Subspaces — four fundamental subspaces of a matrix: Worked examples

4.5.44 · D3 · Maths › Linear Algebra (Full) › Subspaces — four fundamental subspaces of a matrix

Scenario matrix

Har matrix in case classes mein se kisi ek mein aata hai. Columns woh questions hain jo answer badal dete hain; last column us example ka naam hai jo us case ko cover karta hai.

#	Case class	Shape $m \times n$	Rank $r$	Kya special hai	Covered by
1	Wide, rank-deficient	$2 \times 3$	$r < m, r < n$	free variables aur unreachable outputs	Ex 1
2	Tall, full column rank	$3 \times 2$	$r = n < m$	$N (A) = {0}$ , bada left null space	Ex 2
3	Square invertible	$n \times n$	$r = n = m$	saare subspaces ya to trivial ya sab kuch	Ex 3
4	Zero matrix	$2 \times 2$	$r = 0$	degenerate limit	Ex 4
5	Repeated / dependent rows	$3 \times 3$	$r < m, r < n$	dono null spaces nontrivial	Ex 5
6	Word problem (networks)	$3 \times 3$	$r = 2$	left null space = ek conservation law	Ex 6
7	Exam twist: $A x = b$ ki solvability	$3 \times 3$	$r = 2$	$b \in C (A)$ + $N (A^{T})$ use karta hai	Ex 7
8	Single column, single row	$3 \times 1$ , $1 \times 3$	$r = 1$	rank-one limiting shapes	Ex 8

Do "control knobs" hamesha yahi hote hain: kitne pivots ( $r$ ) aur kaun sa dimension bada hai ( $m$ vs $n$ ). Upar ka har cell in dono knobs ki alag setting hai.

Figure — Subspaces — four fundamental subspaces of a matrix

Example 1 — Wide, rank-deficient (cell 1)

Worked example

$A = [122538]$ , size $2 \times 3$ Forecast: pehle $r$ guess karo, phir $dim C (A), dim C (A^{T}), dim N (A), dim N (A^{T})$ — aage padhne se pehle. (Hint: do rows hain, kya woh parallel hain?)

Step 1 — Row reduce karo. $R_{2} \to R_{2} - 2 R_{1}$ se $[102132]$ milta hai, phir $R_{1} \to R_{1} - 2 R_{2}$ se RREF $[1001 - 1 2]$ milta hai. Ye step kyun? RREF pivots expose karta hai, aur pivot count hi rank hai — woh ek number jo charon dimensions fix karta hai. Yahan pivots columns 1 aur 2 mein hain, isliye $r = 2$ .

Step 2 — Column space $C (A) \subseteq R^{2}$ . Pivot columns 1 aur 2 hain, isliye original columns $(1, 2)$ aur $(2, 5)$ lo: $C (A) = span {(1, 2), (2, 5)} = R^{2}$ . Ye step kyun? Pivot columns independent hote hain aur column space span karte hain. $R^{2}$ mein do independent vectors use puri fill kar dete hain: har output reachable hai.

Step 3 — Row space $C (A^{T}) \subseteq R^{3}$ . Do nonzero RREF rows: $span {(1, 0, - 1), (0, 1, 2)}$ , dimension $r = 2$ .

Step 4 — Null space $N (A) \subseteq R^{3}$ . RREF se: $x_{1} = x_{3}$ , $x_{2} = - 2 x_{3}$ . Free variable $x_{3}$ . $x_{3} = 1$ set karo: special solution $(1, - 2, 1)$ . To $N (A) = span {(1, - 2, 1)}$ , dimension $n - r = 3 - 2 = 1$ . Ye step kyun? Har free variable ek independent direction deta hai jise machine zero kar deti hai.

Step 5 — Left null space $N (A^{T}) \subseteq R^{2}$ . Dimension $m - r = 2 - 2 = 0$ , isliye $N (A^{T}) = {0}$ .

Verify: dims sahi se sum hote hain: $r + (n - r) = 2 + 1 = 3 = n$ ✓ aur $r + (m - r) = 2 + 0 = 2 = m$ ✓. Orthogonality: $(1, - 2, 1) \cdot (1, 0, - 1) = 1 - 1 = 0$ ✓ aur $(1, - 2, 1) \cdot (0, 1, 2) = - 2 + 2 = 0$ ✓ — null vector har row ke perpendicular hai.

