Change of basis matrix: Visual walkthrough — Hinglish deep dive

4.5.20 · D2 · Maths › Linear Algebra (Full) › Change of basis matrix

Hum flat plane mein kaam karte hain jo aap kaagaz par draw kar sakte ho — ek fixed dot (the origin) se shuru hone wale tamam arrows ka set. Is set ko $R^{2}$ kehte hain. Sab kuch generalise hota hai, lekin aapko pehle 2D mein hote dekha chahiye.

Step 1 — Arrow real hai; uske numbers nahi hain

KYA. Origin se ek arrow $v$ draw karo. Woh bas wahan baith jaata hai. Ab do rulers rakho — do directions jinhein hum measure karne ke liye agree karte hain. Pehli ruler-direction ko $b_{1}$ aur doosri ko $b_{2}$ bolo. Mil ke ${b_{1}, b_{2}}$ ek basis hai: do directions jo (a) ek hi line par nahi hain aur (b) hame unhe combine karke har point tak pahunchne deti hain. (Dekho Basis and Dimension yeh samajhne ke liye ki plane mein exactly do independent directions kyun kaafi hain.)

KYO. Numbers ki ek column ka koi matlab nahi jab tak hum nahi kahte "numbers kiska, kaise measure kiya." Basis woh "kaise measure kiya" hai. Rulers badlo aur numbers badal jaate hain — lekin arrow ek millimetre bhi nahi hilta.

PICTURE. Black arrow $v$ fixed hai. Red rulers ka pair $b_{1}, b_{2}$ hamaari choice hai.

Step 2 — Do dost ek hi arrow describe karte hain

KYA. Ab ek doosra basis laao $C = {c_{1}^{'}, c_{2}^{'}}$ — rulers ka ek alag pair. (Main $C$ ke vectors ko primes ke saath likhta hoon, $c_{1}^{'}, c_{2}^{'}$ , taaki woh kabhi coordinate numbers $c_{1}, c_{2}$ se confuse na hon.) Usi arrow $v$ ke ab do coordinate columns hain: $[v]_{B}$ aur $[v]_{C}$ .

KYO. Hamara poora goal ek aisi machine hai jo ek column ko doosre mein badal de. Machine design karne ke liye pehle dono columns chahiye jo ek hi arrow ko dekh rahe hon.

PICTURE. Ek black arrow $v$ . $B$ -rulers ka thin black grid; ek akela red cheez hai naya ruler pair $C$ . Same tip, do grids ise read kar rahe hain.

Step 3 — Arrow ko PURANI language mein likho (jo diya gaya hai wahi)

KYA. Hame $[v]_{B} = (c_{1}, c_{2})$ diya gaya hai. Literally unpack karo iska matlab: $v = steps along b_{1} c_{1} b_{1} + steps along b_{2} c_{2} b_{2} .$

KYO. Yahi ek cheez hamare haath mein hai. Hum abhi $C$ use nahi kar sakte — hame abhi tak pata nahi ki $C$ kuch bhi kaise dekhta hai. Jo diya hai wahan se shuru karo.

PICTURE. Red arrow $v$ ek staircase ki tarah bana: $b_{1}$ ki $c_{1}$ copies chalo, phir $b_{2}$ ki $c_{2}$ copies chalo, tip par pahuncho.

Term by term: $c_{1} b_{1}$ staircase ki pehli leg hai, $c_{2} b_{2}$ doosri leg hai, aur unka vector sum $v$ origin se tip tak diagonal hai.

Step 4 — Machine ko EK word ek time par sikhao

KYA. Har purana ruler $b_{j}$ khud bas ek arrow hai, isliye uski bhi ek $C$ -description hai. $C$ -dost se pucho: "tum $b_{1}$ kaise likhte ho? tum $b_{2}$ kaise likhte ho?" Jawab hain columns $[b_{1}]_{C} = [p_{11} p_{21}], [b_{2}]_{C} = [p_{12} p_{22}] .$ Symbol $p_{ij}$ padhte hain "row $i$ , column $j$ ": $j$ -ve $B$ -ruler ko reproduce karne ke liye $i$ -ve $C$ -ruler ke along kitne steps chahiye.

KYO. Yahi translation ka dil hai. Dictionary poori sentences ke liye nahi banti — word by word banti hai. Words hain basis vectors $b_{1}, b_{2}$ . Agar main har ruler ko nai language mein bol sakta hoon, toh main unka koi bhi combination bol sakta hoon, kyunki combining sirf addition aur stretching hai, jo coordinates respect karte hain.

