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

4.5.12 · D3 · Maths › Linear Algebra (Full) › Rank of a matrix — definition, row rank = column rank theore

Kisi bhi computation se pehle, ek reminder plain words mein.

Cell	Case class	Kya cheez tricky banati hai	Example
A	Square, full rank ( $r = n$ )	har direction survive karta hai	Ex 1
B	Square, rank-deficient ( $r < n$ )	ek hidden dependency	Ex 2
C	Wide rectangle $m < n$	columns zyada hain space se	Ex 3
D	Tall rectangle $m > n$	rank columns se cap hota hai	Ex 4
E	Zero rows / zero columns present	nonzero-row trap	Ex 5
F	All-zero (degenerate) matrix	limiting case $r = 0$	Ex 6
G	Rank-1 witness (outer product)	column rank instantly dikh jaata hai	Ex 7
H	Real-world word problem (data table)	translate karo phir rank nikalo	Ex 8
I	Exam twist — rank of a product $A B$	bound $rank (A B) \leq min$	Ex 9
J	Parameter case — rank ek symbol $k$ par depend karta hai	limiting/degenerate values	Ex 10

$3 \times 3$ matrix jahan kuch collapse nahi hota $A = 102210031$ Forecast: aage padhne se pehle rank guess karo. Teen rows hain, koi obvious copy nahi dikh rahi doosri ki — toh ek accha guess hai maximum, $3$ . Verify karte hain.

Step 1 — column 1 ko pivot ke neeche clear karo. $R_{3} \to R_{3} - 2 R_{1}$ . Yeh step kyun? Entry $2$ pehle pivot ke neeche $0$ banni chahiye taaki staircase cleanly descend kar sake. $\to 100 21 - 4 031$ Step 2 — column 2 ko uske pivot ke neeche clear karo. $R_{3} \to R_{3} + 4 R_{2}$ . Kyun? Same reason: expose karo ki row 3 independent hai ya secretly pehle dono ka mix hai. $\to 1002100313$ Step 3 — pivots count karo. Positions $(1, 1), (2, 2), (3, 3)$ : teen pivots. $rank (A) = 3$ . Yeh kyun finish karta hai? Teen non-zero staircase rows ⇒ row space $3$ -dimensional hai ⇒ full rank.

Verify: square matrix par full rank ka matlab hai $det A \neq = 0$ . Yahan $det A = 13 \neq = 0$ (pivots ke product ke equal $1 \cdot 1 \cdot 13$ ). Invertible Matrix Theorem ke hisaab se, $A$ invertible hai — exactly wahi jo full rank promise karta hai. ✔️

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

Cell B — square, rank-deficient

Worked example Ex 2 · Ek square matrix jisme ek dependency chupi hai

$A = 121241361$ Forecast: rows 1 aur 2 dekho — doosri exactly $2 \times$ pehli hai. Toh kam se kam ek row "nayi nahi" hai. Predict karo $rank = 2$ , $3$ nahi.

Step 1 — copy ko khatam karo. $R_{2} \to R_{2} - 2 R_{1}$ . Kyun? Row 2 hai $2 R_{1}$ ; subtract karne se pata chalta hai ki usme zero nayi information hai. $\to 101201301$ Step 2 — row 3 mein column 1 clear karo. $R_{3} \to R_{3} - R_{1}$ . Kyun? Staircase ka standard descent. $\to 100 20 - 1 30 - 2$ Step 3 — clean staircase ke liye swap karo. $R_{2} \leftrightarrow R_{3}$ . Kyun? Zero row neeche baithhni chahiye; row swap kabhi row space nahi badalta, sirf order. $\to 100 2 - 1 0 3 - 2 0$ Step 4 — pivots count karo. $(1, 1)$ aur $(2, 2)$ : do pivots. $rank = 2$ .

