Exercises — Rank-nullity theorem — proof

KYA karte hain: $\mathrm{rank}+\mathrm{nullity}=\dim V$ mein plug karte hain. KYUN: theorem teen slots wali equation hai; do diye hue hain toh teesra automatically aata hai. Domain ${\mathbb{R}}^7$ hai isliye $\dim V=7$. Tab $\mathrm{rank}(T)=\dim V-\mathrm{nullity}(T)=7-3=\boxed{4}.$ Dekho codomain ${\mathbb{R}}^4$ kabhi kaam nahi aaya — humne domain ${\mathbb{R}}^7$ ke against balance kiya.

KYA karte hain: rank ko dono taraf se bound karte hain. Image codomain ke andar baith ti hai, isliye $\mathrm{rank}\le \dim W=9$. Lekin image sirf $5$ input directions se feed hoti hai, isliye $\mathrm{rank}=\dim V-\mathrm{nullity}\le \dim V=5$. Kyunki $7>5$, yeh impossible hai. Jawab: False. Sabse tight cap hai $\mathrm{rank}\le \min (\dim V,\dim W)=\min (5,9)=5$.

KYA karte hain: pivots count karne ke liye row-reduce karo (woh count rank hai — dekho Rank of a Matrix (row vs column space)). KYUN: har pivot ek independent surviving direction of the column space = image mark karta hai. Row 1 ko Row 3 se subtract karo: $R_3\to R_3-R_1=(0,1,1,1)$. Yeh row 2 ke barabar hai, isliye $R_3\to R_3-R_2=(0,0,0,0)$. $\left(\begin{array}{cccc}1& 2& 0& 1\\ 0& 1& 1& 1\\ 0& 0& 0& 0\end{array}\right).$ Do nonzero pivot rows $\Rightarrow \mathrm{rank}=2$. Theorem se, $\dim V=4$ ke saath: $\mathrm{nullity}=4-2=\boxed{2},\mathrm{rank}=\boxed{2}.$

Kernel: $p^{\prime }=0$ ka matlab hai $p$ constant hai, isliye $\ker T=\{\text{constants}\}$, ek $1$-dimensional space $\Rightarrow \mathrm{nullity}=1$. Image: basis $\{1,x,x^2,x^3,x^4\}$ ko differentiate karne pe $\{0,1,2x,3x^2,4x^3\}$ milta hai, jo $P_3$ (degree $\le 3$) ko span karta hai. Isliye image $P_3$ hai, $\dim =4\Rightarrow \mathrm{rank}=4$. Check: $4+1=5=\dim P_4$. ✓ Theorem abstract spaces pe bhi hold karta hai, sirf ${\mathbb{R}}^n$ pe nahi.

KYA karte hain: nullity $0$ ko injectivity mein convert karo, phir surjectivity ke liye rank–nullity use karo. Nullity $0\Rightarrow$ injective: $\ker T=\{0\}$ (dimension $0$ ka matlab sirf zero vector). Agar $T(a)=T(b)$ toh $T(a-b)=0$ isliye $a-b\in \ker T=\{0\}$, deta hai $a=b$. Injective. Injective $\Rightarrow$ surjective yahan: theorem se $\mathrm{rank}=\dim V-0=n$. Image $n$-dimensional codomain ${\mathbb{R}}^n$ ka ek $n$-dimensional subspace hai, isliye image poora ${\mathbb{R}}^n$ hai. Surjective. Injective + surjective $=$ bijective, isliye invertible. Dekho Invertible Linear Maps and Isomorphisms. $■$ Key idea: square situation ($\dim V=\dim W$) mein, injective $\iff$ surjective $\iff$ invertible — rank–nullity inhe ek saath jodhta hai.

KYA karte hain: $Ax=0$ ke solution space ko $\ker T$ ki tarah identify karo jahan $T:{\mathbb{R}}^6\to {\mathbb{R}}^3$. Unknowns ki sankhya $6=\dim V$ hai. Pivots $=\mathrm{rank}=2$, isliye $\text{free variables}=\dim V-\mathrm{rank}=6-2=\boxed{4}.$ Solution space exactly kernel hai, isliye uski dimension nullity $=4$ hai (dekho Solving Linear Systems — free variables count). Free-variable count aur nullity same number hain.

