Exercises — Coordinate vectors — change of basis

KYA: coordinate vector ek recipe hai. Iska matlab hai: ${\mathbf{b}}_1$ ki $5$ copies lo aur ${\mathbf{b}}_2$ ki $-2$ copies lo. KYO: definition se hi $\mathbf{v}=c_1{\mathbf{b}}_1+c_2{\mathbf{b}}_2$. $\mathbf{v}=5\left[\begin{array}{c}2\\ 0\end{array}\right]+(-2)\left[\begin{array}{c}0\\ 3\end{array}\right]=\left[\begin{array}{c}10\\ 0\end{array}\right]+\left[\begin{array}{c}0\\ -6\end{array}\right]=\left[\begin{array}{c}10\\ -6\end{array}\right].$ Equivalently $\mathbf{v}=B[\mathbf{v}{]}_{\mathcal{B}}$ jahan $B=\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]$. Same answer: $\boxed{\mathbf{v}=\left[\begin{array}{c}10\\ -6\end{array}\right]}$.

KYA: hum jaanna chahte hain ki kitne ${\mathbf{b}}_1$'s aur ${\mathbf{b}}_2$'s milkar $\mathbf{w}$ banate hain. KYO inspection kaam karta hai yahan: ${\mathbf{b}}_1$ sirf $x$-slot ko touch karta hai, ${\mathbf{b}}_2$ sirf $y$-slot ko (ek diagonal basis). Toh ye dono interfere nahi karte.

• $x$: $2c_1=8\Rightarrow c_1=4$.
• $y$: $3c_2=9\Rightarrow c_2=3$. $\boxed{[\mathbf{w}{]}_{\mathcal{B}}=\left[\begin{array}{c}4\\ 3\end{array}\right]}.$ Check: $4{\mathbf{b}}_1+3{\mathbf{b}}_2=\left[\begin{array}{c}8\\ 9\end{array}\right]$. ✓

KYA: $B[\mathbf{v}{]}_{\mathcal{B}}=\mathbf{v}$ solve karo, matlab $[\mathbf{v}{]}_{\mathcal{B}}=B^{-1}\mathbf{v}$. KYO $B^{-1}$ aur $B$ nahi: $B$ recipe→plain karta hai; humhare paas plain hai aur recipe chahiye, toh hum ise ulta chalate hain inverse se. $B$ invertible hai kyunki uske columns linearly independent hain (woh ek doosre ke multiples nahi hain). $B=\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right],\det B=1\cdot 1-3\cdot 2=-5.$ $B^{-1}=\frac{1}{-5}\left[\begin{array}{cc}1& -3\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{5}& \frac{3}{5}\\ \frac{2}{5}& -\frac{1}{5}\end{array}\right].$ $[\mathbf{v}{]}_{\mathcal{B}}=B^{-1}\left[\begin{array}{c}5\\ 5\end{array}\right]=\left[\begin{array}{c}-1+3\\ 2-1\end{array}\right]=\left[\begin{array}{c}2\\ 1\end{array}\right].$ Check: $2{\mathbf{b}}_1+1{\mathbf{b}}_2=\left[\begin{array}{c}2\\ 4\end{array}\right]+\left[\begin{array}{c}3\\ 1\end{array}\right]=\left[\begin{array}{c}5\\ 5\end{array}\right]$. ✓ $\boxed{[\mathbf{v}{]}_{\mathcal{B}}=\left[\begin{array}{c}2\\ 1\end{array}\right]}$.

Neeche figure mein dikhaya hai ki $\mathbf{v}$ tak pahunchne ke liye ${\mathbf{b}}_1$ ke $2$ steps aur ${\mathbf{b}}_2$ ka $1$ step chala.

