Worked examples — Span — definition

4.5.16 · D3 · Maths › Linear Algebra (Full) › Span — definition

Shuru karne se pehle, teen words jo hum baar baar use karenge, har ek apne dimag mein ek picture se juda hua:

Span un saare landing spots ka collection hai jo tum is tarah produce kar sakte ho.

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Scenario matrix

Har span ka sawaal in cells mein se kisi ek mein aata hai. Hum har ek ko ek worked example ke saath cover karenge.

Cell	Situation	Kya galat ho sakta hai / kya dhyaan rakhna hai	Example
A	1 vector, nonzero	Hamesha origin se guzarne wali ek line	Ex 1
B	1 vector, zero vector (degenerate)	Span ek single point tak collapse ho jaata hai	Ex 2
C	2 vectors, independent, ${\mathbb{R}}^2$ mein	Poora plane fill karte hain — Determinant se test karo	Ex 3
D	2 vectors, dependent (ek doosre ka multiple hai)	Ek line tak collapse ho jaata hai, plane nahi	Ex 4
E	${\mathbb{R}}^3$ mein 2 vectors	Origin se guzarne wala ek tilted plane, poora space nahi	Ex 5
F	${\mathbb{R}}^3$ mein 3 vectors, independent	Poora ${\mathbb{R}}^3$ fill karte hain	Ex 6
G	${\mathbb{R}}^3$ mein 3 vectors, ek redundant	Phir bhi sirf ek plane — real directions gino	Ex 7
H	Word problem (real-world reachability)	"Amounts" ko scalars mein translate karo	Ex 8
I	Exam twist: negative-scalar / "positive only" trap	Scalars saare $\mathbb{R}$ pe range karte hain	Ex 9

Scalars ke signs ($+$, $-$, $0$) examples ke andar cover kiye gaye hain kyunki span hamesha teeno allow karta hai — yehi cheez use origin ke through symmetric banati hai.

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Cell A — ek nonzero vector: ek line

Figure mein kya dikhta hai: ek single cyan seedhi line jo origin se guzarti hai direction $(3,1)$ ke along, upper-right aur lower-left dono taraf jaati hai. White arrow $v=(3,1)$ ko mark karta hai; amber dot $(6,2)$ pe exactly line pe baitha hai, jabki red dot $(3,-1)$ saaf taur pe usse door hai.

1. Span likho. Ek vector ke saath ek scalar $c$ hota hai: $\mathrm{span}\{v\}=\left\{c\left[\begin{array}{c}3\\ 1\end{array}\right]:c\in \mathbb{R}\right\}.$ Ye step kyun? Definition kehti hai span = saare linear combinations; ek generator ke saath ek linear combination bas $cv$ hota hai.
2. Shape dekho. Jaise $c$ har real number se guzarta hai, $cv$ tip ko seedhi direction $(3,1)$ mein slide karta hai. Positive $c$ ek taraf jaata hai, negative $c$ use flip karta hai doosri taraf, $c=0$ origin pe baitha hai. Figure mein cyan line dekho (upar describe ki gayi) — woh origin se dono directions mein guzarti hai. Toh span ek line through the origin hai.
3. $(6,2)$ test karo. Chahiye $c(3,1)=(6,2)$: pehle slot se $3c=6\Rightarrow c=2$; doosra slot check karo $1\cdot 2=2$ ✔. Toh $(6,2)$ span mein hai (figure mein amber tip).
4. $(3,-1)$ test karo. Chahiye $3c=3\Rightarrow c=1$, lekin phir doosra slot deta hai $1\cdot 1=1\ne -1$. Koi consistent $c$ nahi. Toh $(3,-1)$ span mein nahi hai.

Verify: $2\cdot (3,1)=(6,2)$ ✔. $(3,-1)$ ke liye: donon slots simultaneously $c=1$ aur $c=-1$ maangti hain — impossible, toh correctly reject kiya gaya.

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Cell B — zero vector: ek single point

1. Definition apply karo. $\mathrm{span}\{0\}=\{c0:c\in \mathbb{R}\}$. Ye step kyun? Wahi rule hamesha ki tarah — saare scalar multiples.
2. Multiples compute karo. Har scalar $c$ ke liye, $c\cdot (0,0)=(0,0)$. Kuch bhi nahi ko stretch karna kuch nahi deta; kuch bhi nahi ko flip karna kuch nahi deta. Ye step kyun? $c\cdot 0=0$ kisi bhi real $c$ ke liye, toh "saare landing spots ka set" mein exactly ek member hai.
3. Conclusion. $\mathrm{span}\{0\}=\{0\}$ — ek single point, origin.