KYA karte hain: $T\circ S$ pehle $\mathrm{im}S$ mein land karta hai (${\mathbb{R}}^4$ ke andar ek $3$-dim space), phir $T$ usp e act karta hai. Upper bound: composite ki image $T$ ko $\mathrm{im}S$ pe apply karne se aati hai, jo $\mathrm{im}S$ ya $\mathrm{im}T$ se badi nahi ho sakti. Isliye $\mathrm{rank}(T\circ S)\le \min (\mathrm{rank}S,\mathrm{rank}T)=\min (3,2)=2.$ Lower bound: $T$ ko $3$-dimensional $\mathrm{im}S$ tak restrict karke, is restriction pe rank–nullity deta hai $\mathrm{rank}(T\circ S)\ge \mathrm{rank}(S)-\mathrm{nullity}(T)=3-(4-2)=3-2=1.$ Isliye $\boxed{1\le \mathrm{rank}(T\circ S)\le 2}$, aur dono values achievable hain — data akele ise pin nahi kar sakta. (Prompt mein claim ek deliberate misdirection hai; honest answer range hai.)

KYA karte hain: $T^2=0$ ka matlab hai $T$ ka har output agli $T$ se kill ho jaata hai, yaani $\mathrm{im}T\subseteq \ker T$. KYUN: kisi bhi $v$ ke liye, $T(T(v))=0$, isliye $T(v)\in \ker T$; yahi exactly $\mathrm{im}T\subseteq \ker T$ hai. Nested subspaces ki dimensions lete hain: $\mathrm{rank}(T)=\dim (\mathrm{im}T)\le \dim (\ker T)=\mathrm{nullity}(T).$ Ab rank–nullity apply karo, $\mathrm{rank}+\mathrm{nullity}=4$. Kyunki $\mathrm{rank}\le \mathrm{nullity}$, $2\mathrm{rank}\le \mathrm{rank}+\mathrm{nullity}=4\Rightarrow \mathrm{rank}\le 2.■$ Kyunki $T\ne 0$ isliye bhi $\mathrm{rank}\ge 1$, toh $\mathrm{rank}\in \{1,2\}$.

KYA karte hain: ek clever linear map banao jiska kernel aur image wahi pieces hon jo hum chahte hain. Define karo $T:U\times W\to V,T(u,w)=u-w.$ Yeh linear hai, aur iska domain $U\times W$ ki dimension $\dim U+\dim W$ hai ($U$ ka ek basis aur $W$ ka ek basis concatenate karo; dekho Dimension of a Vector Space). $T$ ki Image: saare differences $u-w$ jahan $u\in U,w\in W$. $U+W$ ka har element $u+w=u-(-w)$ likha ja sakta hai aur $-w\in W$, isliye $\mathrm{im}T=U+W$. Iss tarah $\mathrm{rank}T=\dim (U+W)$. $T$ ka Kernel: $u-w=0$ ka matlab $u=w$ hai, isliye common vector $U\cap W$ mein hai. Kernel hai $\{(x,x):x\in U\cap W\}$, jo $U\cap W$ se isomorphic hai, deta hai $\mathrm{nullity}T=\dim (U\cap W)$. $T$ pe rank–nullity apply karo: $\underset{\text{rank}}{\underbrace{\dim (U+W)}}+\underset{\text{nullity}}{\underbrace{\dim (U\cap W)}}=\underset{\dim (U\times W)}{\underbrace{\dim U+\dim W}}.$ Formula ke liye rearrange karo. $■$ Lesson: dimension identity prove karne ke liye, ek map engineer karo taaki identity rank + nullity = domain ban jaaye.

$\dim (U+W)=\dim U+\dim W-\dim (U\cap W)=9-\dim (U\cap W)$ se. Sum $U+W$ ${\mathbb{R}}^6$ mein rehta hai, isliye $\dim (U+W)\le 6$. Isliye $9-\dim (U\cap W)\le 6\Rightarrow \dim (U\cap W)\ge 3.$ Sabse chhoti intersection dimension $\boxed{3}$ hai, jo tab achieve hoti hai jab $U+W={\mathbb{R}}^6$. $T$ se cross-check: domain dimension $=4+5=9$, image $\le \dim {\mathbb{R}}^6=6$, isliye $\mathrm{nullity}=9-\mathrm{rank}\ge 9-6=3$, match karta hai $\dim (U\cap W)\ge 3$ se.