Basis — definition, uniqueness of representation: Visual walkthrough

4.5.17 · D2 · Maths › Linear Algebra (Full) › Basis — definition, uniqueness of representation

Step 1 — Vector kya hota hai, aur "combine" ka matlab kya hai?

KYA. Ek vector bas ek aisa arrow hai jiska tail origin par hai aur tip plane mein kahin hai. Neeche ki pictures mein plane sunset-coloured grid hai; origin woh dot hai jo centre mein $O$ label ke saath hai.

KYUN. Isse pehle ki hum baat kar sakein ki ek vector ko doosre vectors se "build" karna kya hota hai, hume un do moves par agree karna hoga jo humein allowed hain:

ek arrow ko scale karna (use stretch ya shrink karo, flip karo agar number negative ho),
do arrows ko add karna (doosre ki tail pehle ki tip par lagao — "tip-to-tail" rule).

PICTURE. Neeche, violet arrow $v_{1}$ aur orange arrow $v_{2}$ hamare do building blocks hain. Magenta dashed arrow $2 v_{1}$ dikhata hai (do guna lamba). Navy arrow $v_{1} + v_{2}$ ko tip-to-tail build karte hue dikhata hai.

Step 2 — "Spanning": kya hum har jagah pahunch sakte hain?

KYA. Hum poochte hain: apne building blocks $v_{1}, v_{2}$ se shuru karke, $c_{1}, c_{2}$ ko sab numbers par range karne dete hue, hum kaun si tips $w$ ko hit kar sakte hain?

KYUN. Ek coordinate system bekar hai agar kuch points ka koi address hi nahi hai. Spanning bilkul yahi promise hai ki "har point ka kam se kam ek address hai." Hume pehle yeh check karna hoga kyunki aap kisi aise sawaal ka unique jawaab nahi de sakte jiska koi jawaab hi nahi hai.

PICTURE. Jaise $c_{1}$ aur $c_{2}$ slide karte hain, tip $w = c_{1} v_{1} + c_{2} v_{2}$ shaded region ko sweep karti hai. Jab $v_{1}, v_{2}$ sach mein alag directions mein point karte hain, woh region poore plane ko fill kar deta hai — har target (teen magenta stars) reachable hai.

Step 3 — Khatre ki baat: ek redundant building block

KYA. Ab ek teesra arrow $v_{3}$ add karo jo sach mein naya nahi hai — maano $v_{3} = v_{1} + v_{2}$ . Hum ab bhi plane ko span karte hain, lekin dekhte hain addresses ka kya hota hai.

KYUN. Hum expose karna chahte hain ki sirf spanning kyun kaafi nahi hai. Parent note ka mistake box yeh shabdon mein kehta hai (" ${(1, 0), (0, 1), (1, 1)}$ spans karta hai lekin do addresses hain"); yahaan hum ek point ke liye do addresses dekhte hain.

PICTURE. Woh single navy target point do alag tareekon se reach hota hai:

green path: $2 v_{1} + 3 v_{2}$ (koi $v_{3}$ nahi),
magenta path: $1 v_{1} + 2 v_{2} + 1 v_{3}$ . Same destination, do honest recipes. Redundancy ne uniqueness barbad kar di.

Step 4 — "Independence": zero tak pahunchne ka ek aur sirf ek tarika

KYA. Hum apne set ko ek sabse important target ke khilaaf test karte hain: zero vector $0$ (woh arrow jo kahin nahi jaata, tail aur tip dono $O$ par).

KYUN. Yahan ek clever move hai. Har point par uniqueness check karne ke bajaye (infinitely many), hum ise ek khaas point $0$ par check karte hain. Set linearly independent hai agar wahi recipe jo aapko origin par wapas le jaaye woh do-nothing recipe $c_{1} = \dots = c_{n} = 0$ ho.

PICTURE. Left panel: ek redundant set ke liye, ek non-zero recipe ( $1 v_{1} + 1 v_{2} - 1 v_{3}$ ) $O$ par wapas ek closed loop banati hai — "sneaky path to zero." Right panel: ek independent set ke liye, arrows ke $O$ tak cancel hone ka ek hi tarika hai — bilkul na hilna.

Step 5 — Basis = dono promises saath mein

KYA. Ek basis woh set hai jo dono spans (Step 2) bhi karta hai aur independent (Step 4) bhi hai.

KYUN. Dono promises ko milake:

Spanning $\Rightarrow$ kam se kam ek address (existence).
Independence $\Rightarrow$ zyada se zyada ek address (uniqueness — aage prove hoga).
Saath mein $\Rightarrow$ exactly ek address.

PICTURE. Ek saaf sunset grid jahan $v_{1}, v_{2}$ do axes hain. Plane mein har point exactly ek violet grid-line aur ek orange grid-line ke crossing par baith ta hai — uska unique coordinate pair.

