq = 0$ (equal to the product of the pivots $1 \cdot 1 \cdot 13$ ). By the Invertible Matrix Theorem, $A$ is invertible — exactly what full rank promises. ✔️

Figure — Rank of a matrix — definition, row rank = column rank theorem

Cell B — square, rank-deficient

Worked example Ex 2 · A square matrix hiding one dependency

$A = 121241361$ Forecast: look at rows 1 and 2 — the second is exactly $2 \times$ the first. So at least one row is "not new". Predict $rank = 2$ , not $3$ .

Step 1 — kill the copy. $R_{2} \to R_{2} - 2 R_{1}$ . Why? Row 2 is $2 R_{1}$ ; subtracting reveals it carries zero new information. $\to 101201301$ Step 2 — clear column 1 of row 3. $R_{3} \to R_{3} - R_{1}$ . Why? Standard descent of the staircase. $\to 100 20 - 1 30 - 2$ Step 3 — swap to make a clean staircase. $R_{2} \leftrightarrow R_{3}$ . Why? A zero row must sit at the bottom; a row swap never changes the row space, only its order. $\to 100 2 - 1 0 3 - 2 0$ Step 4 — count pivots. $(1, 1)$ and $(2, 2)$ : two pivots. $rank = 2$ .

Verify: rank-deficient square ⇒ $det A = 0$ . Indeed $det A = 0$ (row 2 $= 2 \cdot$ row 1 forces it). Matches the deficiency. ✔️

Cell C — wide rectangle ( $m < n$ )

Cell D — tall rectangle ( $m > n$ )

Worked example Ex 4 · Four rows, two columns — rank capped by columns

$A = 12 - 1 0 24 - 2 5$ Forecast: four rows but only $2$ columns, so $rank \leq min (4, 2) = 2$ . Rows 1,2,3 are all multiples of $(1, 2)$ ; row 4 is different. Predict $2$ .

Step 1 — eliminate below the first pivot. $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} + R_{1}$ . Why? Rows 2 and 3 start proportional to row 1; clearing exposes whether they vanish. $\to 10002005$ Step 2 — move the live row up. $R_{2} \leftrightarrow R_{4}$ . Why? Put the non-zero row $(0, 5)$ where a second pivot can appear; row swaps preserve the row space. $\to 10002500$ Step 3 — count pivots: $(1, 1)$ and $(2, 2)$ ⇒ two. $rank = 2$ .

Verify: column rank must equal $2$ too. Columns are $(1, 2, - 1, 0)^{T}$ and $(2, 4, - 2, 5)^{T}$ ; the second is not $2 \times$ the first (the last entry $5 \neq = 0$ ), so the two columns are independent ⇒ column rank $= 2 =$ row rank. The theorem holds again. ✔️

Cell E — zero row / zero column present (the nonzero-row trap)

Cell F — the degenerate all-zero matrix (limiting case $r = 0$ )

Cell G — rank-1 witness (outer product)

Cell H — real-world word problem

Worked example Ex 8 · A price table that isn't as informative as it looks

A shop records three receipts. Each row is $(\text{apples},\ \text{bananas},\ \text{total cost in }\$ )$: $A = 2411235108$ Question: how many genuinely independent pieces of pricing information are here?

Forecast: receipt 2 is exactly double receipt 1 (same shopping, twice) — it teaches nothing new. So among $3$ receipts, expect only $2$ independent facts. Predict $rank = 2$ .

Step 1 — translate. "Independent pieces of information" $=$ rank of the receipt table. Why? Each dependent row is redundant data; rank counts the non-redundant ones. Step 2 — reduce. $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - \frac{1}{2} R_{1}$ . $\to 200 10 \frac{5}{2} 50 \frac{11}{2}$ Step 3 — swap zero row down. $R_{2} \leftrightarrow R_{3}$ . $\to 200 1 \frac{5}{2} 0 5 \frac{11}{2} 0$ Step 4 — count pivots: $(1, 1)$ and $(2, 2)$ ⇒ two. $rank = 2$ .

