Coordinate vectors — change of basis: Worked examples

4.5.19 · D3 · Maths › Linear Algebra (Full) › Coordinate vectors — change of basis

Yeh page Coordinate vectors — change of basis ki workout room hai. Parent note ne machinery banayi thi: coordinate vectors, matrix $P$ jiske columns $[b_{j}]_{C}$ hain, aur $R^{n}$ shortcut $P_{C \leftarrow B} = C^{- 1} B$ . Yahan hum har tarah ka case cover karenge jo is topic mein aa sakta hai, ek example per "cell".

Symbols kuch bhi karne se pehle, picture yaad karo: ek vector ek arrow hai jo kabhi hilta nahi; ek basis do rulers ki jodi hai; coordinates woh numbers hain jo un rulers pe read karte ho. Change of basis = rulers badalna, arrow frozen.

Definition Symbol

$C \leftarrow B P$ padhna Stacked notation $C \leftarrow B P$ sirf single letter $P$ hai jiske neeche ek chhota label likha hai (bas itna hi LaTeX ka \underset karta hai — woh text ko symbol ke neeche park karta hai; yeh fraction nahi, exponent nahi, subscript multiplication nahi). Agar tumhara reader kabhi is notation ko tedious lage, toh mentally $P_{C \leftarrow B}$ padho — same object hai, label side mein push ho gaya. Label batata hai matrix ka kaam kya hai.

Arrow $C \leftarrow B$ ko " $B$ se $C$ ki taraf" padho (arrow destination ki taraf point karta hai). Yeh matrix old basis $B$ mein likha coordinate vector consume karti hai aur wohi vector ke coordinates new basis $C$ mein produce karti hai: $[v]_{C} = C \leftarrow B P [v]_{B} .$ Usi tarah $B \leftarrow C P$ dusri direction mein jaata hai, aur (parent note) dono inverses hain. Aur jab bhi hum ek bare capital jaise $B$ likhte hain, hum woh matrix mean karte hain jiske columns basis vectors $b_{1}, \dots, b_{n}$ side by side rakhe hain.

Scenario matrix

Har change-of-basis problem inhi cells mein se ek hai. Hum sab cover karenge.

Cell	Kya special hai	Example
A Standard → non-standard	entries padho, phir $B^{- 1}$ lagao	Ex 1
B Non-standard → standard	bas $B$ se multiply karo	Ex 2
C Non-standard → non-standard	full $C^{- 1} B$	Ex 3
D Negative / mixed-sign coordinates	numbers zero se neeche ja sakte hain	Ex 4
E Round-trip (inverse check)	$P_{B \leftarrow C} P_{C \leftarrow B} = I$	Ex 5
F Non- $R^{n}$ space (polynomials)	koi "entries" nahi padhne — system solve karo	Ex 6
G Degenerate input (zero vector)	limiting/edge case	Ex 7
H Real-world word problem	rotated map / two observers	Ex 8
I Exam twist (basis reordered)	basis ka order matter karta hai	Ex 9

Figure — Coordinate vectors — change of basis

Cell A — standard → non-standard

Worked example Example 1 (Cell A)

Maano $B = {b_{1}, b_{2}}$ jahan $b_{1} = [21]$ , $b_{2} = [- 1 1]$ . Vector $v = [15]$ standard coordinates mein diya gaya hai. $[v]_{B}$ nikalo.

Forecast: compute karne se pehle andaaza lagao — kya $v$ zyada " $b_{2}$ ke along" (jo upar-left point karta hai) hai ya " $b_{1}$ ke along"? Yeh upar ki taraf jhukta hai, toh $b_{2}$ ka share acha-khaasa hoga.

