Worked examples — Change of basis matrix

4.5.20 · D3 · Maths › Linear Algebra (Full) › Change of basis matrix

Agar koi symbol unfamiliar lage, toh woh Change of basis matrix mein build kiya gaya tha — yeh page sirf wohi page assume karta hai.

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Scenario matrix

Har change-of-basis problem in mein se ek shape ka hota hai. Hum inhe enumerate karte hain taaki kuch bhi surprise na kare.

Cell	Isko us case ka kya banata hai	Kaun si direction	Example
A	Old = standard, new = non-standard	$E\to C$ (inverse chahiye)	Ex 1
B	Old = non-standard, new = standard	$B\to E$ (easy, no inverse)	Ex 2
C	Dono non-standard	$B\to C$ (full $C^{-1}B$)	Ex 3
D	New basis old ka scaled version hai	$E\to C$	Ex 4
E	New basis old ka rotated version hai, $\det >0$ (geometry)	$E\to C$	Ex 5
F	Existing translator ko reverse karo	use $P^{-1}$	Ex 6
G	Basis in aur out same hai (identity)	$B\to B$	Ex 7
H	Bada space, $3\times 3$	$E\to C$ in ${\mathbb{R}}^3$	Ex 8
I	Near-degenerate basis (limiting/breaking)	kyun $C^{-1}$ blows up karta hai	Ex 9
J	Real-world word problem	modelling	Ex 10
K	Reflected basis, $\det <0$ (orientation flip)	$E\to C$	Ex 11

Is topic ke liye "signs aur quadrants" ki number line actually determinant ka sign hai aur kya $C^{-1}$ exist karta hai ya nahi. Cell E mein $\det >0$ hai (orientation preserved), cell K mein $\det <0$ hai (orientation flipped, ek mirror), aur cell I degenerate boundary $\det \to 0$ hai jahan $C^{-1}$ exist karna band kar deta hai.

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Example 1 — Cell A: standard → non-standard (inverse chahiye)

Forecast: Compute karne se pehle guess karo — kya pehla $C$-coordinate $4$ se zyada hoga ya kam? (Pehla $C$-vector $e_1$ se twice as long hai, isliye tumhe iska kam use karna padega.)

1. $C$ ka matrix banao. $C=\left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right].$ Yeh step kyun? $C$ ke columns, new basis vectors hain jo standard coordinates mein likhe hain — yeh parent note ka standard-basis shortcut hai.

2. Humein $P_{C\leftarrow E}=C^{-1}$ chahiye. Yeh step kyun? Hum standard-land se $C$-land mein ja rahe hain. Subscript padho: $P_{C\leftarrow E}$ $E$-coords leta hai, $C$-coords deta hai. Parent formula deta hai $P_{C\leftarrow E}=C^{-1}E=C^{-1}$ (kyunki $E=I$).

3. $C$ ko invert karo. $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ ka inverse hai $\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$. Yahan $\det =2\cdot 3-1\cdot 0=6$: $C^{-1}=\frac{1}{6}\left[\begin{array}{cc}3& -1\\ 0& 2\end{array}\right].$ Yeh step kyun? $\det =6$ se divide karna aur entries swap/negate karna $2\times 2$ inversion rule hai — yeh woh translator hai jo standard coords padhta hai aur unhe $C$ mein report karta hai.

4. $(4,6{)}^T$ par apply karo. $[v{]}_C=\frac{1}{6}\left[\begin{array}{cc}3& -1\\ 0& 2\end{array}\right]\left[\begin{array}{c}4\\ 6\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}12-6\\ 0+12\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right].$ Yeh step kyun? Translator ko input column se multiply karna hi translation hai — output hai $[v{]}_C$.

Verify: Arrow reconstruct karo: $1\cdot (2,0)+2\cdot (1,3)=(2+2,0+6)=(4,6)=v$. ✓ Forecast confirm hua: pehla coordinate $1$ hai, $4$ se kaafi kam, exactly isliye kyunki $(2,0)$ ek lamba ruler hai.

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Example 2 — Cell B: non-standard → standard (easy direction)

Forecast: Kya yahan inverse chahiye? (Nahi — standard mein decode karna hamesha multiply-by-$B$ direction hota hai.)

1. $B$ likho. $B=\left[\begin{array}{cc}3& -1\\ 1& 2\end{array}\right]$. Yeh step kyun? Iske columns pehle se standard coordinates mein hain, isliye $P_{E\leftarrow B}=B$ directly milta hai — koi inverse nahi.