Step 6 — Uniqueness proof, subtraction ke roop mein drawn

KYA. Maano, contradiction ke liye, ki kisi vector $w$ ke do addresses hain: $w = address A c_{1} v_{1} + \dots + c_{n} v_{n}, w = address B d_{1} v_{1} + \dots + d_{n} v_{n} .$

KYUN subtract karo? Hum independence use karna chahte hain, lekin independence sirf un combinations ke baare mein ek statement hai jo $0$ ke barabar hon. To hum do barabar expressions subtract karke ek " $= 0$ " statement manufacture karte hain: $0 = w - w = difference in slot 1 (c_{1} - d_{1}) v_{1} + \dots + difference in slot n (c_{n} - d_{n}) v_{n} .$

PICTURE. Geometrically: address A ke saath chalna aur phir address B ke saath ulta chalna aapko origin par wapas le jaata hai — ek closed loop. Step 4 se, ek independent set mein koi sneaky loop to zero nahi hai, to loop mein har coefficient $0$ hona chahiye: $c_{i} - d_{i} = 0 ⟹ c_{i} = d_{i} har i ke liye.$ Dono addresses shuru se hi identical the. Uniqueness. ∎

Step 7 — Degenerate & edge cases (reader ko kabhi stranded mat chhodna)

KYA. Hum har woh scenario check karte hain jo picture ko "break" karta hai, taaki baad mein kuch surprise na kare.

KYUN. Ek trustworthy coordinate system sab inputs par defined hona chahiye, awkward ones bhi include karke.

PICTURE. Char panels:

Bahut kam arrows — plane mein ek arrow $v_{1}$ : yeh sirf apni line tak pahunchta hai. Line se bahar ke points (magenta star) ka koi address nahi hai. Spanning fail. (parent: ${e_{1}, e_{2}}$ in $R^{3}$ .)
Bahut zyada arrows — plane mein teen arrows: har point ko multiple addresses milte hain. Independence fail.
Building block ke roop mein zero vector — agar koi $v_{i} = 0$ hai, to $1 \cdot v_{i}$ ek sneaky non-trivial path to $0$ hai; set automatically dependent hai, kabhi basis nahi.
Coincident arrows — $v_{2}$ parallel to $v_{1}$ : yeh sirf ek line span karte hain, plane nahi. 2D mein spanning fail.

Ek-picture summary

KYA. Upar sab kuch, compressed: existence spanning se aata hai, uniqueness independence se aati hai, aur unka overlap basis ki single-address duniya hai.

Recall Feynman: poora walkthrough plain words mein

Imagine karo tum arrows use karke ek warm sunset grid par directions de rahe ho. Pehle (Step 1–2) tum kuch arrows choose karte ho aur check karte ho ki tum unhe stretch aur stack karke sheher ke har kone tak pahunch sako — yahi spanning hai, matlab har jagah ka kam se kam ek set of directions hai. Lekin (Step 3) agar tum ek extra arrow chupke se le aate ho jo bas "arrow 1 aur arrow 2 mila ke jaao" wala hai, to usi ghar ke ab do sets of directions ho jaate hain — confusing! Ise rokne ke liye, tum demand karte ho (Step 4) ki tumhare arrows ke cancel hokar wapas start tak pahunchne ka ek hi tarika ho — bilkul na hilna — sneaky loops ghar nahi. Yahi independence hai, aur yeh har jagah ke liye zyada se zyada ek set of directions force karta hai. Dono promises saath mein lagao (Step 5) aur har ghar ka exactly ek honest address hoga. Yeh prove karna ki yeh sach mein unique hai (Step 6) bas yahi hai: agar kisi ghar ke do addresses hote, pehle se chalo aur doosre se wapas aao — tum start par loop karte ho, lekin independence koi bhi loop-to-start forbid karta hai sivaaye empty wale ke, to dono addresses secretly same the. Aakhir mein (Step 7) hum broken cases sanity-check karte hain: bahut kam arrows sheher ke kuch hisse tak pahunch nahi paate, bahut zyada double-addresses dete hain, aur ek bekar "stay-put" arrow ek instant sneaky loop hai. Sweet spot — exactly utne arrows jitne sheher mein dimensions hain — ek basis hai.

Connections

Linear Independence — Steps 3, 4, 6: "no sneaky loop to zero" promise jo uniqueness force karta hai.
Span and Spanning Sets — Step 2: "har jagah pahuncho" promise jo existence force karta hai.
Dimension — Step 7: exact number of arrows jo tumhe use karne chahiye.
Coordinate Vectors and Change of Basis — Step 5: address $[w]_{B}$ jo har vector ko milta hai.
Subspaces — $O$ se guzarne wali line ya plane apne chhhote basis se span hoti hai.
Matrix of a Linear Map — basis arrows ko ek map se feed karne se uske matrix columns bante hain.

Visual walkthrough — Basis — definition, uniqueness of representation