PICTURE. Akela red ruler $b_{1}$ $C$ -grid ke against decomposed dikhaya gaya: uska shadow $p_{11}$ along $c_{1}^{'}$ aur $p_{21}$ along $c_{2}^{'}$ .

Step 5 — Coordinates add aur stretch hote hain (linear law)

KYA. Step 3 ki equation ke dono sides ke $C$ -coordinates lo. Map "arrow $\mapsto$ uska $C$ -column", likha $[\cdot]_{C}$ , do rules follow karta hai: $[u + w]_{C} = [u]_{C} + [w]_{C}, [k u]_{C} = k [u]_{C} .$ Yeh do rules mil ke $[\cdot]_{C}$ ko ek linear map banate hain (dekho Linear Transformations). Inhe apply karte hain: $[v]_{C} = [c_{1} b_{1} + c_{2} b_{2}]_{C} = c_{1} [b_{1}]_{C} + c_{2} [b_{2}]_{C} .$

KYO. Kyunki coordinates unique hain (har arrow ka har basis mein ek hi recipe), arrow ko double karne se har step-count double ho jaata hai, aur arrows jodhne se step-counts jud jaate hain. Isliye hum sum ko term by term alag kar sakte hain — "poore arrow ka translate karo" ko "har word translate karo, phir wahi weights $c_{1}, c_{2}$ se recombine karo" mein badal dete hain.

PICTURE. Step 3 jaisa staircase, lekin ab har rung $C$ -grid par read ho rahi hai: red total column $[v]_{C}$ literally pehle dictionary column ka $c_{1}$ aur doosre ka $c_{2}$ hai.

Dhyaan do: weights $c_{1}, c_{2}$ nahi bade — woh abhi bhi purane $B$ -coordinates hain. Sirf woh directions jinhe weight diya ja raha hai switch huin arrows $b_{j}$ se unke $C$ -columns $[b_{j}]_{C}$ mein.

Step 6 — Sum mein chhupa hua matrix pehchano

KYA. Expression $c_{1} [b_{1}]_{C} + c_{2} [b_{2}]_{C}$ exactly wahi hai jo "matrix times column" matlab hai: do dictionary columns ko side by side stack karo, $(c_{1}, c_{2})$ feed karo.

[v]_{C} = P_{C \leftarrow B} [[b_{1}]_{C} [b_{2}]_{C}] [c_{1} c_{2}] = P_{C \leftarrow B} [v]_{B} .

KYO. Matrix–vector product $M x$ define hota hai " $x_{1}$ times column 1 plus $x_{2}$ times column 2 …" ke roop mein. Step 5 ne exactly woh shape produce ki. Toh jo machine hum chahte the woh real hai, aur uske columns forced hain ki woh dictionary entries $[b_{j}]_{C}$ hon — aur kuch fit nahi karta.

PICTURE. Do red dictionary columns ek box labelled $P_{C \leftarrow B}$ ke do columns ki tarah slot ho rahe hain; input column $[v]_{B}$ daayein taraf, output $[v]_{C}$ neeche nikal raha hai.

Step 7 — Standard-basis shortcut (kyun $C^{- 1} B$ )

KYA. $R^{2}$ mein ek free ruler pair hai: standard basis $E = {(1, 0), (0, 1)}$ , axes khud. Uske liye, $[b_{j}]_{E} = b_{j}$ literally (ek vector ke standard coordinates bas uski entries hain). Toh "dictionary $B \to E$ " ke liye koi kaam nahi: $P_{E \leftarrow B} = [b_{1} ∣ b_{2}] =: B (woh matrix jiske columns basis vectors hain) .$ Ulti direction jaane ke liye, $E \to B$ , hum us machine ko undo karte hain — inverse: $P_{B \leftarrow E} = B^{- 1} .$ Ab do translations chain karo, $B \to E \to C$ : $P_{C \leftarrow B} = P_{C \leftarrow E} P_{E \leftarrow B} = C^{- 1} B .$

KYO. Sab kuch standard axes ke through translate karne se hame kabhi haath se "how does $C$ see $b_{j}$ " solve nahi karna padta — matrix inversion ek shot mein kar deta hai. Right-to-left padhte hain: pehle $B$ $B$ -coords ko standard tak le jaata hai, phir $C^{- 1}$ standard ko $C$ -coords tak le jaata hai.

