Exercises — Abstract vector spaces — axioms, examples beyond ℝⁿ

Zero vector woh element hai jo har vector ko unchanged chhod de jab add karo. Yeh har box mein alag hota hai.

• (a) $0=(0,0,0)$. Toh $(x,y,z)+(0,0,0)=(x,y,z)$. ✓
• (b) $0=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$, yeh zero matrix hai. Isse entry-by-entry add karo toh kuch nahi badalega. ✓
• (c) $0=$ woh zero function $z(x)=0$ jo sabhi $x$ ke liye $0$ hai. Toh $(f+z)(x)=f(x)+0=f(x)$. ✓

Lesson yeh hai: $0$ ek role hai, digit $0$ nahi.

• (a) $\dim P_2=3$, basis $\{1,x,x^2\}$ — teen independent "directions" (constant, linear, quadratic).
• (b) $\dim M_{2\times 3}=2\cdot 3=6$, basis woh chhe elementary matrices $E_{ij}$ hain (slot $ij$ mein ek $1$, baaki jagah zeros).
• (c) $\dim =2$, basis $\{\sin x,\cos x\}$: har solution $a\sin x+b\cos x$ hota hai.

Addition aur scaling coefficient-wise hoti hai (yahi wajah hai ki $P_2$ bilkul ${\mathbb{R}}^3$ jaisa behave karta hai). $2p=4+6x-2x^2.$ $2p-q=(4-(-1))+(6-1)x+(-2-4)x^2=5+5x-6x^2.$ Coordinate vector: $\left(\begin{array}{c}5\\ 5\\ -6\end{array}\right)$.

$\mathbb{R}$ ke upar: scalars real hain, toh $z=3\cdot 1+(-2)\cdot i$ se coordinates milte hain $\left(\begin{array}{c}3\\ -2\end{array}\right)$. $\mathbb{C}$ ke upar: scalars complex hain, ek basis vector $\{1\}$ kaafi hai — $z=(3-2i)\cdot 1$, coordinate $\left(\begin{array}{c}3-2i\end{array}\right)$. Scalar field badal deta hai ki "ek step" kya count hota hai, isliye ${\dim }_{\mathbb{R}}\mathbb{C}=2$ lekin ${\dim }_{\mathbb{C}}\mathbb{C}=1$. (Dekho Fields and scalars.)

Teen-part subspace test chalao.

1. Zero: $(0,0,0)$ satisfy karta hai $0+0+0=0$. ✓ $0\in W$.
2. $+$ ke liye closed: lo $u=(x_1,y_1,z_1),v=(x_2,y_2,z_2)$ jahan har ek ka coordinate-sum $0$ ho. Tab $(x_1+x_2)+(y_1+y_2)+(z_1+z_2)=(x_1+y_1+z_1)+(x_2+y_2+z_2)=0+0=0.$ Toh $u+v\in W$. ✓
3. Scaling ke liye closed: $a\cdot (x,y,z)=(ax,ay,az)$ aur $ax+ay+az=a(x+y+z)=a\cdot 0=0$. ✓ Teeno pass ho gaye, toh $W$ subspace hai. (Geometrically yeh origin se guzarne wala ek plane hai; $\dim W=2$.)

Nahi. Sabse jaldi disqualifier zero test hai: $(0,0,0)$ deta hai $0+0+0=0\ne 1$, toh $0\notin U$. Yeh akela kaafi hai (Axiom 3 $U$ ke andar fail hota hai). Failure ko aur clearly dikhane ke liye, closure bhi fail hoti hai: $(1,0,0)\in U$ aur $(0,1,0)\in U$ hain, lekin unka sum $(1,1,0)$ ka coordinate-sum $2\ne 1$ hai, toh $(1,1,0)\notin U$. Figure mein, plane $U$ (coral) origin se hata hua hai; plane $W$ (mint) origin se guzarta hai. Sirf origin se guzarne wale planes hi subspace ho sakte hain.

Haan — yeh disguise mein ek genuine vector space hai. Maange gaye axioms check karo.