Verify: rank-deficient square ⇒ $det A = 0$ . Indeed $det A = 0$ (row 2 $= 2 \cdot$ row 1 force karta hai yeh). Deficiency se match karta hai. ✔️

Cell C — wide rectangle ( $m < n$ )

Worked example Ex 3 · Do rows, chaar columns — rank rows se cap hota hai

$A = (1001 2 - 1 13)$ Forecast: span kiya hua space length $4$ ke vectors mein rehta hai, lekin sirf $2$ rows hain. Rank $min (2, 4) = 2$ se zyada nahi ho sakta. Rows independent lagte hain, toh exactly $2$ predict karo.

Step 1 — dono rows ki independence check karo. Kya koi doosre ka multiple hai? Yeh step kyun? Sirf do rows ke saath, rank $2$ hai jab tak ek doosre ka scalar multiple na ho. Row 1 start hoti hai $(1, 0, \dots)$ se, row 2 start hoti hai $(0, 1, \dots)$ se — koi scalar ek ko doosre mein nahi badal sakta. Step 2 — yeh already echelon form mein hai pivots columns $1$ aur $2$ mein. Count: do pivots. $rank = 2$ .

Verify (bound se): $rank \leq min (m, n) = min (2, 4) = 2$ , aur humne $2$ achieve kiya. Bound tight hai. ✔️ Har wide matrix apni chhoti dimension se cap hoti hai.

Cell D — tall rectangle ( $m > n$ )

Worked example Ex 4 · Chaar rows, do columns — rank columns se cap hota hai

$A = 12 - 1 0 24 - 2 5$ Forecast: chaar rows hain lekin sirf $2$ columns, toh $rank \leq min (4, 2) = 2$ . Rows 1, 2, 3 sab $(1, 2)$ ke multiples hain; row 4 alag hai. Predict karo $2$ .

Step 1 — pehle pivot ke neeche eliminate karo. $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} + R_{1}$ . Kyun? Rows 2 aur 3 row 1 ke proportional hain; clearing expose karta hai ki woh vanish hote hain ya nahi. $\to 10002005$ Step 2 — live row ko upar laao. $R_{2} \leftrightarrow R_{4}$ . Kyun? Non-zero row $(0, 5)$ ko wahan rakho jahan doosra pivot aa sake; row swaps row space preserve karte hain. $\to 10002500$ Step 3 — pivots count karo: $(1, 1)$ aur $(2, 2)$ ⇒ do. $rank = 2$ .

Verify: column rank bhi $2$ hona chahiye. Columns hain $(1, 2, - 1, 0)^{T}$ aur $(2, 4, - 2, 5)^{T}$ ; doosra $2 \times$ pehla nahi hai (last entry $5 \neq = 0$ ), toh do columns independent hain ⇒ column rank $= 2 =$ row rank. Theorem phir se hold karta hai. ✔️

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

Cell F — 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 · Ek price table jo utni informative nahi jitni lagti hai

Ek shop teen receipts record karta hai. Har row hai $(\text{apples},\ \text{bananas},\ \text{total cost in }\$ )$: $A = 2411235108$ Question: yahan kitne genuinely independent pricing information ke pieces hain?

Forecast: receipt 2 exactly double hai receipt 1 ke (wahi shopping, do baar) — yeh kuch naya nahi sikhata. Toh $3$ receipts mein se, sirf $2$ independent facts expect karo. Predict karo $rank = 2$ .

Step 1 — translate karo. "Independent pieces of information" $=$ receipt table ka rank. Kyun? Har dependent row redundant data hai; rank non-redundant wale count karta hai. Step 2 — reduce karo. $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - \frac{1}{2} R_{1}$ . $\to 200 10 \frac{5}{2} 50 \frac{11}{2}$ Step 3 — zero row neeche swap karo. $R_{2} \leftrightarrow R_{3}$ . $\to 200 1 \frac{5}{2} 0 5 \frac{11}{2} 0$ Step 4 — pivots count karo: $(1, 1)$ aur $(2, 2)$ ⇒ do. $rank = 2$ .