Pehle woh matrix naam do jo hum use karenge: $B = [21 - 1 1]$ woh matrix hai jiske columns $b_{1}$ aur $b_{2}$ hain. Ab $[v]_{B} = [c_{1} c_{2}]$ ka matlab likhte hain: yeh satisfy karta hai $c_{1} b_{1} + c_{2} b_{2} = v$ , jo exactly matrix product $B [v]_{B} = v$ hai. Yeh step kyun? Standard basis mein ek vector ke coordinates uske entries hi hote hain, isliye right side mein $v$ already "standard numbers" mein hai. $B$ , $B$ -numbers ko standard numbers mein turn karta hai, isliye hume $B$ ko undo karna hoga.
Toh $[v]_{B} = B^{- 1} v$ . Yahan $det B = 2 (1) - (- 1) (1) = 3$ . Yeh step kyun? $B$ invertible hai exactly isliye kyunki uske columns ek basis hain (independent), toh $B^{- 1}$ exist karta hai; aur pehle $det B$ compute karna batata hai inverse exist karta hai ( $det \neq = 0$ ) aur woh scaling factor deta hai jo hume chahiye.
$B^{- 1} = \frac{1}{3} [1 - 1 12]$ . Yeh step kyun? $2 \times 2$ matrix $[a c b d]$ ka inverse $\frac{1}{a d - b c} [d - c - b a]$ hota hai — diagonal swap karo, off-diagonal negate karo, determinant se divide karo.
$[v]_{B} = \frac{1}{3} [1 - 1 12] [15] = \frac{1}{3} [69] = [23]$ . Yeh step kyun? Row-times-column: top row $1 (1) + 1 (5) = 6$ , bottom row $- 1 (1) + 2 (5) = 9$ , phir $\frac{1}{3}$ dono entries scale karta hai. Yeh $\frac{1}{3}$ exactly isliye aaya kyunki step 2 mein $det B$ compute ki thi.

Verify: $2 b_{1} + 3 b_{2} = [42] + [- 3 3] = [15] = v$ ✓. $b_{2}$ share (3), $b_{1}$ share (2) se zyada hai — forecast sahi nikla.

Cell B — non-standard → standard

Worked example Example 2 (Cell B)

Example 1 waali $B$ hai. Koi tumhe $[w]_{B} = [3 - 2]$ deta hai. Standard coordinates mein $w$ kya hai?

Forecast: $- 2$ wala $- b_{2}$ direction mein (down-right) kheencha hai, toh strongly rightward result expect karo.

Yeh direction easy wali hai: $w = B [w]_{B}$ , jahan $B$ same columns-are-basis matrix hai Example 1 se. Yeh step kyun? $[w]_{B}$ literally coefficients $c_{1}, c_{2}$ list karta hai; $c_{1} b_{1} + c_{2} b_{2}$ mein plug karna bas matrix product $B [w]_{B}$ hai. Koi inverse nahi chahiye — isliye standard ki taraf jaana sasta hai.
$w = [21 - 1 1] [3 - 2] = [6 + 2 3 - 2] = [81]$ . Yeh step kyun? Row-times-column phir se: top $2 (3) + (- 1) (- 2) = 6 + 2 = 8$ , bottom $1 (3) + 1 (- 2) = 3 - 2 = 1$ . Koi determinant nahi aata kyunki koi inverse nahi hai scale karne ke liye.

Verify: Example 1 ki machine se vapas run karo: $B^{- 1} w = \frac{1}{3} [1 - 1 12] [81] = \frac{1}{3} [9 - 6] = [3 - 2]$ ✓ — input pe vapas aa gaye. Forecast ke mutabik rightward.

Cell C — non-standard → non-standard

Worked example Example 3 (Cell C)

$B$ upar jaisi hai. Maano $C = {c_{1}, c_{2}}$ jahan $c_{1} = [11]$ , $c_{2} = [01]$ . $C \leftarrow B P$ banao, phir $[v]_{B} = [23]$ convert karo (Ex 1 wala $v$ ).

Forecast: kyunki $v = [15]$ (Ex 1) aur $c_{1}, c_{2}$ ek easy shear se span karte hain, expect karo ki $[v]_{C}$ ka $c_{1}$ -part $v$ ki first entry ke barabar hoga.

