Visual walkthrough — Coordinate vectors — change of basis

4.5.19 · D2 · Maths › Linear Algebra (Full) › Coordinate vectors — change of basis

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Step 1 — Ek arrow, abhi tak koi numbers nahi

KYA. Blank paper par ek arrow $\mathbf{v}$ draw karo. Woh kisi direction mein point kar raha hai. Bas itna hi hai: ek length aur ek direction. Abhi tak koi numbers us se attached nahi hain.

KYUN. Poora subject ek idea par tika hai jo log bhool jaate hain: arrow pehle se exist karta hai, measure karne se pehle. Numbers kuch aisa hai jo hum baad mein add karte hain rulers choose karke. Agar hum yeh skip kar dein, to baad ke saare formulas magic jaisi lagte hain.

PICTURE. Neeche amber arrow dekho. Deliberately uske peeche koi grid nahi hai. Yeh bas space mein ek cheez hai.

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Step 2 — Ek basis do rulers ka pair hai

KYA. Do arrows ${\mathbf{b}}_1$ aur ${\mathbf{b}}_2$ rakho. Unhe genuinely alag directions mein point karna chahiye (parallel nahi). Yeh pair ek basis hai, jise hum $\mathcal{B}=\{{\mathbf{b}}_1,{\mathbf{b}}_2\}$ kehte hain.

KYUN yeh do conditions. Hume "alag directions" chahiye taaki plane ka har arrow unhe mix karke reach kiya ja sake, aur taaki uss tak pahunchne ki recipe unique ho. Do arrows jo same direction mein point karte hain sirf ek hi line tak pahunch sakte hain — coordinate system ke liye bilkul bekaar. Yeh "genuinely alag" requirement exactly Linear independence hai, aur space ko fill karne ke liye unka kaafi hona Basis and dimension hai.

PICTURE. Neeche, cyan arrows ${\mathbf{b}}_1,{\mathbf{b}}_2$ hamare naye rulers hain. Jo faint white grid unse banti hai woh basis $\mathcal{B}$ ka "graph paper" hai — ek tilted graph paper.

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Step 2b — Woh ek basis jo hum secretly pehle se use karte hain: standard basis

KYA. Ek special pair of rulers hai jo sab log bina bole use karte hain: standard basis $\mathcal{E}=\{{\mathbf{e}}_1,{\mathbf{e}}_2\}$ jisme ${\mathbf{e}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right]\text{(ek step right)},{\mathbf{e}}_2=\left[\begin{array}{c}0\\ 1\end{array}\right]\text{(ek step up)}.$

KYUN yeh yahan matter karta hai. Jab hum ek vector ko $\mathbf{v}=\left[\begin{array}{c}4\\ 2\end{array}\right]$ likhte hain, woh do entries already coordinates hain — matlab hai "$4$ steps right, $2$ steps up", yaani $\mathbf{v}=4{\mathbf{e}}_1+2{\mathbf{e}}_2$. Toh plain column $\left[\begin{array}{c}4\\ 2\end{array}\right]$ kuch nahi hai sirf $[\mathbf{v}{]}_{\mathcal{E}}$ ke, yaani standard basis mein coordinate vector. "Standard graph paper" aage se matlab square grid drawn by ${\mathbf{e}}_1,{\mathbf{e}}_2$.

PICTURE. Square grid jisme ${\mathbf{e}}_1$ right point kar raha hai, ${\mathbf{e}}_2$ upar, aur $\mathbf{v}=(4,2)$ seedha axes se read kiya gaya.

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Step 3 — Coordinates = "har ruler kitna chahiye"

KYA. ${\mathbf{b}}_1$ ke saath kuch amount $c_1$ slide karo, phir ${\mathbf{b}}_2$ ke saath kuch amount $c_2$, jab tak $\mathbf{v}$ ki tip par exactly land na ho jao. Pair $(c_1,c_2)$ coordinate vector $[\mathbf{v}{]}_{\mathcal{B}}$ hai.