2. Multiply karo. $[v{]}_E=B[v{]}_B=\left[\begin{array}{cc}3& -1\\ 1& 2\end{array}\right]\left[\begin{array}{c}2\\ 4\end{array}\right]=\left[\begin{array}{c}6-4\\ 2+8\end{array}\right]=\left[\begin{array}{c}2\\ 10\end{array}\right].$ Yeh step kyun? $[v{]}_B=(2,4)$ ka matlab hai $v=2b_1+4b_2$; matrix–column multiply literally wohi linear combination hai.

Verify: $2(3,1)+4(-1,2)=(6-4,2+8)=(2,10)$. ✓

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Example 3 — Cell C: dono bases non-standard (full $C^{-1}B$)

Forecast: Mnemonic hai "C-inverse Bites B." Kaun sa matrix invert hoga — input basis ya output basis? (Output, yaani $C$.)

1. Dono matrices assemble karo. $B=\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right],C=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right].$ Yeh step kyun? Columns, respective basis vectors ko standard coordinates mein represent karte hain — formula ke liye raw material.

2. $C$ invert karo. $\det C=1$, isliye $C^{-1}=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right].$ Yeh step kyun? $B$-coords pehle standard mein translate karo ($\times B$) phir standard ko $C$ mein ($\times C^{-1}$). Composite hai $C^{-1}B$.

3. $C^{-1}B$ multiply karo. $P_{C\leftarrow B}=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}1-2& 3-1\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right].$ Yeh step kyun? Yeh product ek single translator card hai jo $B$-coords khata hai aur ek hi shot mein $C$-coords deta hai.

4. $[v{]}_B=(1,1{)}^T$ par apply karo. $[v{]}_C=\left[\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 3\end{array}\right].$ Yeh step kyun? Input column ko translator se paas karne par $[v{]}_C$ milta hai.

Verify (Forecast-then-Verify): Asli arrow hai $v=1(1,2)+1(3,1)=(4,3)$. $C$ mein: $1(1,0)+3(1,1)=(1+3,0+3)=(4,3)$. ✓

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Example 4 — Cell D: new basis old ka scaled version ($E\to C$)

Forecast: Ruler stretch karne se measured number shrink hota hai. Compute karne se pehle dono coordinates predict karo.

1. $C$ ka matrix diagonal hai. $C=\left[\begin{array}{cc}5& 0\\ 0& 2\end{array}\right]$. Yeh step kyun? Diagonal isliye kyunki dono rulers interfere nahi karte — pure scaling.

2. Diagonal matrix ka inverse = diagonal par reciprocals. $C^{-1}=\left[\begin{array}{cc}1/5& 0\\ 0& 1/2\end{array}\right].$ Yeh step kyun? Ek diagonal matrix har axis ko independently scale karta hai, isliye undo karne ke liye sirf har scale ka inverse lena hota hai — koi cross terms nahi. Kyunki $B=E$ hai, $P_{C\leftarrow E}=C^{-1}$.

3. Apply karo. $[v{]}_C=\left[\begin{array}{cc}1/5& 0\\ 0& 1/2\end{array}\right]\left[\begin{array}{c}10\\ 6\end{array}\right]=\left[\begin{array}{c}2\\ 3\end{array}\right].$ Yeh step kyun? Standard-coord input ko $C^{-1}$ se paas karne par stretched rulers mein report milta hai.

Verify: $2(5,0)+3(0,2)=(10,6)$. ✓ Numbers shrink gaye ($10\to 2$, $6\to 3$) exactly as forecast — lamba ruler chhoti reading deta hai.

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Example 5 — Cell E: new basis old ka rotated version, $\det >0$ (geometry)

Figure 1 — Faint white arrows purane standard axes hain. Blue arrow naya pehla basis vector $c_1$ hai; red arrow $c_2$ hai. Notice karo ki dono length $1$ ke hain aur right angle par milte hain (ek orientation-preserving rotation, $\det =+1$). Yellow arrow fixed vector $v=(1,1)$ hai — observe karo ki woh $c_1$ ke saath bilkul flat leta hai, exactly isliye uska doosra $C$-coordinate zero aayega. Arrow khud kabhi nahi hila; sirf woh grid ghuma jisse hum ise measure karte hain.

Forecast: Arrow $(1,1)$ exactly naye $c_1$ direction ki taraf point karta hai. Toh ek $C$-coordinate zero hona chahiye. Kaun sa?