$C \leftarrow B P = C^{- 1} B$ use karo, jahan $C = [1101]$ ke columns $c_{1}, c_{2}$ hain aur $B$ pehle jaisa hai. Yeh step kyun? $P$ ke columns $[b_{j}]_{C} = C^{- 1} b_{j}$ hain; un columns ko stack karna exactly $C^{- 1} B$ hai. Right-to-left padho: $B$ , $B$ →standard le jaata hai, $C^{- 1}$ , standard→ $C$ le jaata hai; beech ka "standard" cancel ho jaata hai.
$det C = 1 (1) - 0 (1) = 1$ , toh $C^{- 1} = [1 - 1 01]$ . Yeh step kyun? Same $2 \times 2$ inverse rule (diagonal swap karo, off-diagonal negate karo, $det$ se divide karo); yahan $det = 1$ toh koi fraction nahi aata.
$C \leftarrow B P = C^{- 1} B = [1 - 1 01] [21 - 1 1] = [2 - 1 - 1 2]$ . Yeh step kyun? Column by column multiply karo: first column $[1 - 1 01] [21] = [2 - 1]$ , second $[1 - 1 01] [- 1 1] = [- 1 2]$ .
$[v]_{C} = C \leftarrow B P [v]_{B} = [2 - 1 - 1 2] [23] = [14]$ . Yeh step kyun? Yahi payoff hai — $B$ -numbers $P$ mein feed karo toh $C$ -numbers nikalta hai: top $2 (2) - 1 (3) = 1$ , bottom $- 1 (2) + 2 (3) = 4$ .

Verify: independently, $[v]_{C} = C^{- 1} v = [1 - 1 01] [15] = [14]$ ✓. Aur $1 c_{1} + 4 c_{2} = [11] + [04] = [15] = v$ ✓. First entry match hui — forecast sahi nikla.

Cell D — negative / mixed-sign coordinates

Worked example Example 4 (Cell D)

Ex 1 wali $B$ ke saath, $u = [- 4 - 1]$ ko $B$ -coordinates mein express karo. Signs dhyan se dekhna.

Forecast: $u$ down-left point karta hai, roughly $b_{1}$ ke opposite. Expect karo ek negative $c_{1}$ .

$[u]_{B} = B^{- 1} u = \frac{1}{3} [1 - 1 12] [- 4 - 1]$ . Yeh step kyun? Cell-A machinery phir se — standard→ $B$ matlab $B^{- 1}$ lagao; negatives se kuch nahi toot ta — coordinates koi bhi real number ho sakte hain, positive ya negative.
$= \frac{1}{3} [- 4 - 1 4 - 2] = \frac{1}{3} [- 5 2] = [- \frac{5}{3} \frac{2}{3}]$ . Yeh step kyun? Row-times-column: top $1 (- 4) + 1 (- 1) = - 5$ , bottom $- 1 (- 4) + 2 (- 1) = 4 - 2 = 2$ , phir $\frac{1}{3}$ se scale karo. Fraction theek hai — coordinates integers hone zaruri nahi.

Verify: $- \frac{5}{3} b_{1} + \frac{2}{3} b_{2} = [- \frac{10}{3} - \frac{5}{3}] + [- \frac{2}{3} \frac{2}{3}] = [- 4 - 1] = u$ ✓. $c_{1} < 0$ forecast ke mutabik — coordinates positive hone zaruri nahi, aur integers bhi nahi.

Cell E — round-trip (inverse check)

Worked example Example 5 (Cell E)

Ex 3 wali $B, C$ use karke, confirm karo ki $B \leftarrow C P = (C \leftarrow B P)^{- 1}$ aur isse vapas jaane ke liye use karo.

Forecast: $C \to B \to C$ jaane se tum unchanged vapas aana chahiye, toh dono matrices multiply karke identity $I = [1001]$ deni chahiye.