PICTURE. Ek row mein teen grids — $B$ -grid, phir standard axes, phir $C$ -grid — arrow ka column har hop par relabelled ho raha hai; do hops $B$ aur $C^{- 1}$ red mein marked hain.

Step 8 — Edge aur degenerate cases (kabhi koi unseen scenario mat dekho)

KYA & KYO & PICTURE har corner case ke liye:

(a) $C = B$ (ek language ko khud mein translate karo). Tab $[b_{j}]_{C} = [b_{j}]_{B} = e_{j}$ , toh columns $e_{1}, e_{2}$ hain: machine identity $I$ hai. Kuch karne ki zaroorat nahi — arrow pehle se target language bolti hai.

(b) $C = E$ (target standard hai). Columns $[b_{j}]_{E} = b_{j}$ ban jaate hain, toh $P_{E \leftarrow B} = B$ : bas basis vectors stack karo. Yeh "easy direction" hai — koi inverse nahi. Example: $B = {(2, 1), (- 1, 1)}$ , $[v]_{B} = (4, - 1)^{T}$ deta hai $[v]_{E} = B [v]_{B} = (9, 3)^{T}$ .

(c) $B = E$ (source standard hai). Ab $P_{C \leftarrow E} = C^{- 1}$ : hard direction, inverse zaruri hai. Example: $w = (5, 0)$ , $B = {(2, 1), (- 1, 1)}$ deta hai $[w]_{B} = B^{- 1} w = (5/3, - 5/3)^{T}$ . Check karo $\frac{5}{3} (2, 1) - \frac{5}{3} (- 1, 1) = (5, 0)$ . ✓

(d) Forbidden case: rulers ek line par. Agar $b_{1}, b_{2}$ same (ya opposite) direction mein point karein, woh basis nahi hain — woh line se bahar ke arrows tak nahi pahunch sakte, aur matrix $B$ ka koi inverse nahi hai. Poora construction shuru hone se inkar kar deta hai. Isliye bases independent honi chahiye, aur isliye har genuine change of basis matrix invertible hai.

PICTURE. Left panel — do independent red rulers plane span karte hain (allowed). Right panel — do collinear red rulers ek line par collapse ho rahe hain, shaded "unreachable" region (forbidden).

Ek-picture summary

Is page ki har cheez ek visually bani sentence hai: $P_{C \leftarrow B}$ ke columns purane rulers $b_{j}$ hain jo naye grid par redraw kiye gaye hain; multiply karne par arrow same weights ke saath rebuild ho jaata hai. Figure Steps 3–6 ko ek single flow mein compress karta hai.

Recall Feynman retelling — poora walkthrough kisi dost ko explain karo

Ek park mein khazana rakha hai; woh kabhi nahi hilta. Dost $B$ apne steps mein directions deti hai ( $b_{1}, b_{2}$ ); dost $C$ apne mein ( $c_{1}^{'}, c_{2}^{'}$ ). Mere paas sirf $B$ ki directions hain ( $c_{1}$ uske pehle step ki, $c_{2}$ doosre ki). Convert karne ke liye main khazana dobara nahi dhundta — main sirf $C$ se ek baar puchta hoon: "tum $B$ ke do steps mein se har ek ko apne steps mein kaise bolo?" Woh do jawab mere do dictionary columns hain. Phir main wahi counts $c_{1}, c_{2}$ reuse karta hoon — kyunki walk ko double karne se har step double hoti hai aur walks jodhne se steps jud jaate hain — aur dictionary columns ko ek box mein stack karta hoon. Mere $B$ -counts us box ko feed karne par $C$ -counts nikalta hai jo identical khazane ki taraf point karta hai. Agar kabhi doosri taraf jaana ho toh box ko ulta kar deta hoon (uska inverse). Aur standard axes ke through boxes free hain: $B$ stack karne ke liye, $C^{- 1}$ peel karne ke liye — toh $P_{C \leftarrow B} = C^{- 1} B$ .

Connections

Basis and Dimension — kyun do independent rulers exactly plane cover karte hain.
Coordinate Vectors — woh columns jo hum translate karte hain.
Invertible Matrices — case (d): ek real change of basis hamesha invertible hota hai.
Similar Matrices — $A^{'} = P^{- 1} A P$ yeh machine ek map par reuse karta hai.
Eigenvectors and Diagonalization — eigenbasis mein change karo.
Linear Transformations — Step 5 mein use ki gayi $[\cdot]_{C}$ ki linearity.