Question bank — Coordinate vectors — change of basis

False. Arrow fixed hai, lekin $[\mathbf{v}{]}_{\mathcal{B}}$ chosen basis ke relative ek description hai; basis badlo aur numbers ki list badal jaati hai chahe $\mathbf{v}$ move na ho — exactly jaisa $\left[\begin{array}{c}0\\ 2\end{array}\right]$ vs $\left[\begin{array}{c}1\\ 1\end{array}\right]$ Figure s01 mein.

False. Ek single vector do alag bases mein coincidentally identical coordinates rakh sakta hai (jaise koi bhi $\mathbf{v}$ jo map $P$ se fix ho); sabhi $\mathbf{v}$ ke liye equal coordinates $P=I$ force karenge aur isliye $\mathcal{B}=\mathcal{C}$.

True. Iske columns $[{\mathbf{b}}_j{]}_{\mathcal{C}}$ hain, ek linearly independent set ke coordinate images, to ye independent rehte hain — ek invertible matrix jiska inverse $\underset{\mathcal{B}\leftarrow \mathcal{C}}{P}$ hai.

False. Orthonormality $P$ ko orthogonal banati hai ($P^{-1}=P^{\top }$ $\mathbb{R}$ pe), symmetric nahi; ek rotation matrix orthogonal hota hai lekin symmetric nahi.

False. Ye inverses hain, isliye unke determinants reciprocals hain: $\det (P^{-1})=1/\det (P)$; sirf special case $\det P=\pm 1$ mein equal.

True. $\mathbf{v}=\sum c_i{\mathbf{b}}_i$ mein coefficients $c_i$ arbitrary scalars hain; positivity force nahi hoti — parent note ke Example 3 ne $[p{]}_{\mathcal{C}}$ diya tha jisme $-1$ tha.

True. Similarity exactly ye statement hai ki $A$ aur $A^{\prime }$ do bases mein same linear map represent karte hain, aur $P$ un bases ko relate karne wala change-of-basis matrix hai.

True. $\mathbf{u}+\alpha \mathbf{w}$ ko $\mathcal{B}$ mein expand karne se coefficients termwise add aur scale hote hain, isliye $[\mathbf{u}+\alpha \mathbf{w}{]}_{\mathcal{B}}=[\mathbf{u}{]}_{\mathcal{B}}+\alpha [\mathbf{w}{]}_{\mathcal{B}}$.

Galat. Machine ko trace karo: $P$ ek $\mathcal{B}$-label $\left[\begin{array}{c}{c}_1\\ ⋮\end{array}\right]$ receive karta hai aur iske columns un $c_i$ se scale hote hain; isliye har column woh hona chahiye jo ek $\mathcal{B}$-vector target mein kaisa dikhta hai, yani $[{\mathbf{b}}_j{]}_{\mathcal{C}}$. ${\mathbf{c}}_j$ use karna reverse matrix banayega.

Galat order. Input ko walk through karo: $\mathcal{B}$-label se shuru karo; $B$ use multiply karta hai standard numbers dene ke liye ($\mathcal{B}\to \mathcal{E}$); phir $C^{-1}$ un standard numbers ko $\mathcal{C}$-label mein convert karta hai ($\mathcal{E}\to \mathcal{C}$). Right-to-left apply karne par ye $C^{-1}B$ hai; inner "standard" step cancel ho jata hai. $B^{-1}C$, $\mathcal{C}\to \mathcal{E}\to \mathcal{B}$ chain karta hai — opposite direction.

Galat. Sirf standard basis mein hi vector ki entries uske coordinates ke equal hoti hain; ek tilted $\mathcal{B}$ ke liye (Figure s01) arrow ko tilted steps ki ek alag count chahiye, isliye tumhe $B[\mathbf{v}{]}_{\mathcal{B}}=\mathbf{v}$ solve karna hoga, jo $[\mathbf{v}{]}_{\mathcal{B}}=B^{-1}\mathbf{v}$ deta hai.

Galat. $P$ numbers ki list move karta hai, arrow ko nahi. Figures s01 aur s02 compare karo: arrow ek identical spot pe baith a hai; sirf uska label $\left[\begin{array}{c}0\\ 2\end{array}\right]\to \left[\begin{array}{c}1\\ 1\end{array}\right]$ badlata hai.

Galat. Ek baar ordered basis fix karne ke baad, koi bhi finite-dimensional space (jaise $P_2$) ke coordinate vectors ${\mathbb{R}}^n$ mein hote hain, isliye change of basis ek ordinary matrix hai — Example 3 ne $\{1,t,t^2\}$ use kiya tha.