$\mathbf{v}=\underset{\text{kitna }{\mathbf{b}}_1}{\underbrace{c_1}}\underset{\text{ruler 1}}{\underbrace{{\mathbf{b}}_1}}+\underset{\text{kitna }{\mathbf{b}}_2}{\underbrace{c_2}}\underset{\text{ruler 2}}{\underbrace{{\mathbf{b}}_2}},[\mathbf{v}{]}_{\mathcal{B}}=\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right].$

KYUN yeh numbers ka column hai. Kyunki ek baar rulers fix ho jaayein, do amounts hi ek maatra freedom hai. Aur kyunki rulers independent hain, $\mathbf{v}$ par land karne ka exactly ek hi tarika hai — toh column well-defined hai, kabhi ambiguous nahi.

PICTURE. Dashed amber path do slides dikhata hai: pehle $c_1{\mathbf{b}}_1$, phir $c_2{\mathbf{b}}_2$, $\mathbf{v}$ par pahunchte hue. Parallelogram close hota hai — woh closure hi upar wali equation hai.

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Step 4 — Matrix $B$ arrow ko uske numbers se rebuild karta hai

KYA. Do ruler-arrows ko ek matrix ke columns ke roop mein stack karo aur usse $B$ kaho: $B=\left[{\mathbf{b}}_1{\mathbf{b}}_2\right].$ Tab Step 3 ka slide-recipe exactly ek matrix times coordinate column hai: $\mathbf{v}=B[\mathbf{v}{]}_{\mathcal{B}}=[{\mathbf{b}}_1{\mathbf{b}}_2]\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\underset{\text{col 1}\times c_1}{\underbrace{c_1{\mathbf{b}}_1}}+\underset{\text{col 2}\times c_2}{\underbrace{c_2{\mathbf{b}}_2}}.$

Yahan left side $\mathbf{v}$ actually $[\mathbf{v}{]}_{\mathcal{E}}$ hai (Step 2b): $B$ $\mathcal{B}$-numbers leta hai aur arrow ke standard numbers deta hai.

KYUN matrix aur sirf "add them up" nahi. Kyunki matrix wahi machine hai "har column ko matching input number se multiply karo aur add karo". Yahi slide-recipe hai. Isse $B[\mathbf{v}{]}_{\mathcal{B}}$ likhne se ek picture ek aisi cheez ban jaati hai jise hum baad mein invert kar sakte hain. (Ek matrix jiske columns independent basis hain hamesha invertible hota hai — dekho Invertible matrices.)

PICTURE. Yahan $\mathbf{v}=\left[\begin{array}{c}4\\ 2\end{array}\right]=[\mathbf{v}{]}_{\mathcal{E}}$ standard square grid par, aur $B$ ke do columns cyan arrows hain. Amber arrow wahi $\mathbf{v}$ hai jo $3{\mathbf{b}}_1+1{\mathbf{b}}_2$ ke roop mein rebuild kiya gaya.

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Step 5 — Same arrow par doosra basis $\mathcal{C}$

KYA. Ab ek alag pair of rulers $\mathcal{C}=\{{\mathbf{c}}_1,{\mathbf{c}}_2\}$ laao aur unhe $C=[{\mathbf{c}}_1{\mathbf{c}}_2]$ ke roop mein stack karo. Arrow $\mathbf{v}$ move nahi hua hai. Lekin uski number-list alag hai, use $[\mathbf{v}{]}_{\mathcal{C}}$ kehte hain.

$[\mathbf{v}{]}_{\mathcal{E}}=C[\mathbf{v}{]}_{\mathcal{C}}.$

KYUN. Same arrow, alag rulers ⇒ har ruler ki alag amounts chahiye. Dono equations $[\mathbf{v}{]}_{\mathcal{E}}=B[\mathbf{v}{]}_{\mathcal{B}}$ aur $[\mathbf{v}{]}_{\mathcal{E}}=C[\mathbf{v}{]}_{\mathcal{C}}$ ek fixed arrow ke standard numbers describe karti hain — toh unke right-hand sides equal hone chahiye. Woh equality hi sab kuch ka beej hai.

PICTURE. Ek amber arrow, do overlaid grids — $\mathcal{B}$ ke liye cyan, $\mathcal{C}$ ke liye white-dotted. Same tip, do sets of gridlines usse count karte hue.

