Span — definition: Exercises — Hinglish deep dive

4.5.16 · D4 · Maths › Linear Algebra (Full) › Span — definition

Yahan sab kuch sirf parent se liye gaye toolkit se hai: ek linear combination hai $c_{1} v_{1} + \dots + c_{k} v_{k}$ jisme real scalars $c_{i}$ hain, aur span un saari combinations ka set hai. Membership test karne ke liye $\sum c_{i} v_{i} = w$ solve karte hain.

L1 — Recognition

Ye check karte hain ki tum definition ko bina zyada compute kiye padh sako.

Problem 1.1. Kya zero vector $0 = (0, 0)$ $span {(2, 1)}$ mein hai? Woh scalar do jo ise prove kare.

Recall Solution 1.1

KYA karte hain: ek scalar $c$ dhoondho jisme $c (2, 1) = (0, 0)$ ho. Reasoning: $2 c = 0$ aur $1 c = 0$ dono $c = 0$ force karte hain. $c = 0$ choose karna kaam karta hai. Answer: Haan, $c = 0$ ke saath. Ye parent se general fact hai: har scalar ko zero lena hamesha origin par land karta hai, isliye $0$ har span mein hota hai.

Problem 1.2. True ya false: $span {(1, 0), (0, 1)}$ exactly do-element set ${(1, 0), (0, 1)}$ hai.

Recall Solution 1.2

False. Span saari linear combinations $a (1, 0) + b (0, 1) = (a, b)$ ka set hai jisme $a, b$ poore $R$ par range karte hain. Wo poora plane $R^{2}$ hai — infinitely many points, sirf do generators nahi.

Problem 1.3. Bina compute kiye, $span {(3, 7), (- 1, 2), (5, 5)}$ origin se kyun zaroor guzarna chahiye?

Recall Solution 1.3

Har scalar ko $0$ set karo: $0 (3, 7) + 0 (- 1, 2) + 0 (5, 5) = (0, 0)$ . All-zero combination hamesha valid hoti hai, isliye $0$ hamesha reachable hai. Har span mein origin hota hai — chahe kitne bhi generators hon ya kaun se bhi hon.

L2 — Application

Ab membership recipe chalao: $\sum c_{i} v_{i} = w$ set up karo aur solve karo.

Problem 2.1. Kya $w = (6, 3)$ $span {(2, 1)}$ mein hai? Kya $u = (2, 3)$ hai?

Recall Solution 2.1

KYA/KYUN: membership ka matlab hai "kya ek scalar $c$ hai jisme $c (2, 1) = w$ ho?" $w = (6, 3)$ ke liye: $2 c = 6 \Rightarrow c = 3$ ; doosra coord check karo $1 \cdot 3 = 3$ ✔. Haan, $c = 3$ . $u = (2, 3)$ ke liye: $2 c = 2 \Rightarrow c = 1$ ; check karo $1 \cdot 1 = 1 \neq = 3$ ✗. Nahi — koi single scalar dono coordinates satisfy nahi karta.

Problem 2.2. Kya $w = (3, - 7, 0)$ $span {(1, 0, 0), (0, 1, 0)}$ mein hai? Kya $u = (0, 0, 5)$ hai?

Recall Solution 2.2

Ek combination hai $a (1, 0, 0) + b (0, 1, 0) = (a, b, 0)$ : teesra coordinate forced hai $0$ par. $w = (3, - 7, 0)$ : $a = 3, b = - 7$ lo; teesra coord $0$ hai ✔. Haan. $u = (0, 0, 5)$ : teesra coord $= 5$ chahiye, lekin woh sirf $0$ ho sakta hai. Nahi. Geometrically ye span 3D space ke andar baithe $x y$ -plane hai (neeche figure dekho).

Problem 2.3. Kya $w = (4, 5)$ $span {(1, 2), (3, 1)}$ mein hai? Scalars dhoondho.

Recall Solution 2.3

Set up: $c_{1} (1, 2) + c_{2} (3, 1) = (4, 5)$ se system milta hai $c_{1} + 3 c_{2} = 4, 2 c_{1} + c_{2} = 5.$ Solve: pehli se, $c_{1} = 4 - 3 c_{2}$ . Substitute karo: $2 (4 - 3 c_{2}) + c_{2} = 5 \Rightarrow 8 - 6 c_{2} + c_{2} = 5 \Rightarrow - 5 c_{2} = - 3 \Rightarrow c_{2} = \frac{3}{5}$ . Phir $c_{1} = 4 - 3 \cdot \frac{3}{5} = 4 - \frac{9}{5} = \frac{11}{5}$ . Answer: Haan, $c_{1} = \frac{11}{5}, c_{2} = \frac{3}{5}$ ke saath. (Check: $\frac{11}{5} + 3 \cdot \frac{3}{5} = \frac{11 + 9}{5} = 4$ ✔, $2 \cdot \frac{11}{5} + \frac{3}{5} = \frac{22 + 3}{5} = 5$ ✔.)

L3 — Analysis

Structure ke baare mein reason karo: kya vectors add karne se kuch change hota hai?

