Exercises — Inner product spaces — dot product generalization

Teeno axioms ko order mein check karo.

• Symmetry: $⟨\mathbf{u},\mathbf{v}⟩=u_1{v}_1+u_2{v}_2$ aur $⟨\mathbf{v},\mathbf{u}⟩=v_1{u}_1+v_2{u}_2$. Numbers kisi bhi order mein multiply hote hain, toh ye equal hain. ✓
• Linearity in slot 1: $⟨c\mathbf{u}+\mathbf{w},\mathbf{v}⟩=(c{u}_1+w_1)v_1+(c{u}_2+w_2)v_2=c(u_1{v}_1+u_2{v}_2)+(w_1{v}_1+w_2{v}_2)=c⟨\mathbf{u},\mathbf{v}⟩+⟨\mathbf{w},\mathbf{v}⟩$. ✓
• Positive-definite: $⟨\mathbf{v},\mathbf{v}⟩=v_1^2+v_2^2\ge 0$, aur yeh $0$ sirf tab hoga jab $v_1=v_2=0$, yaani $\mathbf{v}=0$. ✓

Teeno hold karte hain, toh haan, yeh ek inner product hai.

Symmetry ✓ aur linearity ✓ (L1.1 jaisi hi algebra, bas minus sign ke saath). Lekin positivity $\mathbf{v}=(0,1)$ se check karo: $⟨(0,1),(0,1)⟩=0^2-1^2=-1<0.$ Ek "length-squared" negative aaya. Positive-definiteness fail ho gayi, toh yeh inner product nahi hai.

Symmetry: $\mathbf{u},\mathbf{v}$ swap karne par $v_1{u}_2+v_2{u}_1$ milta hai — same cheez, ✓. Linearity ✓. Positivity: $⟨\mathbf{v},\mathbf{v}⟩=v_1{v}_2+v_2{v}_1=2v_1{v}_2$. $\mathbf{v}=(1,-1)$ lo: $2(1)(-1)=-2<0.$ Inner product nahi — positivity phir se fail.

Inner product se norm: $\Vert \mathbf{v}\Vert =\sqrt{⟨\mathbf{v},\mathbf{v}⟩}$ (dekho Norms and Distance). $⟨(1,1),(1,1)⟩=3(1)(1)+2(1)(1)=3+2=5,\Vert (1,1)\Vert =\sqrt{5}.$

Distance difference ki norm hoti hai: $d(\mathbf{u},\mathbf{v})=\Vert \mathbf{u}-\mathbf{v}\Vert$. $\mathbf{u}-\mathbf{v}=(2-0,1-(-1))=(2,2).$ $⟨(2,2),(2,2)⟩=3(4)+2(4)=12+8=20,d=\sqrt{20}=2\sqrt{5}.$

$⟨x,x⟩={\int }_0^1{x}^{2}dx=[\frac{x^3}{3}{]}_0^1=\frac{1}{3},\Vert x\Vert =\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}.$

Angle formula: $\cos \theta =\frac{⟨\mathbf{u},\mathbf{v}⟩}{\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert }$. $⟨\mathbf{u},\mathbf{v}⟩=1\cdot 1+0\cdot 1=1,\Vert \mathbf{u}\Vert =1,\Vert \mathbf{v}\Vert =\sqrt{2}.$ $\cos \theta =\frac{1}{1\cdot \sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow \theta =45^{\circ }.$ Figure s01 yeh dikhata hai: $\mathbf{u}=(1,0)$ x-axis ke saath flat draw kiya hai (lavender), $\mathbf{v}=(1,1)$ diagonal par upar-daayein point karta hai (coral), aur dono ke beech mint arc $45^{\circ }$ ka opening hai. Arrow $\mathbf{v}$ har ek unit daayein ke liye ek unit upar jaata hai, toh corner exactly aadha karta hai — woh aadha-right-angle hi $45^{\circ }$ hai, aur $\cos 45^{\circ }=1/\sqrt{2}$ unit-scaled diagonal ki horizontal shadow hai.