Example 2 — Tall, full column rank (cell 2)

Worked example

$A = 111012$ , size $3 \times 2$ Forecast: columns clearly independent hain (multiples nahi). To $r = 2$ . "Missing" dimension kahan jaati hai — $N (A)$ mein ya $N (A^{T})$ mein?

Step 1 — Rank. Columns $(1, 1, 1)$ aur $(0, 1, 2)$ independent hain, isliye $r = 2 = n$ . Ye step kyun? Full column rank ka matlab koi bhi input waste nahi hota.

Step 2 — Null space. $dim N (A) = n - r = 2 - 2 = 0$ , isliye $N (A) = {0}$ : kuch bhi crush nahi hota.

Step 3 — Column space $C (A) \subseteq R^{3}$ . $span {(1, 1, 1), (0, 1, 2)}$ , $R^{3}$ mein origin se gujarta ek plane, dimension $2$ .

Step 4 — Left null space $N (A^{T}) \subseteq R^{3}$ . Dimension $m - r = 3 - 2 = 1$ . $A^{T} y = 0$ solve karo: $y_{1} + y_{2} + y_{3} = 0$ aur $y_{2} + 2 y_{3} = 0$ . Doosre se $y_{2} = - 2 y_{3}$ ; pehle se $y_{1} = - y_{2} - y_{3} = 2 y_{3} - y_{3} = y_{3}$ . $y_{3} = 1$ set karo: $(1, - 2, 1)$ . To $N (A^{T}) = span {(1, - 2, 1)}$ .

Verify: ye $(1, - 2, 1)$ pure column space (plane) ke perpendicular hona chahiye. $(1, - 2, 1) \cdot (1, 1, 1) = 0$ ✓ aur $(1, - 2, 1) \cdot (0, 1, 2) = - 2 + 2 = 0$ ✓. To $N (A^{T})$ exactly plane $C (A)$ ki normal line hai — ye hai " $N (A^{T}) ⊥ C (A)$ in $R^{m}$ " ka real matalab, aur ye woh vector hai jo least squares use karta hai.

Example 3 — Square invertible (cell 3)

Example 4 — The zero matrix (degenerate limit, cell 4)

Example 5 — Dependent rows, dono null spaces nontrivial (cell 5)

Worked example

$A = 121120121$ , size $3 \times 3$ Forecast: row 2, row 1 ka $2 \times$ hai. To saari rows independent nahi ho sakti — $r$ guess karo, phir charon dims.

Step 1 — Row reduce karo. $R_{2} \to R_{2} - 2 R_{1}$ se row 2 zero ho jaati hai. $R_{3} \to R_{3} - R_{1} = (0, - 1, 0)$ . Reorder/clean karke RREF $100010100$ milta hai. Do pivots $\Rightarrow r = 2$ .

Step 2 — Row space $C (A^{T}) \subseteq R^{3}$ . $span {(1, 0, 1), (0, 1, 0)}$ , dimension $2$ .

Step 3 — Null space $N (A) \subseteq R^{3}$ . RREF se: $x_{1} = - x_{3}$ , $x_{2} = 0$ , $x_{3}$ free. $x_{3} = 1$ set karo: $(- 1, 0, 1)$ . To $N (A) = span {(- 1, 0, 1)}$ , dimension $n - r = 1$ .

Step 4 — Column space $C (A) \subseteq R^{3}$ . Original $A$ ke pivot columns 1,2: $(1, 2, 1)$ aur $(1, 2, 0)$ . Dimension $2$ .

Step 5 — Left null space $N (A^{T}) \subseteq R^{3}$ . Dimension $m - r = 1$ . $y_{1} (1, 1, 1) + y_{2} (2, 2, 2) + y_{3} (1, 0, 1) = 0$ column-wise solve karo (yaani $A^{T} y = 0$ ): dependency row-2 $= 2 \times$ row-1 se $y_{1} + 2 y_{2} = 0$ aur $y_{3} = 0$ milta hai. $y_{2} = 1$ set karo: $y = (- 2, 1, 0)$ . To $N (A^{T}) = span {(- 2, 1, 0)}$ .