Galat. Ek basis ordered hoti hai; ${\mathbf{b}}_1,{\mathbf{b}}_2$ ko permute karne se entries $c_1,c_2$ correspondingly permute hoti hain — ek permutation change-of-basis matrix ke through ek alag coordinate vector.

Galat. $P$ purely do bases se build hota hai; wahi $P$ har $\mathbf{v}$ ke liye kaam karta hai, jo exactly woh reason hai ki ise matrix kehna deserve karta hai.

Independence guarantee karta hai ki expansion $\mathbf{v}=\sum c_i{\mathbf{b}}_i$ unique hai; do alag expansions subtract hoke $0$ ke equal ek nontrivial combination denge, jo independence ko contradict karega.

Kyunki $\mathbf{v}=\sum c_i{\mathbf{b}}_i$ aur $\mathcal{C}$-coordinates lena (ek linear map) deta hai $[\mathbf{v}{]}_{\mathcal{C}}=\sum c_i[{\mathbf{b}}_i{]}_{\mathcal{C}}$ — ek matrix jiske columns $[{\mathbf{b}}_i{]}_{\mathcal{C}}$ hain, $c_i$ se multiply hoke.

$\mathcal{C}$ mein translate karna aur immediately $\mathcal{B}$ mein wapas aana har coordinate vector ko untouched chhod deta hai, isliye composite $I$ hai — matlab do matrices inverses hain.

$\mathcal{E}$ mein ek vector ke coordinates uski literal entries hoti hain, isliye $\underset{\mathcal{E}\leftarrow \mathcal{B}}{P}=B$ (basis vectors as columns) ko koi solving nahi chahiye — mushkil kaam sirf doosri taraf jaane ke liye invert karna hai.

Tum $\mathcal{B}\to \mathcal{C}$ directly chain nahi kar sakte; tum standard basis ke through route karte ho: $B$, $\mathcal{B}$-numbers ko standard mein le jaata hai, $C^{-1}$ standard ko $\mathcal{C}$-numbers mein le jaata hai, jo $C^{-1}B$ deta hai.

Dono coordinate lists ko same $\mathbf{v}$ reconstruct karna chahiye; $B[\mathbf{v}{]}_{\mathcal{B}}=\mathbf{v}=C[\mathbf{v}{]}_{\mathcal{C}}$ equate karo aur $[\mathbf{v}{]}_{\mathcal{C}}$ solve karo to exactly $P=C^{-1}B$ milta hai.

Ye har basis mein zero column $\left[\begin{array}{c}0\\ ⋮\\ 0\end{array}\right]$ hai, kyunki ek independent set se $0$ dene wala ek-maatra combination all-zero coefficients ka hai.

Identity $I$: ek vector ko us basis mein re-describe karna jisme woh pehle se hai, kuch nahi badlata, isliye har column $[{\mathbf{b}}_j{]}_{\mathcal{B}}$ ek standard unit vector hai.

Haan — agar ek "basis" actually dependent hai, to $B$ ya $C$ singular hai, $C^{-1}$ exist nahi karta, aur koi change-of-basis matrix define nahi hoti; independence ek hard prerequisite hai.

$P=\frac{1}{k}I$: lambe basis rulers matlab unki kam zaroorat, isliye har coordinate $1/k$ se shrink hoti hai — ek degenerate lekin instructive diagonal case.

Tumhe standard unit column $e_2$ milta hai (slot 2 mein $1$, baki zeros), kyunki ${\mathbf{b}}_2=0\cdot {\mathbf{b}}_1+1\cdot {\mathbf{b}}_2+\cdots$.

Ek single nonzero scalar mein: $\mathcal{B}=\{b\}$, $\mathcal{C}=\{c\}$ aur $b=\lambda c$ ke saath, "matrix" $1\times 1$ number $\lambda$ hai, sirf ek coordinate ki rescaling.

Nahi. Ek complex vector space pe inner product conjugate-symmetric hai, isliye orthonormal-to-orthonormal change of basis unitary hai: $P^{-1}=P^{\ast }={\overline{P}}^{\top }$, yaani conjugate-transpose, plain transpose nahi. Real orthogonal case sirf woh special case hai jahan entries real hain aur conjugation kuch nahi karta.

Numerically haan — wahi array $M$ ko "fixed vector ko re-describe karo" (change of basis) ya "vector ko actually move karo" (ek transformation) ki tarah padha ja sakta hai; interpretation, numbers nahi, alag hota hai.