$⟨\mathbf{u},\mathbf{v}⟩=3(1)(1)+2(0)(1)=3.$ $\Vert \mathbf{u}\Vert =\sqrt{3(1)+2(0)}=\sqrt{3},\Vert \mathbf{v}\Vert =\sqrt{3(1)+2(1)}=\sqrt{5}.$ $\cos \theta =\frac{3}{\sqrt{3}\cdot \sqrt{5}}=\frac{3}{\sqrt{15}}=\sqrt{\frac{9}{15}}=\sqrt{\frac{3}{5}}\approx 0.7746\Rightarrow \theta \approx 39.2^{\circ }.$ Alag kyun hai? Weights axes ko stretch karte hain, toh "angle" ek alag ruler se measure hota hai. Paper par same arrows, lekin geometry alag. Figure s02 yeh dikhata hai: wahi arrows $\mathbf{u}=(1,0)$ (lavender) aur $\mathbf{v}=(1,1)$ (coral) ek baar draw hain, lekin corner mein do arcs hain — mint arc jis par likha hai "dot: $45^{\circ }$" (L3.1 ka ordinary ruler) aur uske andar tucked hua chota butter-coloured arc jis par likha hai "weighted: $39.2^{\circ }$". Kyunki $x$-part ka weight ($3$) $y$-part ke weight ($2$) se bada hai, weighted ruler horizontal agreement zyada count karta hai, toh dono arrows "zyada aligned" lagte hain aur angle $45^{\circ }$ se ghatkar lagbhag $39.2^{\circ }$ ho jaata hai.

Pehle cross term: $⟨p,q⟩=p(0)q(0)+p(1)q(1)=(0)(1)+(1)(0)=0.$ Kyunki $⟨p,q⟩=0$, woh is inner product ke under orthogonal hain (dekho Orthogonality and Gram-Schmidt). Ab norms, taaki $\cos \theta$ mein division valid ho. Har polynomial ko $x=0$ aur $x=1$ par evaluate karo: $⟨p,p⟩=p(0{)}^2+p(1{)}^2=0^2+1^2=1\Rightarrow \Vert p\Vert =1,$ $⟨q,q⟩=q(0{)}^2+q(1{)}^2=1^2+0^2=1\Rightarrow \Vert q\Vert =1.$ Dono norms nonzero hain, toh $\cos \theta =\frac{⟨p,q⟩}{\Vert p\Vert \Vert q\Vert }=\frac{0}{1\cdot 1}=0,$ jo safely $[-1,1]$ ke andar hai, toh Cauchy–Schwarz hold karta hai ($\theta =90^{\circ }$ is inner product mein).

$\Vert \mathbf{w}{\Vert }^2=⟨\mathbf{w},\mathbf{w}⟩$ se shuru karo aur dono slots mein linearity se expand karo (FOIL) — har cross term explicitly dikhate hue: $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2=⟨\mathbf{u}+\mathbf{v},\mathbf{u}+\mathbf{v}⟩=⟨\mathbf{u},\mathbf{u}⟩+⟨\mathbf{u},\mathbf{v}⟩+⟨\mathbf{v},\mathbf{u}⟩+⟨\mathbf{v},\mathbf{v}⟩.$ Ab symmetry lagao ($⟨\mathbf{v},\mathbf{u}⟩=⟨\mathbf{u},\mathbf{v}⟩$): dono middle terms equal hain, toh combine hokar $2⟨\mathbf{u},\mathbf{v}⟩$ ban jaate hain: $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2+2⟨\mathbf{u},\mathbf{v}⟩+\Vert \mathbf{v}{\Vert }^2.$ Usi tarah, difference ke liye (dono cross terms $⟨\mathbf{u},-\mathbf{v}⟩$ aur $⟨-\mathbf{v},\mathbf{u}⟩$ hain, dono $=-⟨\mathbf{u},\mathbf{v}⟩$): $\Vert \mathbf{u}-\mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2-2⟨\mathbf{u},\mathbf{v}⟩+\Vert \mathbf{v}{\Vert }^2.$ Dono add karo: $\pm 2⟨\mathbf{u},\mathbf{v}⟩$ cross terms cancel ho jaate hain, bachta hai $2\Vert \mathbf{u}{\Vert }^2+2\Vert \mathbf{v}{\Vert }^2.■$ Iska matlab: ek parallelogram ke dono diagonals (sum $\mathbf{u}+\mathbf{v}$ aur difference $\mathbf{u}-\mathbf{v}$) milke exactly wutna energy carry karte hain jitni uski chaar sides.

L4.1 mein establish ki gayi dono FOIL expansions use karo, jahan har "$2⟨\mathbf{u},\mathbf{v}⟩$" do equal cross terms $⟨\mathbf{u},\mathbf{v}⟩+⟨\mathbf{v},\mathbf{u}⟩$ se aayi thi jo symmetry se collapse hui. Unhe subtract karo: $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2-\Vert \mathbf{u}-\mathbf{v}{\Vert }^2=(\Vert \mathbf{u}{\Vert }^2+2⟨\mathbf{u},\mathbf{v}⟩+\Vert \mathbf{v}{\Vert }^2)-(\Vert \mathbf{u}{\Vert }^2-2⟨\mathbf{u},\mathbf{v}⟩+\Vert \mathbf{v}{\Vert }^2).$ $\Vert \mathbf{u}{\Vert }^2$ aur $\Vert \mathbf{v}{\Vert }^2$ terms cancel ho jaate hain, aur $2⟨\mathbf{u},\mathbf{v}⟩-(-2⟨\mathbf{u},\mathbf{v}⟩)=4⟨\mathbf{u},\mathbf{v}⟩$: $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2-\Vert \mathbf{u}-\mathbf{v}{\Vert }^2=4⟨\mathbf{u},\mathbf{v}⟩.$ $4$ se divide karo. $■$ Kyun yeh remarkable hai: inner product (angles!) sirf norm (lengths!) se completely reconstruct ho sakti hai. Length secretly angle jaanta hai.