$B \leftarrow C P = B^{- 1} C$ compute karo. Yeh step kyun? Shortcut $C^{- 1} B$ mein $B$ aur $C$ ki roles swap karo. Right-to-left padho: $C$ , $C$ →standard bhejta hai, $B^{- 1}$ , standard→ $B$ bhejta hai.
$B^{- 1} C = \frac{1}{3} [1 - 1 12] [1101] = \frac{1}{3} [2112]$ . Yeh step kyun? Column-by-column: first column $\frac{1}{3} [1 - 1 12] [11] = \frac{1}{3} [21]$ , second $\frac{1}{3} [1 - 1 12] [01] = \frac{1}{3} [12]$ .
Ex 3 ke $C \leftarrow B P = [2 - 1 - 1 2]$ ke saath product check karo: $\frac{1}{3} [2112] [2 - 1 - 1 2] = \frac{1}{3} [3003] = [1001] = I .$ Yeh step kyun? Yahi poora reason hai "back-and-forth = kuch nahi"; woh matrices jo ek dusre ko undo karti hain inverses hoti hain. Top-left entry hai $\frac{1}{3} (2 \cdot 2 + 1 \cdot (- 1)) = \frac{1}{3} (3) = 1$ , aur off-diagonals cancel hokar $0$ dete hain.
$[v]_{C} = [14]$ pe sanity round-trip: $B \leftarrow C P [14] = \frac{1}{3} [2 + 4 1 + 8] = [23] = [v]_{B}$ ✓. Yeh step kyun? Yeh confirm karta hai ki inverse actually Ex-3 ke starting numbers vapas laata hai, sirf abstractly nahi balki hamare concrete vector pe bhi.

Verify: hum exactly $[v]_{B} = [23]$ Ex 3 se recover kar liye. Circle band ho gaya.

Cell F — ek aise space mein jahan "entries" nahi hain (polynomials)

Worked example Example 6 (Cell F)

$V = P_{2}$ (degree 2 tak polynomials) mein, $B = {1, t, t^{2}}$ aur $C = {1, 1 + t, (1 + t)^{2}}$ . $q (t) = 3 - 4 t + 2 t^{2}$ ke liye $[q]_{C}$ nikalo.

Forecast: $(1 + t)^{2} = 1 + 2 t + t^{2}$ , toh iska $t^{2}$ -term 1 hai. Sirf $c_{3}$ mein $t^{2}$ hai, toh expect karo ki uska coordinate $q$ ke $t^{2}$ -coefficient ke barabar hoga, yani $2$ .

Koi "entries nahi padhne" — ek polynomial numbers ka column nahi hai jab tak tum $B$ choose nahi karte. Toh system solve karo: $a \cdot 1 + b (1 + t) + c (1 + t)^{2} = 3 - 4 t + 2 t^{2}$ . Yeh step kyun? Coordinates define hote hain basis expansion mein unique coefficients ke roop mein; unhe find karna literally us linear equation ko solve karna hai. Uniqueness guarantee hai $C$ ki Linear independence se.
Sabse highest power pehle match karo: $t^{2}$ sirf $c (1 + t)^{2}$ mein aata hai, deta hai $c = 2$ . Yeh step kyun? Top-down kaam karne se har nayi equation mein sirf ek naya unknown hota hai — ek triangular system turant solve ho jaata hai.
$t$ -term: $b (1 + t) + c (1 + 2 t)$ se $t$ -coefficient hai $b + 2 c = - 4$ , toh $b = - 4 - 4 = - 8$ . Yeh step kyun? $c$ already $2$ pin ho gaya, toh is equation mein sirf $b$ bacha hai.
Constant term: $a + b + c = 3$ , toh $a = 3 - (- 8) - 2 = 9$ . Yeh step kyun? Constant equation last hai kyunki isme sab teen hain — par ab $b, c$ already pata hain, toh $a$ nikal aata hai.
$[q]_{C} = 9 - 8 2$ .