1. $C$ apne columns se banao. $C=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right].$ Yeh step kyun? Columns standard coords mein rotated basis vectors hain.

2. Rotation ke liye, inverse transpose hota hai. $C^{-1}=C^T=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right].$ Yeh step kyun? Rotation matrices orthogonal hote hain — unke columns unit-length aur perpendicular hote hain (Figure 1 mein blue aur red arrows dekho) — isliye $C^{T}C=I$, matlab $C^{-1}=C^T$. Koi fraction-heavy inversion nahi chahiye.

3. $(1,1{)}^T$ par apply karo. $[v{]}_C=C^T\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\end{array}\right]=\left[\begin{array}{c}\sqrt{2}\\ 0\end{array}\right].$ Yeh step kyun? Standard coords ko $C^T=C^{-1}$ se paas karne par wohi arrow rotated grid mein report hota hai.

Verify: $\sqrt{2}\cdot c_1+0\cdot c_2=\sqrt{2}\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)=(1,1)=v$. ✓ Doosra coordinate $0$ hai, as forecast — arrow $c_1$ ke along leta hai (Figure 1 mein yellow arrow, blue arrow par).

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Example 6 — Cell F: existing translator ko reverse karo ($P^{-1}$)

Forecast: Tumhare paas $C$-coords hain lekin $B$-coords chahiye. Kaun sa matrix tumhe backwards le jaata hai? ($P_{B\leftarrow C}=P_{C\leftarrow B}^{-1}$ — translator card ko ulta palatao.)

1. Diye gaye matrix ko invert karo. $\det =(-1)(1)-(2)(2)=-5$: $P_{B\leftarrow C}=(P_{C\leftarrow B}{)}^{-1}=\frac{1}{-5}\left[\begin{array}{cc}1& -2\\ -2& -1\end{array}\right]=\left[\begin{array}{cc}-1/5& 2/5\\ 2/5& 1/5\end{array}\right].$ Yeh step kyun? Change of basis ki direction reverse karna exactly inverse lena hai — parent note ka inverse relation.

2. $(1,3{)}^T$ par apply karo. $[w{]}_B=\left[\begin{array}{cc}-1/5& 2/5\\ 2/5& 1/5\end{array}\right]\left[\begin{array}{c}1\\ 3\end{array}\right]=\left[\begin{array}{c}-1/5+6/5\\ 2/5+3/5\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right].$ Yeh step kyun? $C$-coord input ko reversed translator se paas karne par $B$-coords milte hain.

Verify: Yeh Ex 3 undo karna chahiye, jahan $[v{]}_B=(1,1{)}^T$ ne $[v{]}_C=(1,3{)}^T$ diya tha. $(1,3{)}^T$ wapas feed karne par $(1,1{)}^T$ milta hai. ✓

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Example 7 — Cell G: same basis in aur out (identity)

Forecast: Ek language ko apne aap mein translate karna kuch nahi badalti — toh yeh matrix kaun si honi chahiye?

1. Formula $C=B$ ke saath use karo. $P_{B\leftarrow B}=B^{-1}B=I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right].$ Yeh step kyun? $C^{-1}B$ jab $C=B$ ho toh identity mein collapse ho jaata hai — degenerate lekin essential "no change" case.

2. Apply karo. $I(2,4{)}^T=(2,4{)}^T$. Unchanged, as it must be. Yeh step kyun? Identity kisi bhi input column ko untouched chhod deti hai — yeh confirm karta hai ki "same basis in and out" kuch nahi karta.

Verify: $P_{B\leftarrow B}$ ka column $j$, $[b_j{]}_B=e_j$ hona chahiye — aur wakai $[b_1{]}_B=(1,0{)}^T$, $[b_2{]}_B=(0,1{)}^T$, jo stack hoke $I$ deta hai. ✓

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Example 8 — Cell H: bada space, $3\times 3$ ($E\to C$)

Forecast: $C$ mein "har vector ek aur $1$ add karta hai." Iska inverse ek simple difference pattern hona chahiye. Top coordinate guess karo.

1. $C$ ka matrix. $C=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right].$ Yeh step kyun? Columns = new basis vectors in standard coords.