Problem 3.1. $v_{1} = (1, 1)$ , $v_{2} = (2, 2)$ . $span {v_{1}, v_{2}}$ kaisa dikhta hai, aur kyun woh $R^{2}$ nahi hai?

Recall Solution 3.1

Observe karo: $v_{2} = 2 v_{1}$ , isliye koi bhi combination $c_{1} v_{1} + c_{2} v_{2} = c_{1} v_{1} + 2 c_{2} v_{1} = (c_{1} + 2 c_{2}) v_{1}$ hai. $t = c_{1} + 2 c_{2}$ rename karo (ek single free real number), span hai ${t (1, 1) : t \in R}$ — sirf direction $(1, 1)$ mein origin se guzarne wali line. $R^{2}$ kyun nahi: dono vectors ek hi direction mein point karte hain (dependent), isliye doosra koi naya direction nahi deta. Plane fill karne ke liye tumhe do genuinely alag directions chahiye. (Figure dekho: dono arrows ek hi line par hain.)

Problem 3.2. Kya $span {(1, 2), (3, 1)} = R^{2}$ hai? Determinant use karke justify karo.

Recall Solution 3.2

Vectors ko columns ki tarah stack karo: $M = [1231]$ . Phir $det M = 1 \cdot 1 - 3 \cdot 2 = 1 - 6 = - 5$ . Kyunki $det M \neq = 0$ , $M$ invertible hai, isliye $M c = (x, y)$ ka har target $(x, y)$ ke liye ek (unique) solution $c$ hai. Har vector reachable hai $\Rightarrow$ haan, woh $R^{2}$ span karte hain.

Problem 3.3. Tumhare paas $span {(1, 0)}$ ( $x$ -axis) hai. Inme se kaun sa, add karne par, span ko enlarge karega: $(3, 0)$ , $(0, 1)$ , $(- 7, 0)$ ?

Recall Solution 3.3

Current span hai $x$ -axis ${(t, 0)}$ . Ek naya vector span ko enlarge karta hai sirf tab jab woh already usme nahi hota.

$(3, 0)$ : already $x$ -axis par hai $\Rightarrow$ koi change nahi.
$(- 7, 0)$ : already $x$ -axis par hai $\Rightarrow$ koi change nahi.
$(0, 1)$ : $x$ -axis par nahi hai (doosra coord nonzero hai) $\Rightarrow$ woh $y$ -direction add karta hai, aur span poore $R^{2}$ tak jump kar jaata hai. Sirf $(0, 1)$ span ko enlarge karta hai.

L4 — Synthesis

Membership, dependence, aur determinant reasoning combine karo.

Problem 4.1. $k$ ki kaun si value(s) par $v_{1} = (1, 2)$ aur $v_{2} = (2, k)$ $R^{2}$ span karne mein fail karte hain?

Recall Solution 4.1

Woh exactly tab fail karte hain jab columns dependent hon, yaani $det [12 2 k] = 0$ . $det = 1 \cdot k - 2 \cdot 2 = k - 4$ . $k - 4 = 0$ set karne se $k = 4$ milta hai. $k = 4$ par: $v_{2} = (2, 4) = 2 (1, 2) = 2 v_{1}$ , isliye woh sirf ek line span karte hain — span $R^{2}$ nahi ban paata. Har doosre $k$ ke liye, $det \neq = 0$ aur woh $R^{2}$ span karte hain.

Problem 4.2. Kya $w = (1, 1, 1)$ $span {(1, 0, 1), (0, 1, 1)}$ mein hai?

Recall Solution 4.2

Set up: $a (1, 0, 1) + b (0, 1, 1) = (a, b, a + b)$ . $(1, 1, 1)$ se match karo: $a = 1, b = 1, a + b = 1.$ Pehli do se $a = 1, b = 1$ milta hai, lekin phir $a + b = 2 \neq = 1$ hai. System inconsistent hai. Answer: Nahi — $(1, 1, 1)$ span mein nahi hai. (Span hai plane ${(a, b, a + b)}$ ; point $(1, 1, 1)$ usse door hai.)

Problem 4.3. Dikhao ki $span {(1, 0, 0), (0, 1, 0), (1, 1, 0)}$ equals $span {(1, 0, 0), (0, 1, 0)}$ ( $x y$ -plane) hai, isliye teesra vector redundant hai.

Recall Solution 4.3

Teesra vector already chhoti span ke andar hai: $(1, 1, 0) = 1 \cdot (1, 0, 0) + 1 \cdot (0, 1, 0)$ . Isliye woh koi naya direction nahi deta, aur "enlarge only if outside" rule se span unchanged rehti hai. Formally, dono spans $x y$ -plane ${(x, y, 0)}$ hain:

Teeno ka koi bhi combo teesra coord $0$ rakhta hai (teeno vectors ka teesra coord $0$ hai), isliye woh $x y$ -plane mein land karta hai.
Ulta, koi bhi $(x, y, 0) = x (1, 0, 0) + y (0, 1, 0)$ sirf pehle do use karke reachable hai. Isliye dono sets ka exactly same span hai — teesra generator redundant hai. Ye ek minimal spanning set, yaani Basis, ka seed hai.

