Question bank — Change of basis matrix

Jhooth. Arrow $v$ kabhi nahi hiltaa; sirf uski coordinate description $B$-language se $C$-language mein badlti hai. Yeh passive view hai — linear transformation ke bilkul ulta, jo actually arrows ko relocate karta hai.

Sach. Basis by definition ek linearly independent spanning set hoti hai, isliye dono $B$ aur $C$ (basis vectors ki matrices) invertible hain, aur $P_{C\leftarrow B}=C^{-1}B$ invertibles ka product hai — isliye invertible hai.

Jhooth. Columns purane basis vectors $b_j$ hain jo naye coordinates mein likhe gaye hain, yaani $[b_j{]}_C$. Output $C$-land mein rehta hai, isliye har column pehle se hi ek $C$-coordinate vector hona chahiye.

Sach. Ek language ko khud mein translate karne se kuch nahi badlta: har $[b_j{]}_C=[b_j{]}_B=e_j$, isliye columns $e_1,\dots ,e_n$ hain — identity.

Jhooth. Ye inverses hain, same matrix flip nahi: $P_{B\leftarrow C}=(P_{C\leftarrow B}{)}^{-1}$. Translation ko reverse karna matlab use undo karna hai, jo inverse operation hai.

Jhooth — woh direction easy wala hai: $[v{]}_E=B[v{]}_B$, bas $B$ se multiply karo jiske columns pehle se standard coords mein hain. Inverse ($B^{-1}$) ulti trip $E\to B$ ke liye chahiye.

Sach. Diagonal $P$ ka matlab hai ki $C$ mein express kiya gaya har $b_j$ sirf $c_j$ use karta hai: $b_j=p_{jj}{c}_j$. Toh har purana vector corresponding naye ka scalar multiple hai — same directions, rescaled.

Jhooth. Ek fixed basis ke relative coordinates unique hote hain, aur alag bases ek fixed arrow ko alag tarah se read karti hain. Alag bases mein same coordinates alag arrows describe karenge (jab tak bases $v$ banane mein ittifaq na rakhein).

Sach. Kyunki $P$ hamesha invertible hai, uska determinant nonzero hai. Zero determinant ka matlab hoga "basis" vectors dependent hain — jo ki basis hi nahi hai.

Error: formula hai $P_{C\leftarrow B}=C^{-1}B$. Reasoning: $B$-coords $\to$ standard (multiply by $B$), phir standard $\to C$-coords (multiply by $C^{-1}$); matrices right-to-left apply hoti hain, isliye $C^{-1}B$ hai.

Error: $P_{C\leftarrow B}$ $B$-coords leta hai aur $C$-coords deta hai. $C$-coords input leke $B$-coords output dene ke liye tumhe $P_{B\leftarrow C}=(P_{C\leftarrow B}{)}^{-1}$ chahiye. Subscript arrow ko "input on the right" ki tarah padho.

Error: $[b_1{]}_C\mid \cdots \mid [b_n{]}_C$ stack karo — naye language mein likhe purane $b_j$. Test: $e_j$ feed karne par $[b_j{]}_C$ milna chahiye, jo tabhi kaam karta hai jab column $j$ ho $[b_j{]}_C$.

Error: arrow — aur uski asli length — fixed hai. Coordinates sirf ek description hain; jab tak naya basis orthonormal na ho, coordinate numbers length ko reflect bhi nahi karte. Physically kuch nahi hila.

Setup mein error hai: $P_{C\leftarrow E}=C^{-1}E=C^{-1}$ hai, $C$ nahi. $P_{E\leftarrow C}=C$ hai jo $C$ ke vectors ki matrix ke barabar hai. Subscript direction flip tha.

Error: ye similar hain, generally equal nahi. Same underlying linear map alag basis mein dekha gaya — same eigenvalues/determinant/trace, lekin alag entries.

Kyunki basis mein expansion unique hai: vectors ko add karne se unke coordinate columns add hote hain aur vector ko scale karne se uska column scale hota hai. Yahi linearity hai jo translation ko ek single matrix mein collapse karne deti hai.

Column $e_j=[b_j{]}_B$ feed karo: product column $j$ select karta hai, $[b_j{]}_C$ return karta hai. Toh $P$ har building block ko sahi tarike se map karta hai, aur linearity se har combination ko bhi sahi tarike se map karta hai.

Kyunki koi bhi basis woh reference frame nahi hai jismein ${\mathbb{R}}^n$ built hai — standard basis $E$ hai. Hum $E$ se route karte hain: $B$ ko standard mein decode karo ($B$), phir standard ko $C$ mein encode karo ($C^{-1}$).

Diagonalizing eigenvectors ka basis choose karta hai. Us basis mein map sirf har axis ko stretch karta hai, isliye $P^{-1}AP=D$ diagonal hai — same map, apni sabse natural language mein describe kiya gaya.

Same space $V$ ke dono bases mein elements ki same number hoti hai — dimension $n$. Toh $P$, $n$ coordinates ko $n$ coordinates mein map karta hai, jo $n\times n$ shape force karta hai.

Kyunki "translate phir wapas translate" ko undo karne se identity milni chahiye: $P_{B\leftarrow C}{P}_{C\leftarrow B}=I$, jo inverse ki definition hai. Transpose sirf orthogonal bases ke liye inverse se coincide karta hai — ek special case, rule nahi.

Yeh identity $I$ hoga: $E^{-1}E=I$. Standard coordinates ko standard coordinates mein translate karne se kuch nahi badlta.

Nahi. Zero column ka matlab hoga koi $[b_j{]}_C=0$, yaani $b_j=0$ — lekin basis mein zero vector nahi ho sakta. Isliye har column nonzero hai.

Toh woh basis nahi hai — set dependent hai, matrix $B$ singular hai ($\det =0$), aur $C^{-1}B$ ya $B^{-1}$ exist nahi karta. Change of basis undefined hai; invertibility guarantee toot jaati hai.

Ek $1\times 1$ single nonzero number. Agar $B=\{b\}$ aur $C=\{c\}$ aur $b=kc$, toh $P_{C\leftarrow B}=[k]$ — ek coordinate ka pure rescaling, aur uska inverse $[1/k]$ hai.

Zaroor nahi. $P_{C\leftarrow B}=C^{-1}B$; chahe $C$ orthonormal ho ($C^{-1}=C^T$), product tabhi orthogonal hoga jab $B$ bhi orthonormal ho. Ek orthonormal side kaafi nahi hai.

Nahi — $P\cdot 0=0$ kisi bhi matrix ke liye, toh $[v{]}_C=0$ bhi hoga. Zero arrow ke har basis mein zero coordinates hote hain; translate karne ke liye kuch hai hi nahi.