2. Humein $C^{-1}$ chahiye. Yeh upper-triangular "sab diagonal aur upar ones" matrix finite-difference matrix mein invert hota hai: $C^{-1}=\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ 0& 0& 1\end{array}\right].$ Yeh step kyun? Kyunki $B=E$ hai, $P_{C\leftarrow E}=C^{-1}$; aur "next component add karna" undo karna matlab usse subtract karna, jo yeh difference pattern deta hai. Tum directly $C^{-1}C=I$ check kar sakte ho.

3. $(3,5,2{)}^T$ par apply karo. $[v{]}_C=\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}3\\ 5\\ 2\end{array}\right]=\left[\begin{array}{c}3-5\\ 5-2\\ 2\end{array}\right]=\left[\begin{array}{c}-2\\ 3\\ 2\end{array}\right].$ Yeh step kyun? Standard-coord input ko $C^{-1}$ se paas karne par woh $C$-basis mein report hota hai.

Verify: $-2(1,0,0)+3(1,1,0)+2(1,1,1)=(-2+3+2,3+2,2)=(3,5,2)$. ✓

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Example 9 — Cell I: near-degenerate basis (kyun toot ta hai)

Figure 2 — Blue arrow $c_1=(1,0)$ aur red arrow $c_2=(1,0.1)$ almost coincide karte hain: dono rulers lagbhag ek hi direction mein point karte hain. Yellow arrow woh vector $v=(1,1)$ hai jise hum measure karna chahte hain. Kyunki rulers mein bahut kam farq hai, $v$ tak pahunchne ke liye ek bada forward push aur ek bada cancelling pull chahiye, isliye coordinates $(-9,10)$ chhote arrow ke muqable mein bahut bade hain — yeh near-degenerate basis ki visual signature hai.

Forecast: Jaise dono basis vectors almost parallel ho jaate hain (Figure 2: red aur blue almost overlap), kya kisi vector ko measure karna aasaan hoga ya dangerously sensitive?

1. Determinant. $\det C_{\epsilon }=1\cdot \epsilon -1\cdot 0=\epsilon$. Yeh step kyun? Inverse determinant se divide karta hai. Jaise $\epsilon \to 0$, hum almost zero se divide karte hain — coordinates explode ho jaate hain. Yeh puri topic ki limiting/degenerate boundary hai.

2. Inverse. $C_{\epsilon }^{-1}=\frac{1}{\epsilon }\left[\begin{array}{cc}\epsilon & -1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& -1/\epsilon \\ 0& 1/\epsilon \end{array}\right].$ Yeh step kyun? $2\times 2$ rule apply karne par $1/\epsilon$ entries dikhti hain — translator ke numbers blow up hote hain jaise basis degenerate hoti hai.

3. $\epsilon =0.1$ ke saath $(1,1{)}^T$ par apply karo. $[v{]}_{C_{0.1}}=\left[\begin{array}{cc}1& -10\\ 0& 10\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}1-10\\ 10\end{array}\right]=\left[\begin{array}{c}-9\\ 10\end{array}\right].$ Yeh step kyun? Input ko exploded translator se paas karne par Figure 2 mein dikhne wale enormous coordinates milte hain.

Verify: $-9(1,0)+10(1,0.1)=(-9+10,1.0)=(1,1)$. ✓

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Example 10 — Cell J: real-world word problem

Forecast: Speed magnitude hai $\sqrt{8^2+6^2}=10$ m/s. Jo bhi coordinates milein, unki length bhi $10$ honi chahiye (axes rotate karne se drone ki speed nahi badalti). Ise apna sanity anchor maano.

1. Pilot ka basis $C=\{f,\ell \}$. $C=\left[\begin{array}{cc}0.6& -0.8\\ 0.8& 0.6\end{array}\right].$ Yeh step kyun? Columns, runway axes ko standard (East, North) coordinates mein represent karte hain.

2. Yeh ek rotation hai ⇒ $C^{-1}=C^T$ (orthogonal, jaise Ex 5). $C^T=\left[\begin{array}{cc}0.6& 0.8\\ -0.8& 0.6\end{array}\right].$ Yeh step kyun? $f,\ell$ perpendicular unit vectors hain, isliye koi messy inverse nahi — transpose kaafi hai, aur $P_{C\leftarrow E}=C^{-1}=C^T$.