Verify: $9 (1) - 8 (1 + t) + 2 (1 + 2 t + t^{2}) = 9 - 8 - 8 t + 2 + 4 t + 2 t^{2} = 3 - 4 t + 2 t^{2} = q$ ✓. Top coordinate $= 2$ forecast ke mutabik; middle coordinate strongly negative gaya — Cell D waala behaviour phir se dikh raha hai.

Cell G — degenerate input (zero vector)

Worked example Example 7 (Cell G — edge case)

Kisi bhi basis $B$ ke liye $[0]_{B}$ kya hoga, aur change-of-basis matrix usse kya karta hai?

Forecast: zero vector origin hai — koi rulers nahi, koi measuring nahi — toh guess karo "sab zeros" chahe $B$ kuch bhi ho.

$c_{1} b_{1} + \dots + c_{n} b_{n} = 0$ dhundo. Yeh step kyun? Yahi $[0]_{B}$ ki defining equation hai.
Linear independence se, single solution hai $c_{1} = \dots = c_{n} = 0$ . Yeh step kyun? "Linearly independent" ka matlab hi yahi hai ki $0$ dene wala ek hi combination hai — all-zero wala. Yahi uniqueness coordinates ko well-defined banati hai.
Toh $[0]_{B} = 0$ , aur $C \leftarrow B P 0 = 0$ har $P$ ke liye. Yeh step kyun? Koi bhi matrix zero column se multiply hoke zero column deta hai — origin ko har linear map fix karta hai.

Verify: Ex 3 ke $C \leftarrow B P = [2 - 1 - 1 2]$ ke saath, $P [00] = [00]$ ✓. Origin woh ek point hai jis par har observer agree karta hai.

Cell H — real-world word problem

Worked example Example 8 (Cell H)

Do surveyors ek dabbe ko mark karte hain. Ann East–North axes $E = {e_{1}, e_{2}}$ use karti hai. Bob ka map $9 0^{\circ}$ counter-clockwise rotate hai, toh uske rulers hain $f_{1} = [01]$ (North point karta hai) aur $f_{2} = [- 1 0]$ (West point karta hai). Ann dabbe ko $[v]_{E} = [34]$ pe padhti hai (3 East, 4 North). Bob kya likhega?

Forecast: axes ko $+ 9 0^{\circ}$ rotate karne se vector ke numbers $- 9 0^{\circ}$ rotate hote hain. "3 East, 4 North" wala point Bob ke frame mein " $f_{1}$ ke along 4, $f_{2}$ ke along 3" hona chahiye (sign sort out karna padega). Kuch $[4 - 3]$ jaisa guess karo.

Bob ki basis-as-columns matrix hai $F = [01 - 1 0]$ . Hume chahiye $[v]_{F} = F^{- 1} [v]_{E}$ . Yeh step kyun? $F$ , Bob-numbers ko Ann ke standard East–North numbers mein turn karta hai; dusri taraf jaane ke liye hum invert karte hain. Yeh exactly Cell-A move hai $F$ ke saath $B$ ki jagah, kyunki Ann ka frame yahan standard basis hai.
$det F = 0 \cdot 0 - (- 1) (1) = 1$ , toh $F^{- 1} = [0 - 1 10]$ . Yeh step kyun? Same $2 \times 2$ inverse rule (diagonal swap karo, off-diagonal negate karo, $det = 1$ se divide karo); ek rotation ka inverse dusri taraf ka rotation hota hai, jo yeh sign pattern dikhata hai.
$[v]_{F} = [0 - 1 10] [34] = [4 - 3]$ . Yeh step kyun? Row-times-column: top $0 (3) + 1 (4) = 4$ , bottom $- 1 (3) + 0 (4) = - 3$ . Yeh Bob ki reading hai.

Verify: $4 f_{1} + (- 3) f_{2} = 4 [01] - 3 [- 1 0] = [34]$ = Ann ka chest ✓. Bob $[4 - 3]$ likhta hai — forecast match kiya. Same chest, alag numbers; rotation matrix bas relabel karta hai.