Verify: span abhi bhi ek valid Subspace hai (isme $0$ hai, $+$ aur scaling ke under closed hai). Iska Dimension $0$ hai. Ye sabse chota possible span hai.

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Cell C — do independent vectors ${\mathbb{R}}^2$ fill karte hain

Figure mein kya dikhta hai: cyan dots ka ek lattice — points $i{v}_1+j{v}_2$ integer $i,j$ ke liye — poore plane mein evenly phela hua (ek tilted grid). Amber arrow $v_1=(1,2)$ hai, white arrow $v_2=(3,1)$ hai, aur ek red dot target $(0,1)$ ko mark karta hai jahan grid pahunchta hai.

1. Reachability question set up karo. "Span ${\mathbb{R}}^2$" ka matlab hai ki har target $(x,y)$ hit hota hai: $c_1\left[\begin{array}{c}1\\ 2\end{array}\right]+c_2\left[\begin{array}{c}3\\ 1\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right].$ Ye step kyun? Span mein membership hamesha scalars $c_1,c_2$ mein linear equations ka system hota hai.
2. Ise matrix ke roop mein padho. Vectors ko columns ki tarah rakho: $M=\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right]$. System hai $M\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$. Ye step kyun? Vectors ko columns ki tarah stack karna exactly wahi hai jo linear combination $c_1{v}_1+c_2{v}_2$ ko single product $M\mathbf{c}$ banata hai — toh "kya $(x,y)$ reachable hai?" standard solvability question "kya $M\mathbf{c}=(x,y)$ ka solution hai?" mein badal jaata hai, jise phir determinant ek number mein answer kar sakta hai.
3. Determinant se decisive question poochho. Determinant kyun, kuch aur kyun nahi? Kyunki ek square matrix ke liye, $\det \ne 0$ exactly woh condition hai ki "$M$ ko undo kiya ja sakta hai" (invertible) — matlab har $(x,y)$ ka ek solution hai. $\det M=1\cdot 1-3\cdot 2=1-6=-5\ne 0.$
4. Conclude karo. Nonzero determinant ⇒ invertible ⇒ har $(x,y)$ reachable ⇒ woh ${\mathbb{R}}^2$ span karte hain (full span). Figure mein, do cyan arrows ek genuine "grid" of tip-to-tail parallelograms banate hain jo poore plane ko tile karte hain.

Verify: ek specific hard target ke liye solve karo, maano $(0,1)$. $M^{-1}=\frac{1}{-5}\left[\begin{array}{cc}1& -3\\ -2& 1\end{array}\right]$, toh $\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\frac{1}{-5}\left[\begin{array}{cc}1& -3\\ -2& 1\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\frac{1}{-5}\left[\begin{array}{c}-3\\ 1\end{array}\right]=\left[\begin{array}{c}3/5\\ -1/5\end{array}\right]$. Check: $\frac{3}{5}(1,2)-\frac{1}{5}(3,1)=(\frac{3}{5}-\frac{3}{5},\frac{6}{5}-\frac{1}{5})=(0,1)$ ✔.

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Cell D — do dependent vectors: ek line tak collapse

Figure mein kya dikhta hai: origin se guzarne wali ek single cyan line. Amber arrow $v_1=(2,-1)$ aur white arrow $v_2=(-4,2)$ dono usi line ke along hain — $v_2$ bas opposite direction mein point karta hai aur do guna lamba hai. Ek red dot $(1,1)$ pe line se door baitha hai, dikhata hai ki woh unreachable hai.

1. Hidden multiple check karo. Kya $v_2=\lambda v_1$ hai? $(-4,2)=\lambda (2,-1)$ ko pehle slot se $\lambda =-2$ chahiye; doosra check karo: $-2\cdot (-1)=2$ ✔. Toh $v_2=-2v_1$ — woh linearly dependent hain (dekho Linear independence).
2. Linear combination simplify karo. Har combination collapse ho jaata hai: $c_1{v}_1+c_2{v}_2=c_1{v}_1+c_2(-2v_1)=(c_1-2c_2)v_1.$ Ye step kyun? $v_2=-2v_1$ substitute karna do free scalars ko ek single effective scalar $(c_1-2c_2)$ mein badal deta hai, jo abhi bhi koi bhi real number ho sakta hai.
3. Conclude karo. Span hai $\{t{v}_1:t\in \mathbb{R}\}$ — bas $(2,-1)$ se guzarne wali line, plane nahi. Figure mein dono arrows ek hi cyan line pe hain (ek bas opposite direction mein point karta hai).
4. Determinant se confirm karo. $\det \left[\begin{array}{cc}2& -4\\ -1& 2\end{array}\right]=2\cdot 2-(-4)(-1)=4-4=0$. Zero determinant ⇒ matrix invertible nahi ⇒ woh full span achieve nahi karte (sirf ek line cover karte hain, poora ${\mathbb{R}}^2$ nahi), hamare finding se match karta hai.