• Closure: positives ka product positive hota hai; koi bhi positive number kisi bhi real power mein positive hota hai. ✓
• Axiom 3 (zero vector): humein $e$ chahiye jiske liye $x\oplus e=x$ ho, yaani $x\cdot e=x$, toh $e=1$. Zero vector hai $0=1$. ✓
• Axiom 4 (inverse): chahiye $x\oplus x^{\prime }=1$, yaani $x\cdot x^{\prime }=1$, toh $x^{\prime }=1/x$ (positive). Isliye yahan $-v$ ka matlab hai $1/v$. ✓
• Axiom 5 (scalar identity): $1\odot x=x^1=x$. ✓
• Axiom 8 (scalar sum ke upar distribute karna): $(a+b)\odot x=x^{a+b}=x^a{x}^b=(a\odot x)\oplus (b\odot x)$. ✓

(Baaki axioms bhi similarly check ho jaate hain.) Actually map $x\mapsto \ln x$ $\oplus \to +$ aur $\odot \to$ ordinary scaling mein turn karta hai, jo $\mathbb{R}$ ke saath ek isomorphism hai — toh $\dim V=1$, basis $\{e\}$ (kyunki $\ln e=1$).

Dono identities ko ek dusre ke against khelo. $0=0+0^{\prime }(\text{kyunki }{0}^{\prime }\text{ ek zero hai, ise }v=0\text{ par use karo})$ $0+0^{\prime }=0^{\prime }+0(\text{Axiom 1, commutativity})$ $0^{\prime }+0=0^{\prime }(\text{kyunki }0\text{ ek zero hai, ise }v=0^{\prime }\text{ par use karo})$ Chaining: $0=0^{\prime }$. $■$ Proof mein sirf axioms use hue, toh yeh har vector space mein valid hai.

(a) Har basis vector ko differentiate karo aur $\{1,x,x^2\}$ mein coordinates padhko: $D(1)=0,D(x)=1,D(x^2)=2x.$ Columns in outputs ke coordinate vectors hain: $[D]=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 2\\ 0& 0& 0\end{array}\right).$ (b) $D(p)=0$ ka matlab hai $p^{\prime }=0$, toh $p$ constant hai: $\ker D=\{a\cdot 1\}=\mathrm{span}\{1\}$, $\dim \ker D=1$. (c) Outputs $p\in P_2$ ke liye $p^{\prime }$ hain; degree $\le 2$ ke derivatives se degree $\le 1$ ke saare polynomials milte hain. Toh $\mathrm{im}D=P_1=\mathrm{span}\{1,x\}$, $\dim \mathrm{im}D=2$. (d) Rank–nullity: $\dim \ker D+\dim \mathrm{im}D=1+2=3=\dim P_2$. ✓

Orthogonality: $⟨\sin x,\cos x⟩={\int }_{-\pi }^{\pi }\sin x\cos xdx$. $\sin x\cos x=\frac{1}{2}\sin 2x$ use karke, ${\int }_{-\pi }^{\pi }\frac{1}{2}\sin 2xdx=\frac{1}{2}{\left[-\frac{\cos 2x}{2}\right]}_{-\pi }^{\pi }=\frac{1}{2}\cdot \left(-\frac{\cos 2\pi -\cos (-2\pi )}{2}\right)=0.$ Toh $\sin x\perp \cos x$. ✓ (Iske alawa, integrand ek odd function hai, jo $0$ force karta hai.) Norm-squared: ${\sin }^{2}x=\frac{1-\cos 2x}{2}$ use karke, $⟨\sin x,\sin x⟩={\int }_{-\pi }^{\pi }\frac{1-\cos 2x}{2}dx=\frac{1}{2}[x-\frac{\sin 2x}{2}{]}_{-\pi }^{\pi }=\frac{1}{2}(2\pi -0)=\pi .$ Toh $\Vert \sin x\Vert =\sqrt{\pi }$. Yeh orthogonal functions bilkul wahi building blocks hain Fourier series ke.