FOIL expansion $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2+2⟨\mathbf{u},\mathbf{v}⟩+\Vert \mathbf{v}{\Vert }^2$ se, middle term $2\cdot 0=0$ hai, toh $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2+\Vert \mathbf{v}{\Vert }^2.■$ Yeh har right-triangle picture ka abstract core hai.

Symmetry aur linearity har $c$ ke liye hold karti hai. Akela obstacle positive-definiteness hai: $⟨\mathbf{v},\mathbf{v}⟩=v_1^2+c{v}_2^2.$ Iske liye ki yeh sabhi $\mathbf{v}$ ke liye $\ge 0$ ho aur sirf $\mathbf{v}=0$ par zero ho, $\mathbf{v}=(0,1)$ lo: hume $c>0$ chahiye.

• Agar $c>0$: $v_1^2+c{v}_2^2\ge 0$, zero force karta hai $v_1=0$ aur $v_2=0$. ✓
• Agar $c=0$: $\mathbf{v}=(0,1)$ se $⟨\mathbf{v},\mathbf{v}⟩=0$ lekin $\mathbf{v}\ne 0$ — positivity fail.
• Agar $c<0$: $\mathbf{v}=(0,1)$ se negative value — fail.

Answer: $\boxed{c>0}$.

$⟨1,\cos (2\pi x)⟩={\int }_0^1\cos (2\pi x)dx=[\frac{\sin (2\pi x)}{2\pi }{]}_0^1=\frac{\sin 2\pi -\sin 0}{2\pi }=0.$ Orthogonal ✓. $[0,1]$ par ek full-period cosine integrate karna use zero mein wash out kar deta hai — constant aur wave overlap nahi karte.

$f$ ka $g$ par projection hai ${\mathrm{proj}}_{g}f=\frac{⟨f,g⟩}{⟨g,g⟩}g$. $⟨f,g⟩={\int }_0^{1}x\cdot 1dx=\frac{1}{2},⟨g,g⟩={\int }_0^{1}1dx=1.$ ${\mathrm{proj}}_{g}f=\frac{1/2}{1}\cdot 1=\frac{1}{2}.$ $[0,1]$ par $x$ ka best constant approximation constant $\frac{1}{2}$ hai — literally uski average value. Figure s03 yeh dikhata hai: coral line $f(x)=x$ hai jo $(0,0)$ se $(1,1)$ tak chadhti hai; height $\frac{1}{2}$ par flat lavender line projection hai. Chote mint segments vertical gaps $x-\frac{1}{2}$ hain (residual) — woh right half mein positive aur left half mein negative hain, aur interval par cancel ho jaate hain, jo exactly isliye hai ki $\frac{1}{2}$ balanced (average) height hai.

Teeno pieces integrals ki tarah compute karo. Cross term: $⟨f,g⟩={\int }_0^{1}x\cdot x^{2}dx={\int }_0^1{x}^{3}dx=[\frac{x^4}{4}{]}_0^1=\frac{1}{4}=\mathrm{0.25.}$ Norms: $\Vert f{\Vert }^2={\int }_0^1{x}^{2}dx=\frac{1}{3},\Vert g{\Vert }^2={\int }_0^1{x}^{4}dx=\frac{1}{5},$ $\Vert f\Vert \Vert g\Vert =\sqrt{\frac{1}{3}\cdot \frac{1}{5}}=\sqrt{\frac{1}{15}}\approx \mathrm{0.2582.}$ Compare: $|⟨f,g⟩|=0.25\le 0.2582=\Vert f\Vert \Vert g\Vert .✓$ Cauchy–Schwarz hold karta hai, aur inequality strict hai ($0.25<0.2582$) kyunki $x$ aur $x^2$ ek doosre ke scalar multiples nahi hain — Cauchy–Schwarz mein equality sirf tab hoti hai jab dono vectors parallel hon.

L1 axioms test karo → L2 norm aur distance formulas use karo → L3 angle aur orthogonality use karo → L4 axioms se identities prove karo → L5 nayi inner products banao aur functions project karo. Har rung sirf wahi reuse karta hai jo neeche ke rung ne earn kiya.