Verify: $N (A) ⊥$ rows: $(- 1, 0, 1) \cdot (1, 0, 1) = 0$ ✓, $(- 1, 0, 1) \cdot (0, 1, 0) = 0$ ✓. $N (A^{T}) ⊥$ columns: $(- 2, 1, 0) \cdot (1, 2, 1) = - 2 + 2 = 0$ ✓, $(- 2, 1, 0) \cdot (1, 2, 0) = - 2 + 2 = 0$ ✓. Sums: $2 + 1 = 3 = n$ ✓, $2 + 1 = 3 = m$ ✓.

Example 6 — Word problem: ek network conservation law (cell 6)

Worked example Ek triangle mein currents. Teen nodes, teen wires. Edges

$1 \to 2, 2 \to 3, 1 \to 3$ ke liye incidence matrix hai $A = - 1 0 - 1 1 - 1 0 011 .$ Har row kehti hai us edge ke liye " $(head par value) - (tail par value)$ ". Hum dhundh rahe hain: kaun se node-potential vectors har edge par zero voltage dete hain, aur left null space physically kya batata hai. Forecast: saare teen nodes ek same potential par hone se har jagah zero hona chahiye — isliye $N (A)$ kam se kam ek line hai. $r$ guess karo.

Step 1 — Rank. Jodo: row3 $=$ row1 $+$ row2 (check karo: $(- 1, 1, 0) + (0, - 1, 1) = (- 1, 0, 1)$ ✓). To rows dependent hain, $r = 2$ . Ye step kyun? Ek dependent row ka matlab ek edge ki voltage baaki dono se force hoti hai — yahi loop law hai.

Step 2 — Null space $N (A) \subseteq R^{3}$ (potentials). $A x = 0$ ka matlab har edge ke dono endpoints ka potential equal hai, yaani $x_{1} = x_{2} = x_{3}$ . To $N (A) = span {(1, 1, 1)}$ , dimension $n - r = 1$ . Physical meaning: saare potentials ko ek constant se shift karne se koi voltage nahi badlta.

Step 3 — Left null space $N (A^{T}) \subseteq R^{3}$ (edge currents). Dimension $m - r = 1$ . $A^{T} y = 0$ solve karo: iska matlab currents $y$ har node par Kirchhoff's current law satisfy karte hain. Dependency row3=row1+row2 se $y = (1, 1, - 1)$ milta hai (loop ke around current).

Verify: $x = (1, 1, 1)$ ke liye $A x = 0$ : har row apna $\pm 1$ pattern zero sum karta hai ✓. $y = (1, 1, - 1)$ ke liye $A^{T} y = 0$ : node 1 column $(- 1, 0, - 1) \cdot (1, 1, - 1) = - 1 + 1 = 0$ ✓; node 2 $(1, - 1, 0) \cdot (1, 1, - 1) = 1 - 1 = 0$ ✓; node 3 $(0, 1, 1) \cdot (1, 1, - 1) = 1 - 1 = 0$ ✓. To left null space hi conservation law hai.

Example 7 — Exam twist: kya $A x = b$ solvable hai? (cell 7)

Worked example

$A = 101202011$ ke saath, decide karo ki $b = (3, 1, 4)$ aur $b^{'} = (3, 1, 5)$ reachable hain ya nahi. Forecast: $A$ ki row 3 $=$ row 1 $+$ row 2. Ye kisi bhi reachable $b$ par kaun si constraint force karta hai?

Step 1 — Rank aur left null space. row3 $=$ row1 $+$ row2 $\Rightarrow r = 2$ , aur $dim N (A^{T}) = m - r = 1$ . $A^{T} y = 0$ solve karne par dependency vector $y = (1, 1, - 1)$ milta hai. Ye step kyun? $Ax=b$ solvable hai $⟺ b \in C (A) ⟺ b ⊥ N (A^{T})$ . To sirf ek condition hai $y \cdot b = 0$ .

Step 2 — $b = (3, 1, 4)$ test karo. $y \cdot b = (1, 1, - 1) \cdot (3, 1, 4) = 3 + 1 - 4 = 0$ ✓ — solvable hai.

Step 3 — $b^{'} = (3, 1, 5)$ test karo. $y \cdot b^{'} = 3 + 1 - 5 = - 1 \neq = 0$ — solvable nahi, $b^{'} \in / C (A)$ .