KYA: $\underset{\mathcal{C}\leftarrow \mathcal{B}}{P}=C^{-1}B$ use karo. KYO yeh order: right to left padho — $B$ $\mathcal{B}$-numbers ko standard numbers mein badalta hai, phir $C^{-1}$ standard ko $\mathcal{C}$-numbers mein badalta hai. Andar ka "standard" cancel ho jaata hai, bas $\mathcal{B}\to \mathcal{C}$ bachta hai. $C=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right],C^{-1}=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right].$ $\underset{\mathcal{C}\leftarrow \mathcal{B}}{P}=C^{-1}B=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}1-2& 3-1\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right].$ Ab $[\mathbf{v}{]}_{\mathcal{B}}=\left[\begin{array}{c}2\\ 1\end{array}\right]$ par apply karo: $[\mathbf{v}{]}_{\mathcal{C}}=\left[\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right]\left[\begin{array}{c}2\\ 1\end{array}\right]=\left[\begin{array}{c}-2+2\\ 4+1\end{array}\right]=\left[\begin{array}{c}0\\ 5\end{array}\right].$ Plain vector $\mathbf{v}=\left[\begin{array}{c}5\\ 5\end{array}\right]$ se sanity check: $0{\mathbf{c}}_1+5{\mathbf{c}}_2=5\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}5\\ 5\end{array}\right]$. ✓ $\boxed{[\mathbf{v}{]}_{\mathcal{C}}=\left[\begin{array}{c}0\\ 5\end{array}\right]}$.

KYA: $a,b,c$ nikalo jahan $a\cdot 1+b(1-t)+c(1-t{)}^2=3-4t+t^2$. KYO top-down match karo: sabse highest-degree basis vector $(1-t{)}^2$ sirf wahi hai jo $t^2$ carry karta hai, toh uska coefficient pehle force hota hai; phir $t$; phir constant. Ye triangular hai — har step mein ek unknown. $(1-t{)}^2=1-2t+t^2$ expand karo. Collect karo: $a+b+c(\text{const}),(-b-2c)t,c{t}^2.$

• $t^2$: $c=1$.
• $t$: $-b-2c=-4\Rightarrow -b-2=-4\Rightarrow b=2$.
• const: $a+b+c=3\Rightarrow a+2+1=3\Rightarrow a=0$. $\boxed{[p{]}_{\mathcal{C}}=\left[\begin{array}{c}0\\ 2\\ 1\end{array}\right]}.$ Check: $0\cdot 1+2(1-t)+1(1-t{)}^2=2-2t+1-2t+t^2=3-4t+t^2$. ✓

KYA: har column ek purana basis vector hai $\mathcal{C}$ mein likha hua. Toh $1,t,t^2$ ko $\{1,1-t,(1-t{)}^2\}$ se express karo. KYO: parent formula se, $\underset{\mathcal{C}\leftarrow \mathcal{B}}{P}$ ka column $j$ hai $[{\mathbf{b}}_j{]}_{\mathcal{C}}$.

• $1=1\cdot 1$, toh $[1{]}_{\mathcal{C}}=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$.
• $t$: solve karo $a+b(1-t)+c(1-t{)}^2=t$. Top-down: $t^2$ se $c=0$; $t$ se $-b=1\Rightarrow b=-1$; const $a+b=0\Rightarrow a=1$. Toh $[t{]}_{\mathcal{C}}=\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]$.
• $t^2$: $t^2=(1-t{)}^2-1+2t$. Pichli line se $2t$ ke liye: $2t=2(1-t)-2+...$ — seedha solve karna zyada clean hai. Set karo $a+b(1-t)+c(1-t{)}^2=t^2$: $t^2$ se $c=1$; $t$ se $-b-2c=0\Rightarrow b=-2$; const $a+b+c=0\Rightarrow a=1$. Toh $[t^2{]}_{\mathcal{C}}=\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]$. $\underset{\mathcal{C}\leftarrow \mathcal{B}}{P}=\left[\begin{array}{ccc}1& 1& 1\\ 0& -1& -2\\ 0& 0& 1\end{array}\right].$ $[p{]}_{\mathcal{B}}=\left[\begin{array}{c}3\\ -4\\ 1\end{array}\right]$ par apply karo: $\left[\begin{array}{ccc}1& 1& 1\\ 0& -1& -2\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}3\\ -4\\ 1\end{array}\right]=\left[\begin{array}{c}3-4+1\\ 4-2\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 1\end{array}\right].$ 3.1 se match karta hai. ✓ Dhyaan do ki matrix upper-triangular hai — exactly isliye kyunki $\mathcal{C}$ ka $j$-th vector degree $j-1$ ka hai, toh low-degree $\mathbf{b}$'s ko high-degree $\mathbf{c}$'s ki kabhi zaroorat nahi padti.