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Step 6 — Dono descriptions ko equal set karo aur solve karo

KYA. Dono matrices same standard column produce karti hain, toh: $\underset{\text{arrow via }\mathcal{C}}{\underbrace{C[\mathbf{v}{]}_{\mathcal{C}}}}=\underset{\text{arrow via }\mathcal{B}}{\underbrace{B[\mathbf{v}{]}_{\mathcal{B}}}}.$ Hume $[\mathbf{v}{]}_{\mathcal{C}}$ akele left par chahiye. Toh dono sides ko left se $C^{-1}$ se multiply karo: $\underset{=I}{\underbrace{C^{-1}C}}[\mathbf{v}{]}_{\mathcal{C}}=C^{-1}B[\mathbf{v}{]}_{\mathcal{B}}⟹[\mathbf{v}{]}_{\mathcal{C}}=(C^{-1}B)[\mathbf{v}{]}_{\mathcal{B}}.$

KYUN $C^{-1}$ se multiply karo aur kisi aur se nahi. Hum "$\mathcal{C}$-numbers in, arrow out" machine $C$ ko undo karne ki koshish kar rahe hain. Woh unique tool jo ek matrix ko undo karta hai uska inverse hai (Step 4 box) — yahi reason hai ki $\mathcal{C}$ ek genuine basis honi chahiye: tabhi $C^{-1}$ exist karta hai (Invertible matrices).

PICTURE. Ek pipeline: $\mathcal{B}$-numbers enter karte hain, $B$ unhe standard column mein turn karta hai, phir $C^{-1}$ us column ko $\mathcal{C}$-numbers mein read karta hai. Beech ka "standard numbers" woh meeting point hai jahan do halves milte hain.

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Step 7 — Picture par numbers (ek full run)

KYA. $\mathcal{B}$ use karo jisme ${\mathbf{b}}_1=\left[\begin{array}{c}1\\ 1\end{array}\right],{\mathbf{b}}_2=\left[\begin{array}{c}1\\ -1\end{array}\right]$ aur $\mathcal{C}$ jisme ${\mathbf{c}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right],{\mathbf{c}}_2=\left[\begin{array}{c}1\\ 2\end{array}\right]$. Tab $C=\left[\begin{array}{cc}1& 1\\ 0& 2\end{array}\right],C^{-1}=\left[\begin{array}{cc}1& -\frac{1}{2}\\ 0& \frac{1}{2}\end{array}\right],P=C^{-1}B=\left[\begin{array}{cc}\frac{1}{2}& \frac{3}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right].$

$[\mathbf{v}{]}_{\mathcal{B}}=\left[\begin{array}{c}3\\ 1\end{array}\right]$ lo (yeh $\mathbf{v}=\left[\begin{array}{c}4\\ 2\end{array}\right]=[\mathbf{v}{]}_{\mathcal{E}}$ hai, Step 4 se). Isse $P$ se push karo: $[\mathbf{v}{]}_{\mathcal{C}}=\left[\begin{array}{cc}\frac{1}{2}& \frac{3}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}3\\ 1\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}+\frac{3}{2}\\ \frac{3}{2}-\frac{1}{2}\end{array}\right]=\left[\begin{array}{c}3\\ 1\end{array}\right].$

Arrow par check: $3{\mathbf{c}}_1+1{\mathbf{c}}_2=3\left[\begin{array}{c}1\\ 0\end{array}\right]+\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}4\\ 2\end{array}\right]=\mathbf{v}$. ✓

KYUN sahi land kiya. Humne arrow ko kabhi touch nahi kiya — sirf uska address re-read kiya. Check $\mathbf{v}$ ko naye numbers aur naye rulers se re-assemble karta hai aur original standard column $(4,2)$ wapas laata hai.

PICTURE. Amber arrow jisme dono address-labels dikhaye gaye hain: "$(3,1)$ in $\mathcal{B}$" aur "$(3,1)$ in $\mathcal{C}$" — yahan woh numerically coincide karte hain in particular bases ki ek coincidence se; gridlines jo tip tak count karti hain alag hain.