Verify (units & meaning): the entries in column 3 are dollars, columns 1–2 are counts — the numbers stay consistent, and the answer " $2$ independent receipts" matches the observation that receipt 2 duplicated receipt 1. The shop effectively has $2$ pricing equations, enough to solve for two unknown prices. ✔️

Cell I — exam twist: rank of a product

Worked example Ex 9 · Predict

$rank (A B)$ before multiplying $A = (1224), B = (111111)$ Forecast: $A$ has rank $1$ (row 2 $= 2 \times$ row 1) and $B$ has rank $1$ (row 2 $=$ row 1). The bound says $rank (A B) \leq min (rank A, rank B) = min (1, 1) = 1$ . Predict $rank (A B) \leq 1$ .

Step 1 — multiply. Why? To see whether the bound is met with equality or the product collapses further (it can drop below the bound). $A B = (1 \cdot 1 + 2 \cdot 1 2 \cdot 1 + 4 \cdot 1 \dots \dots \dots \dots) = (363636)$ Step 2 — rank of the result. Every column is $(3, 6)^{T}$ ; row 2 $= 2 \times$ row 1. One pivot after reducing. $rank (A B) = 1$ .

Verify: $1 \leq min (1, 1) = 1$ ✔️ — the bound holds and is tight here.

Recall Why the product can drop below the bound

Take the same $A$ but $B^{'} = (1 - \frac{1}{2} 00)$ (rank $1$ ). Then the columns of $A B^{'}$ can land in the kernel of $A$ and give $rank (A B^{'}) = 0$ . The bound is an upper limit, never a guarantee of equality.

Cell J — parameter case: rank depends on a symbol

Worked example Ex 10 · When does the rank drop?

$A = (13 2 k)$ for a real number $k$ . Find $rank (A)$ in every case.

Forecast: two rows. They become dependent only when row 2 is a multiple of row 1. Row 1 is $(1, 2)$ ; scaling by $3$ gives $(3, 6)$ . So the danger value is $k = 6$ . Predict: rank $2$ usually, rank $1$ at $k = 6$ .

Step 1 — eliminate. $R_{2} \to R_{2} - 3 R_{1}$ . Why? To isolate the single entry whose vanishing signals dependence. $\to (10 2 k - 6)$ Step 2 — split on the pivot entry $k - 6$ .

If $k - 6 \neq = 0$ (i.e. $k \neq = 6$ ): a second pivot exists ⇒ $rank = 2$ .
If $k - 6 = 0$ (i.e. $k = 6$ ): row 2 vanishes ⇒ only one pivot ⇒ $rank = 1$ . Why cover both? The parent contract: every case, including the degenerate/limiting value $k = 6$ , must be shown.

Verify: $det A = 1 \cdot k - 2 \cdot 3 = k - 6$ . Full rank $⟺ det \neq = 0 ⟺ k \neq = 6$ . At $k = 6$ , $det = 0$ and rank drops to $1$ . Consistent with the pivot analysis. ✔️

Recall Quick self-check across the scenario matrix

Which cell has $rank = 0$ , and why is that allowed? ::: Cell F (all-zero matrix); the bound $0 \leq rank \leq min (m, n)$ permits the lower limit $0$ . A wide $2 \times 5$ matrix — what is its maximum possible rank? ::: $min (2, 5) = 2$ . If $rank A = 1$ and $rank B = 1$ , can $rank (A B) = 0$ ? ::: Yes — the bound is only an upper limit; overlap with $A$ 's kernel can push it lower.

Connections

Parent: Rank definition & theorem
Gaussian Elimination & Echelon Form
Row Space and Column Space
Linear Independence and Basis
Matrix Multiplication as Linear Combination
Rank–Nullity Theorem
Invertible Matrix Theorem

Worked examples — Rank of a matrix — definition, row rank = column rank theorem

The scenario matrix

Cell A — square, full rank