L5 — Mastery

Prove aur generalise karo.

Problem 5.1. Prove karo ki agar $w \in span {v_{1}, \dots, v_{k}}$ , toh $span {v_{1}, \dots, v_{k}, w} = span {v_{1}, \dots, v_{k}}$ . (Jo vector already span mein hai use add karne se kuch change nahi hota.)

Recall Solution 5.1

Maano $S = span {v_{1}, \dots, v_{k}}$ aur $S^{'} = span {v_{1}, \dots, v_{k}, w}$ . Hum $S = S^{'}$ dikhate hain ye prove karke ki dono ek doosre ko contain karte hain.

$S \subseteq S^{'}$ : koi bhi $x = \sum c_{i} v_{i}$ ko $\sum c_{i} v_{i} + 0 \cdot w$ likha ja sakta hai, badi list ka ek combination. Isliye $x \in S^{'}$ . ✔

$S^{'} \subseteq S$ : koi bhi $y = \sum c_{i} v_{i} + d w \in S^{'}$ lo. Kyunki $w \in S$ , $w = \sum a_{i} v_{i}$ likho. Substitute karo: $y = \sum c_{i} v_{i} + d \sum a_{i} v_{i} = \sum (c_{i} + d a_{i}) v_{i},$ sirf $v_{i}$ ki linear combination $\Rightarrow y \in S$ . ✔

Dono inclusions hold karte hain, isliye $S = S^{'}$ . Redundant vector kabhi span enlarge nahi karta — yahi reason hai ki ek Basis ko independent vectors tak trim kiya ja sakta hai.

Problem 5.2. Prove karo ki $R^{2}$ mein koi bhi single nonzero vector $v$ ek line (1-dimensional subspace) span karta hai, kabhi poora $R^{2}$ nahi.

Recall Solution 5.2

Span ek line hai: $span {v} = {c v : c \in R}$ . Jab $c$ , $R$ par sweep karta hai, point $c v$ direction $v$ mein origin se guzarne wali straight line trace karta hai — ek 1-dimensional flat. Ye poora $R^{2}$ nahi hai: koi bhi vector $u$ choose karo jo $v$ ke parallel nahi hai (2D mein ek aisa milta hai — jaise $v$ ko $9 0^{\circ}$ rotate karo). Agar $u = c v$ kisi $c$ ke liye ho, toh $u$ , $v$ ke parallel hai, contradiction. Isliye $u \in / span {v}$ , matlab span $R^{2}$ ke kam se kam ek point ko miss karta hai. Isliye ye ek proper subset hai — Dimension $1$ ki ek line. $■$

Problem 5.3. Maano $A$ woh $2 \times 2$ matrix hai jiske columns $v_{1}, v_{2} \in R^{2}$ hain. Prove karo: $span {v_{1}, v_{2}} = R^{2} ⟺ det A \neq = 0$ .

Recall Solution 5.3

Span $A$ ka Column space hai, aur " $w$ reachable hai" ka matlab hai $A c = w$ solvable hai. ( $\Leftarrow$ ) Agar $det A \neq = 0$ , toh $A$ invertible hai, isliye $c = A^{- 1} w$ , $A c = w$ ko har $w \in R^{2}$ ke liye solve karta hai. Har vector reachable hai $\Rightarrow$ span $= R^{2}$ . ( $\Rightarrow$ ) Contrapositive: agar $det A = 0$ , columns dependent hain, isliye ek doosre ka multiple hai; unka span ek line (ya point) hai, jo $R^{2}$ ka proper subset hai $\Rightarrow$ span $\neq = R^{2}$ . Dono directions combine karke equivalence milti hai. Ye exactly woh determinant test hai jo poore parent note mein use hua hai. $■$

Figures

$R^{3}$ ke andar ek span ke roop mein $x y$ -plane (Problem 2.2).

Dependent vectors ek line mein collapse ho jaate hain; independent ones plane fill kar dete hain (Problems 3.1–3.2).

Active recall

Membership ke liye ek candidate scalar-tuple ko kya satisfy karna chahiye?

\sum c_{i} v_{i} = w

ki har coordinate equation simultaneously, sirf ek nahi.

Span mein already maujood ek vector add karne se span ka kya hota hai?

Kuch nahi — span unchanged rehta hai (use already woh vector contain karna chahiye).

R^{2}

mein do vectors poora plane exactly tab span karte hain jab?

Unka determinant (matrix columns ke roop mein) nonzero ho.

S = S^{'}

prove karna do-part kaam kyun hai?

Tumhe

S \subseteq S^{'}

aur

S^{'} \subseteq S

dono dikhana hota hai; akela ek inclusion equality nahi hai.

Connections

Linear combination — har membership test ek linear-combination equation hai.
Linear independence — decide karta hai ki extra vectors span enlarge karte hain ya nahi (L3–L4).
Determinant — square cases mein full-span test (L3.2, L4.1, L5.3).
Column space — ek matrix ke columns ka span (L5.3).
Basis — redundant vectors se trim kiya hua minimal spanning set (L4.3, L5.1).
Dimension — span ka "size" (L5.2).
Subspace — har span ek hoti hai.