KYA: $\mathcal{B}\to \mathcal{C}$ change karo, phir $\mathcal{C}\to \mathcal{D}$, aur check karo ki ye $\mathcal{B}\to \mathcal{D}$ seedha jaane ke barabar hai. KYO ye hona chahiye: translating dictionaries compose karte hain. Numerically har ek $(\text{new}{)}^{-1}(\text{old})$ hai, aur beech wala basis cancel ho jaata hai: $(D^{-1}C)(C^{-1}B)=D^{-1}B$. $B=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right],C=\left[\begin{array}{cc}1& 1\\ 0& 2\end{array}\right],D=\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right].$ $C^{-1}=\left[\begin{array}{cc}1& -\frac{1}{2}\\ 0& \frac{1}{2}\end{array}\right],D^{-1}=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& 1\end{array}\right].$ $\underset{\mathcal{C}\leftarrow \mathcal{B}}{P}=C^{-1}B=\left[\begin{array}{cc}1& -\frac{1}{2}\\ 0& \frac{1}{2}\end{array}\right]\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& \frac{3}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right].$ $\underset{\mathcal{D}\leftarrow \mathcal{C}}{P}=D^{-1}C=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ 0& 2\end{array}\right].$ Product: $\underset{\mathcal{D}\leftarrow \mathcal{C}}{P}\underset{\mathcal{C}\leftarrow \mathcal{B}}{P}=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ 0& 2\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& \frac{3}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ 1& -1\end{array}\right].$ Direct route: $\underset{\mathcal{D}\leftarrow \mathcal{B}}{P}=D^{-1}B=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ 1& -1\end{array}\right].$ Identical. ✓ $\boxed{\underset{\mathcal{D}\leftarrow \mathcal{B}}{P}=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ 1& -1\end{array}\right]}$.

KYA: $T$ ko $\mathcal{B}$-numbers se apply karne ke liye, pehle $\mathcal{B}\to$ standard ($B$) convert karo, $A$ apply karo, phir standard$\to \mathcal{B}$ ($B^{-1}$) convert karo. Right to left padho: $B^{-1}AB$. KYO ye ek similarity hai: $[T{]}_{\mathcal{B}}$ aur $A$ same map ko alag-alag rulers se describe karte hain, toh ye change-of-basis matrix $B$ ke zariye similar hain. $B=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right],B^{-1}=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right].$ $AB=\left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}2& 3\\ 0& 3\end{array}\right].$ $[T{]}_{\mathcal{B}}=B^{-1}(AB)=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}2& 3\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right].$ $\boxed{[T{]}_{\mathcal{B}}=\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]}$ — diagonal! Naya basis eigenvector basis nikla, toh $T$ sirf har axis ko stretch karta hai. (Trace $=5$ aur determinant $=6$ $A$ se preserved hain, jaise similarity guarantee karta hai.)

KYA/KYO, step by step.