Verify: kya $(1,1)$ span mein hai? Chahiye hoga $t(2,-1)=(1,1)$: $2t=1\Rightarrow t=1/2$, lekin phir doosra slot deta hai $-1\cdot \frac{1}{2}=-\frac{1}{2}\ne 1$. Nahi — correctly, plane fill nahi ki gayi. Determinant $=0$ ✔ agree karta hai.

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Cell E — ${\mathbb{R}}^3$ mein do vectors: ek tilted plane

Figure mein kya dikhta hai: ek flat, tilted cyan sheet jo 3D mein origin se guzarti hai — ye plane $x+y-z=0$ hai. Amber arrow $v_1=(1,0,1)$ aur white arrow $v_2=(0,1,1)$ dono sheet mein lie karte hain, aur amber dot $(2,3,5)$ uske bilkul upar hai, reachability confirm karta hai.

1. Ek general point likho. $a(1,0,1)+b(0,1,1)=(a,b,a+b)$. Ye step kyun? Coordinate-by-coordinate add karo pattern dekhne ke liye.
2. Constraint spot karo. Teesra coordinate forced hai: $z=a+b=x+y$. Toh span mein har point $x+y-z=0$ obey karta hai. Do free numbers $a,b$ ek 2-dimensional flat sweep karte hain — origin se guzarne wala ek plane, tilted (coordinate planes mein se ek nahi). Upar describe ki gayi cyan sheet dekho. Poora ${\mathbb{R}}^3$ kyun nahi ho sakta? Do arrows se zyada se zyada do independent directions milte hain; 3D fill karne ke liye teen chahiye. Dekho Dimension.
3. Membership ek plane test ban jaata hai. Ek point span mein hai iff $x+y-z=0$.

Verify: kya $(2,3,5)$ usme hai? Check karo $2+3-5=0$ ✔ — haan, $a=2,b=3$ ke saath $(2,3,5)$ milta hai. Kya $(1,1,1)$ usme hai? $1+1-1=1\ne 0$ — nahi. Dono coordinate formula se agree karte hain.

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Cell F — teen independent vectors poora ${\mathbb{R}}^3$ fill karte hain

1. Columns ki tarah stack karo aur determinant se test karo. Determinant kyun? Ek square $3\times 3$ matrix ke liye, $\det \ne 0$ ⇔ invertible ⇔ har target reachable, exactly Cell C ki tarah lekin ek dimension upar. $M=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right].$
2. Determinant compute karo. $M$ upper-triangular hai, toh $\det M=$ diagonal ka product $=1\cdot 1\cdot 1=1\ne 0$. Ye shortcut kyun? Triangular matrix ke liye determinant bas diagonal product hota hai — koi row reduction ki zaroorat nahi.
3. Conclude karo. $\det \ne 0$ ⇒ teen columns independent hain ⇒ woh poora ${\mathbb{R}}^3$ span karte hain (full span). Woh ek Basis bhi banate hain (spanning aur independent).

Verify: target $(0,0,1)$ hit karo. Solve karo $c_1(1,0,0)+c_2(1,1,0)+c_3(1,1,1)=(0,0,1)$. Teesra slot: $c_3=1$. Doosra: $c_2+c_3=0\Rightarrow c_2=-1$. Pehla: $c_1+c_2+c_3=0\Rightarrow c_1=0$. Check: $0(1,0,0)-1(1,1,0)+1(1,1,1)=(0,0,1)$ ✔.

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Cell G — teen vectors, ek redundant: phir bhi sirf ek plane

1. Redundancy dhundho. Notice karo $(1,1,0)=(1,0,0)+(0,1,0)$. Teesra vector pehle do ka combination hai — woh koi nayi direction nahi add karta. Ye step kyun? Span tabhi barhta hai jab ek naya vector kahi unreachable jagah point kare; yahan woh us plane mein point karta hai jo hamare paas pehle se thi.
2. Essentials tak reduce karo. Toh teeno ka $\mathrm{span}$ $=\mathrm{span}\{(1,0,0),(0,1,0)\}$, jo hai set $\{(a,b,0)\}$ — $xy$-plane, ek 2D flat.
3. Determinant se confirm karo. $\det \left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]=0$ (teesre coordinate ki puri row zero hai). Zero ⇒ full span nahi (sirf ek plane, poora ${\mathbb{R}}^3$ nahi), match karta hai. Genuinely independent columns ki sankhya $2$ hai, toh Dimension $=2$.