Verify (units & meaning): column 3 ki entries dollars hain, columns 1–2 counts hain — numbers consistent rehte hain, aur answer " $2$ independent receipts" us observation se match karta hai ki receipt 2 ne receipt 1 ko duplicate kiya. Shop ke paas effectively $2$ pricing equations hain, do unknown prices solve karne ke liye kaafi. ✔️

Cell I — exam twist: rank of a product

Worked example Ex 9 · Multiply karne se pehle

$rank (A B)$ predict karo $A = (1224), B = (111111)$ Forecast: $A$ ka rank $1$ hai (row 2 $= 2 \times$ row 1) aur $B$ ka rank $1$ hai (row 2 $=$ row 1). Bound kehta hai $rank (A B) \leq min (rank A, rank B) = min (1, 1) = 1$ . Predict karo $rank (A B) \leq 1$ .

Step 1 — multiply karo. Kyun? Dekhne ke liye ki bound equality se meet hota hai ya product aur collapse hota hai (woh bound se neeche jaa sakta hai). $A B = (1 \cdot 1 + 2 \cdot 1 2 \cdot 1 + 4 \cdot 1 \dots \dots \dots \dots) = (363636)$ Step 2 — result ka rank. Har column $(3, 6)^{T}$ hai; row 2 $= 2 \times$ row 1. Reduce karne ke baad ek pivot. $rank (A B) = 1$ .

Verify: $1 \leq min (1, 1) = 1$ ✔️ — bound hold karta hai aur yahan tight bhi hai.

Recall Product bound se neeche kyun jaa sakta hai

Wahi $A$ lo lekin $B^{'} = (1 - \frac{1}{2} 00)$ (rank $1$ ). Toh $A B^{'}$ ke columns $A$ ke kernel mein jaa sakte hain aur $rank (A B^{'}) = 0$ de sakte hain. Bound ek upper limit hai, equality ki guarantee nahi.

Cell J — parameter case: rank ek symbol par depend karta hai

Worked example Ex 10 · Rank kab drop karta hai?

$A = (13 2 k)$ ek real number $k$ ke liye. Har case mein $rank (A)$ nikalo.

Forecast: do rows hain. Woh dependent tab bante hain jab row 2 row 1 ka multiple ho. Row 1 hai $(1, 2)$ ; $3$ se scale karke $(3, 6)$ milta hai. Toh danger value hai $k = 6$ . Predict karo: generally rank $2$ , $k = 6$ par rank $1$ .

Step 1 — eliminate karo. $R_{2} \to R_{2} - 3 R_{1}$ . Kyun? Uss single entry ko isolate karo jiska vanish hona dependence signal karta hai. $\to (10 2 k - 6)$ Step 2 — pivot entry $k - 6$ par split karo.

Agar $k - 6 \neq = 0$ (yaani $k \neq = 6$ ): doosra pivot exist karta hai ⇒ $rank = 2$ .
Agar $k - 6 = 0$ (yaani $k = 6$ ): row 2 vanish ho jaati hai ⇒ sirf ek pivot ⇒ $rank = 1$ . Dono kyun cover karein? Parent contract: har case, including degenerate/limiting value $k = 6$ , dikhana zaroori hai.

Verify: $det A = 1 \cdot k - 2 \cdot 3 = k - 6$ . Full rank $⟺ det \neq = 0 ⟺ k \neq = 6$ . $k = 6$ par, $det = 0$ aur rank $1$ par drop karta hai. Pivot analysis se consistent. ✔️

Recall Scenario matrix par quick self-check

Kaunse cell mein $rank = 0$ hai, aur yeh allowed kyun hai? ::: Cell F (all-zero matrix); bound $0 \leq rank \leq min (m, n)$ lower limit $0$ allow karta hai. Ek wide $2 \times 5$ matrix — uski maximum possible rank kya hai? ::: $min (2, 5) = 2$ . Agar $rank A = 1$ aur $rank B = 1$ , toh kya $rank (A B) = 0$ ho sakta hai? ::: Haan — bound sirf ek upper limit hai; $A$ ke kernel ke saath overlap use aur neeche push kar sakta hai.

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