1. Kisi bhi $\mathbf{v}\in {\mathbb{R}}^n$ ke liye, uska plain (standard) form $\mathcal{B}$-coordinates se recover hota hai: $\mathbf{v}=B[\mathbf{v}{]}_{\mathcal{B}}$. KYO: $B$ ke columns ${\mathbf{b}}_j$ hain, toh $B[\mathbf{v}{]}_{\mathcal{B}}=\sum c_j{\mathbf{b}}_j=\mathbf{v}$, ye coordinate vector ki definition hai.
2. Isi tarah $\mathbf{v}=C[\mathbf{v}{]}_{\mathcal{C}}$.
3. Dono barabar set karo: $C[\mathbf{v}{]}_{\mathcal{C}}=B[\mathbf{v}{]}_{\mathcal{B}}$. KYO hum kar sakte hain: dono ek hi invariant arrow $\mathbf{v}$ express karte hain.
4. Left mein $C^{-1}$ se multiply karo: $[\mathbf{v}{]}_{\mathcal{C}}=C^{-1}B[\mathbf{v}{]}_{\mathcal{B}}$.
5. Ye har $\mathbf{v}$ ke liye sach hai, toh jo matrix $[\mathbf{v}{]}_{\mathcal{B}}\mapsto [\mathbf{v}{]}_{\mathcal{C}}$ bhejti hai woh exactly $C^{-1}B$ hai. Woh matrix definition se $\underset{\mathcal{C}\leftarrow \mathcal{B}}{P}$ hai. $■$

$C$ invertible kyun hai: uske columns ek basis hain, isliye linearly independent hain, isliye $C$ full rank ka hai, isliye invertible hai. $B$ ke liye bhi same.

(a) Andar aur bahar same ruler hone se kuch nahi badalta, toh $\underset{\mathcal{B}\leftarrow \mathcal{B}}{P}=I$. Formula-check: $B^{-1}B=I$. ✓

(b) Standard basis ke liye $E=I$, toh $C^{-1}=I$ aur $\underset{\mathcal{E}\leftarrow \mathcal{B}}{P}=I^{-1}B=B$. Matlab: $\mathcal{B}$-numbers ko ordinary numbers ki tarah padhne ke liye $B$ se multiply karo. Ye parent note ka "standard-basis shortcut" hai.

(c) ${\mathbf{u}}_2=2{\mathbf{u}}_1$, toh dono vectors linearly dependent hain — ye sirf ek line span karte hain, poora ${\mathbb{R}}^2$ nahi. Consequences jo machinery ko todte hain:

• Basis nahi hai: us line se bahar ke vectors ka koi expansion hi nahi, toh $[\mathbf{v}{]}_{\mathcal{U}}$ exist nahi kar sakta.
• Uniqueness nahi: us line par ke vectors ke infinitely many expansions hain, toh coordinate vector well-defined nahi hai.
• $U$ invertible nahi: $\det \left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]=4-4=0$, toh $U^{-1}$ (jo $C^{-1}B$ ke liye chahiye) exist nahi karta.

Teeno ek hi fact ko teen tarah se dekh rahe hain: ek valid basis linearly independent honi chahiye. $■$

• Coordinates se plain vector? ::: $\mathbf{v}=B[\mathbf{v}{]}_{\mathcal{B}}$.
• Plain vector se coordinates? ::: $[\mathbf{v}{]}_{\mathcal{B}}=B^{-1}\mathbf{v}$.
• ${\mathbb{R}}^n$ mein $\mathcal{B}\to \mathcal{C}$ matrix? ::: $C^{-1}B$.
• Chain $\mathcal{B}\to \mathcal{C}\to \mathcal{D}$? ::: multiply karo $\underset{\mathcal{D}\leftarrow \mathcal{C}}{P}\underset{\mathcal{C}\leftarrow \mathcal{B}}{P}=\underset{\mathcal{D}\leftarrow \mathcal{B}}{P}$.
• Naye basis mein map $T$? ::: $[T{]}_{\mathcal{B}}=B^{-1}AB$ (ek similarity hai).
• Basis independent kyun honi chahiye? ::: warna coordinates exist nahi karte ya unique nahi hote, aur $B$ invertible nahi hota.