Verify: kya $(0,0,7)$ span mein hai? Koi bhi combination teesra coordinate $0\cdot (\cdots )=0$ rakhta hai, toh teesra coordinate hamesha $0\ne 7$ rahega. Nahi. Teen vectors ne ${\mathbb{R}}^3$ guarantee nahi kiya.

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Cell H — ek real-world word problem

1. Span mein translate karo. "Packet $P$ ka $c_1$ aur packet $Q$ ka $c_2$" exactly $c_1(1,2)+c_2(3,1)$ hai. Reachable blends hain $\mathrm{span}\{(1,2),(3,1)\}$. Ye step kyun? "Har ingredient ki kuch amount use karo aur combine karo" scaling-and-adding hai — linear combination ki definition.
2. Full span test karo. Ye wahi vectors hain jaise Ex 3 mein, $\det =-5\ne 0$, toh har $(s,c)$ reachable hai — packets ${\mathbb{R}}^2$ span karte hain.
3. Concrete order. Chahiye $\left[\begin{array}{c}s\\ c\end{array}\right]=\left[\begin{array}{c}7\\ 4\end{array}\right]$. Solve karo $c_1(1,2)+c_2(3,1)=(7,4)$: Ex 3 ke inverse se, $\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\frac{1}{-5}\left[\begin{array}{cc}1& -3\\ -2& 1\end{array}\right]\left[\begin{array}{c}7\\ 4\end{array}\right]=\frac{1}{-5}\left[\begin{array}{c}7-12\\ -14+4\end{array}\right]=\frac{1}{-5}\left[\begin{array}{c}-5\\ -10\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right]$.

Verify: $1\cdot (1,2)+2\cdot (3,1)=(1+6,2+2)=(7,4)$ ✔. Units: $1$ packet $P$ + $2$ packets $Q$ → $7$ kg sand, $4$ kg cement. (Physically tum $c_i\ge 0$ restrict karoge; mathematically span saare reals allow karta hai — agla trap dekho.)

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Cell I — exam twist (negative-scalar trap)

1. Exact definition yaad karo. $\mathrm{span}$ mein, scalars saare $\mathbb{R}$ pe range karte hain — positive, negative, aur zero. Ye optional nahi hai; ye definition mein built-in hai (dekho Linear combination). Ye kyun matter karta hai: $c<0$ allow karna exactly wahi hai jo span ko origin ke through symmetric banata hai.
2. Ek negative combination dikhao. $c=-4$ lo: $-4(1,0)=(-4,0)$, jiska $x=-4<0$ hai. Woh point span mein hai, lekin proposed ray mein nahi hai. Ye step kyun? Ek concrete counterexample "for all" claim todne ke liye kaafi hai: "span sirf ray $x\ge 0$ hai" disprove karne ke liye, hum bas ek aisa legitimate span element dikhate hain jo us ray se bahar ho. $(-4,0)$ exactly yahi karta hai.
3. Restricted version ke liye correct object naam do. Agar tum actually negatives forbid karo ($c\ge 0$), tum ek ray paoge (generally ek cone), span nahi. Ye statement ek cone ko span se confuse karti hai.
4. Verdict deliver karo. Statement FALSE hai. Sach wala span hai $\mathrm{span}\{(1,0)\}=\{(x,0):x\in \mathbb{R}\}$ — poora horizontal axis, origin ke through dono directions mein extend karta hua, ek-taraf wali ray nahi.

Verify: $\mathrm{span}\{(1,0)\}=\{(x,0):x\in \mathbb{R}\}$ — poora horizontal axis. Point $(-4,0)$ $c=-4$ se reachable hai ✔, phir bhi $-4<0$ use claimed ray $\{x\ge 0\}$ ke bahar rakhta hai — toh claim correctly reject kiya gaya.

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Active recall

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Connections

• Linear combination — har worked example ek membership question hai jo linear combination ki tarah phrase kiya gaya hai.
• Linear independence — woh hidden pivot jo line vs plane decide karta hai (Cells D, G).
• Determinant — square cases ke liye yes/no full-span test (Cells C, F).
• Subspace — $\{0\}$ (Cell B) bhi ek valid subspace hai.
• Basis — Cell F ke teen independent vectors ek banate hain.
• Dimension — saari cells mein real directions count karta hai.
• Column space — ek matrix ke columns ka span, exactly wahi jo determinant tests probe karte hain.