3. Apply karo. $[v{]}_C=C^T\left[\begin{array}{c}8\\ 6\end{array}\right]=\left[\begin{array}{c}0.6\cdot 8+0.8\cdot 6\\ -0.8\cdot 8+0.6\cdot 6\end{array}\right]=\left[\begin{array}{c}4.8+4.8\\ -6.4+3.6\end{array}\right]=\left[\begin{array}{c}9.6\\ -2.8\end{array}\right].$ Yeh step kyun? Ground-frame velocity ko $C^T$ se paas karne par wohi velocity runway coordinates mein report hoti hai. Toh screen dikhayegi $9.6$ m/s aage aur $-2.8$ m/s left (yaani $2.8$ m/s right).

Verify (units + magnitude): $\sqrt{9.6^2+(-2.8{)}^2}=\sqrt{92.16+7.84}=\sqrt{100}=10$ m/s. ✓ Speed exactly preserved as forecast — drone sirf isliye tez nahi udta kyunki pilot runway axes use karta hai.

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Example 11 — Cell K: reflected basis, $\det <0$ (orientation flip)

Forecast: Rotation (Ex 5) mein $\det =+1$ rehta hai. Ek ruler ko flip karna orientation reverse karta hai, toh predict karo: kya $\det C$ positive hoga ya negative? Aur $v$ ke doosre coordinate ka kya hona chahiye?

1. $C$ ka matrix. $C=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$. Yeh step kyun? Columns, mirror basis vectors ko standard coords mein represent karte hain; $-1$ downward $y$-ruler encode karta hai.

2. Determinant — orientation check karo. $\det C=(1)(-1)-(0)(0)=-1<0$. Yeh step kyun? Negative determinant ek orientation-reversing (mirror) basis ka algebraic fingerprint hai — cell K isi case ke liye exist karta hai. Yeh abhi bhi non-zero hai, isliye $C^{-1}$ exist karta hai.

3. Inverse. Yeh reflection apna khud ka inverse hai: $C^{-1}=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=C$. Yeh step kyun? Do baar mirror karne par tum wapas start par aa jaate ho, isliye $C^2=I$ aur $C^{-1}=C$. Kyunki $B=E$ hai, $P_{C\leftarrow E}=C^{-1}=C$.

4. $(3,4{)}^T$ par apply karo. $[v{]}_C=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{c}3\\ 4\end{array}\right]=\left[\begin{array}{c}3\\ -4\end{array}\right].$ Yeh step kyun? Standard coords ko $C^{-1}$ se paas karne par woh flipped grid mein report hote hain — doosra coordinate sign change karta hai, exactly as forecast.

Verify: $3(1,0)+(-4)(0,-1)=(3,0+4)=(3,4)=v$. ✓ Arrow $(3,4)$ unchanged hai; sirf uski description ne apna doosra component flip kiya kyunki ruler doosri taraf point karta hai.

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Recall

Recall Kis direction mein inverse ki zaroorat NAHI hoti?

Non-standard $\to$ standard ($B\to E$): sirf $B$ se multiply karo, kyunki iske columns pehle se standard coordinates mein hain.

Recall Rotated basis ke liye

$C^{-1}=C^T$ kyun hota hai? Ek rotation ke columns perpendicular unit vectors hote hain (orthogonal matrix), isliye $C^{T}C=I$, hence $C^{-1}=C^T$ — koi fractions nahi chahiye.

Recall

$\det C$ ka sign kya batata hai? $\det >0$: naye axes rotation hain (orientation preserved). $\det <0$: ek reflection hai (orientation flipped, ek mirror). $\det \to 0$: basis degenerate ho jaati hai aur $C^{-1}$ blow up / cease to exist kar deta hai.

Recall Basis degenerate hone par kya galat hota hai?

$\det C\to 0$, isliye $C^{-1}$ (jo $\det$ se divide karta hai) blow up karta hai: coordinates explode hote hain aur errors $1/\det$ se amplify hoti hain.

Recall Same basis in aur out —

$P_{B\leftarrow B}$ kya hai? Identity $I$: kisi language ko apne aap mein translate karna kuch nahi badalti.

Connections

• Change of basis matrix — parent machine jinhe yeh examples exercise karte hain.
• Invertible Matrices — Ex 9 ka degenerate limit exactly non-invertibility hai; Ex 11 ka $\det <0$ abhi bhi invertible hai.
• Coordinate Vectors — yahan har input/output column ek hai.
• Linear Transformations — Ex 5, 10 & 11 passive view hain (axes move karte hain, arrow fixed).
• Basis and Dimension — kyun ek independent spanning set zaroori hai.
• Similar Matrices · Eigenvectors and Diagonalization — jahan yeh translators poore maps par reuse hote hain.