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Step 8 — Degenerate cases (kahan toot ta hai, aur kyun)

KYA. Kya hoga agar do "rulers" parallel hon, e.g. ${\mathbf{c}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right],{\mathbf{c}}_2=\left[\begin{array}{c}2\\ 0\end{array}\right]$?

KYUN fail hota hai — ek hi bimari ke do symptoms. ${\mathbf{c}}_1=(1,0),{\mathbf{c}}_2=(2,0)$ ke saath $\det C=1\cdot 0-0\cdot 2=0$ milta hai, toh $C^{-1}$ exist nahi karta aur $P=C^{-1}B$ ka koi matlab nahi. Geometrically yeh do tarahon se ek saath dikhta hai:

• Koi solution nahi (line se bahar). Yeh rulers sirf horizontal line tak pahunch sakte hain. Koi bhi arrow jisme nonzero vertical part ho — jaise $(1.5,1.5)$ — unreachable hai: uska koi $\mathcal{C}$-address nahi hai.
• Bahut zyada solutions (line par). Ek arrow jo actually line par lie karta hai, jaise $(4,0)$, ke infinitely many addresses hain: $4=4{\mathbf{c}}_1+0{\mathbf{c}}_2$, lekin yeh bhi $4=2{\mathbf{c}}_1+1{\mathbf{c}}_2$, ya $0{\mathbf{c}}_1+2{\mathbf{c}}_2$, … Kyunki rulers overlap karte hain, aap ek ko doosre ke liye freely trade kar sakte hain. Uniqueness — poora reason ki coordinates kuch matlab rakhte hain — destroy ho jaati hai.

Dono symptoms hi reason hain ki ek basis independent honi chahiye: independence exactly woh condition hai "$\det C\ne 0$", jo guarantee karta hai har arrow ke liye exactly ek address.

Zero coordinates aur negatives theek hain, though. Agar $[\mathbf{v}{]}_{\mathcal{C}}$ mein koi zero entry hai, arrow sirf us ruler ka kuch use nahi karta. Agar usmein negative entry hai, aap us ruler ke saath backwards slide karte hain — phir bhi bilkul legal (Example 3 ka $-1$ yaad karo).

PICTURE. Left panel: do parallel cyan arrows, ek off-line target jisme "✗ no address" mark hai, aur ek on-line arrow $(4,0)$ jo do alag amber decompositions ke saath dikhaya gaya ("∞ many addresses"). Right panel: ek valid basis jisme ek negative coordinate hai, backwards slide ek dashed reversed amber arrow ke roop mein draw kiya gaya.

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Ek-picture summary

Sab kuch maps ke ek triangle mein compress ho jaata hai: $\mathcal{B}$-numbers left par, $\mathcal{C}$-numbers right par, standard column $[\mathbf{v}{]}_{\mathcal{E}}$ top apex par, $B$, $C$, aur unke inverses edges par. $\mathcal{B}\to$ apex $\to \mathcal{C}$ walk karna $C^{-1}B$ hai.

Recall Feynman retelling — poora walkthrough plain words mein

Ek khazana ek field mein pada hai (Step 1: arrow, koi measure karne se pehle). Town ka official map pehle se "steps east / steps north" use karta hai — yeh standard basis hai, aur $(4,2)$ khazane ka official address hai (Step 2b). Anna apni khud ki do step-directions prefer karti hai; uska "recipe machine" $B$ uske step-counts ko official address mein turn karta hai (Step 3–4). Ben aur step-directions use karta hai apni machine $C$ ke saath (Step 5). Anna ke counts ko Ben ke counts mein turn karne ke liye: Anna ki machine ko official address tak aage chalaao, phir Ben ki machine backwards ($C^{-1}$, jo exist karta hai sirf isliye kyunki uski directions genuinely alag hain) chalaao aur read karo ki woh use kitne steps mein pace karega (Step 6). Woh combo $C^{-1}B$ hai; numbers ek real example par check out ho gaye (Step 7). Aur agar Ben ki do directions secretly same line par hoti, off-line khazanon ko koi address nahi milta aur on-line khazanon ko infinitely many milte hain — uski machine ko bilkul backwards nahi chalaya ja sakta (Step 8). Khazana kabhi move nahi hua; sirf paperwork badla.