Step 4 — Solvable $b$ ke liye general solution. $[A ∣ b]$ ka RREF deta hai $x_{1} + 2 x_{2} = 3, x_{3} = 1$ . Particular: $(3, 0, 1)$ . Homogeneous part $N (A) = span {(- 2, 1, 0)}$ . To saare solutions: $x = (3, 0, 1) + t (- 2, 1, 0)$ .

Verify: $A (3, 0, 1) = (3, 1, 3 + 1) = (3, 1, 4) = b$ ✓. Aur $A (- 2, 1, 0) = (- 2 + 2, 0, - 2 + 2) = (0, 0, 0)$ ✓ isliye koi bhi $t (- 2, 1, 0)$ add karne par solution raha rehta hai. Exam ka trap — "kaun se right-hand sides kaam karte hain?" — poora left null space se answer hota hai.

Example 8 — Single column aur single row (rank-one limits, cell 8)

Worked example Column vector

$A = 2 - 1 4$ ( $3 \times 1$ ) aur row vector $B = [2 - 1 4]$ ( $1 \times 3$ ). Forecast: dono mein ek hi nonzero entry-pattern hai, isliye dono ke liye $r = 1$ . Kaun si spaces points tak collapse hoti hain, kaun si lines/planes bharti hain?

Step 1 — $A$ ke liye ( $m = 3, n = 1, r = 1$ ). $C (A) = span {(2, - 1, 4)} \subseteq R^{3}$ (ek line). $N (A)$ : $dim = n - r = 1 - 1 = 0$ , isliye $N (A) = {0}$ $R^{1}$ mein. $N (A^{T})$ : $dim = m - r = 3 - 1 = 2$ , $R^{3}$ mein ek plane — $(2, - 1, 4)$ ke saare perpendicular vectors.

Step 2 — $B = A^{T}$ ke liye ( $m = 1, n = 3, r = 1$ ). Symmetry se roles swap hote hain: $C (B^{T}) =$ wahi line $span {(2, - 1, 4)} \subseteq R^{3}$ ; $N (B)$ : $dim = n - r = 3 - 1 = 2$ , $(2, - 1, 4)$ ke perpendicular plane; $C (B) = R^{1}$ ; $N (B^{T}) = {0}$ .

Step 3 — Perpendicular plane ki basis. $2 u_{1} - u_{2} + 4 u_{3} = 0$ ko satisfy karne wale do independent vectors chahiye. $(u_{2}, u_{3}) = (2, 0)$ set karo $\Rightarrow u_{1} = 1$ : $(1, 2, 0)$ . $(u_{2}, u_{3}) = (0, - 1)$ set karo $\Rightarrow 2 u_{1} = 4 \Rightarrow (2, 0, - 1)$ ... check: $2 \cdot 2 - 0 + 4 (- 1) = 0$ ✓. Plane $= span {(1, 2, 0), (2, 0, - 1)}$ .

Verify: $(2, - 1, 4) \cdot (1, 2, 0) = 2 - 2 + 0 = 0$ ✓, $(2, - 1, 4) \cdot (2, 0, - 1) = 4 + 0 - 4 = 0$ ✓. Dims: $A$ ke liye, $1 + 0 = 1 = n$ ✓ aur $1 + 2 = 3 = m$ ✓. Transpose karna input/output worlds ko swap karta hai — $R^{3}$ mein ek line ka orthogonal complement ek plane hota hai, bilkul waise jaise dimension count $n - r$ predict karta hai.

Recall Self-test: shape aur rank se charon dims batao

$A$ ka size $5 \times 3$ hai aur rank $2$ hai. Saare charon dimensions do. $dim C (A)$ ::: $r = 2$ $dim C (A^{T})$ ::: $r = 2$ $dim N (A)$ ::: $n - r = 3 - 2 = 1$ $dim N (A^{T})$ ::: $m - r = 5 - 2 = 3$

Connections

Rank of a Matrix — knob $r$ jo scenario matrix ke har cell ko set karta hai.
Rank–Nullity Theorem — har Verify block mein sum-checks.
RREF and Pivots — Examples 1, 5, 7 mein use hua engine.
Solving Ax=b — Example 7 ka solvability test $b \in C (A)$ ke zariye.
Orthogonal Complements — Examples 2 aur 8 ki normal line/plane.
Least Squares & Projections — Example 2 mein mila left null space direction.
Subspaces — four fundamental subspaces of a matrix — parent jise ye drill karta hai.