Cell I — exam twist (order matters)

Worked example Example 9 (Cell I)

Exam mein $B = {b_{1}, b_{2}}$ diya hai jahan $b_{1} = [10]$ , $b_{2} = [11]$ , aur ek reordered basis $B^{'} = {b_{2}, b_{1}}$ — same vectors, swapped order. $[v]_{B} = [52]$ ke liye $[v]_{B^{'}}$ nikalo.

Forecast: agar sirf order flip hota hai, toh surely coordinates bas swap ho jaayengi: $[25]$ . Chalte hain machinery se confirm karte hain.

Change-of-basis matrix $B^{'} \leftarrow B P$ ke columns $[b_{1}]_{B^{'}}$ aur $[b_{2}]_{B^{'}}$ hain. Yeh step kyun? Same rule hamesha ki tarah — columns old vectors ( $b_{1}, b_{2}$ $B$ ke order mein) hain new coordinates ( $B^{'}$ ) mein likhe hue.
$B^{'}$ mein, pehla ruler $b_{2}$ hai aur doosra $b_{1}$ . Toh $b_{1} = 0 \cdot b_{2} + 1 \cdot b_{1} \Rightarrow [b_{1}]_{B^{'}} = [01]$ , aur $b_{2} = 1 \cdot b_{2} + 0 \cdot b_{1} \Rightarrow [b_{2}]_{B^{'}} = [10]$ . Yeh step kyun? Reordering bas rename karta hai ki kaun sa coordinate slot kis ruler ka hai; pehla slot ab $b_{2}$ ka hai, doosra $b_{1}$ ka.
$B^{'} \leftarrow B P = [0110]$ — swap matrix. Yeh step kyun? Iske do columns exactly step 2 ke coordinate vectors hain, side by side rakhe hue.
$[v]_{B^{'}} = [0110] [52] = [25]$ . Yeh step kyun? Swap matrix se multiply karna dono entries exchange karta hai: top old bottom ban jaata hai ( $2$ ), bottom old top ban jaata hai ( $5$ ).

Verify: $B^{'}$ mein iska matlab hai $2 b_{2} + 5 b_{1}$ ; $B$ mein original tha $5 b_{1} + 2 b_{2}$ — identical sums, same $v$ ✓. Order matters: ek ordered basis data ka hissa hoti hai.

Active recall

Recall Answers cover karo

Q: Standard→ $B$ jaane ke liye, kisse multiply karo? → $B^{- 1}$ (Cell A).
Q: $B$ →standard jaane ke liye, kisse multiply karo? → $B$ (Cell B, no inverse).
Q: $R^{n}$ mein $B$ → $C$ matrix kya hai? → $C^{- 1} B$ (Cell C).
Q: Kya ek coordinate negative ya fractional ho sakta hai? → haan (Cell D).
Q: $[0]_{B}$ kya hai? → zero column, hamesha (Cell G).
Q: Kya basis reorder karne se coordinates badlte hain? → haan, woh permute ho jaate hain (Cell I).

R^{n}

mein standard coordinates ko

B

-coordinates mein convert karne wali matrix kaun si hai?

B^{- 1}

, jahan

B

ke columns basis vectors hain.

Standard basis ki taraf jaate waqt inverse kyun nahi chahiye?

Kyunki

B [v]_{B}

seedha

v

ko uske coefficients se rebuild karta hai.

Polynomials ke liye, matrix inversion ke bina coordinates kaise dhundhte hain?

Expansion equation solve karo, highest powers se constants tak match karte hue.

Kisi bhi basis mein zero vector ke coordinates kya hain?

Sab zeros, kyunki independence kisi bhi non-trivial combination ko

0

ke barabar hone se rokti hai.

Kya basis vectors ka order swap karne se change-of-basis matrix badlti hai?

Haan — tumhe ek permutation (swap) matrix milti hai.