Exercises — Dimension — basis cardinality

Standard basis hai $\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\}$ — 4 vectors, independent aur spanning. Dimension Theorem ke according har basis ka count same hota hai, isliye $\dim {\mathbb{R}}^4=4$.

Basis $\{1,x,x^2,x^3\}$ mein 4 elements hain. Isliye $\dim P_3=4$. Generally $\dim P_n=n+1$ (har power $x^0,\dots ,x^n$ ke liye ek slot).

Nahi. Ek basis independent honi chahiye, aur exchange lemma ke hisaab se ${\mathbb{R}}^2$ mein koi bhi independent set mein zyada se zyada $\dim {\mathbb{R}}^2=2$ vectors ho sakte hain. Teen vectors independent nahi ho sakte, isliye ye basis nahi hai. (Indeed $(2,3)=2(1,0)+3(0,1)$.)

Nahi. Steinitz: independent ≤ spanning. Yahan spanning size $5$ hai, isliye koi bhi independent set mein $\le 5$ vectors honge. Saat $5$ se zyada hai ⇒ woh dependent zaroor honge.

Set karo $a(2,1)+b(1,3)=(0,0)$. Component-wise: $2a+b=0$ aur $a+3b=0$. Pehle se, $b=-2a$; substitute karo: $a+3(-2a)=a-6a=-5a=0\Rightarrow a=0\Rightarrow b=0$. Sirf trivial solution ⇒ independent. Ek 2-dimensional space mein do independent vectors use space ko span bhi karte hain, isliye ye hai ek basis — jo $\dim {\mathbb{R}}^2=2$ confirm karta hai. Neeche figure dekho.

Spanning: hum har standard basis vector recover kar sakte hain: $1=1$; $x=(1+x)-1$; $x^2=(1+x+x^2)-(1+x)$. Kyunki $\{1,x,x^2\}$, $P_2$ span karta hai aur har ek reachable hai, hamaara set bhi karta hai. Count: 3 elements $=\dim P_2$. Ek spanning set jiska size dimension ke barabar ho, automatically ek basis hota hai.

Solve karo $v_1=a{w}_1+b{w}_2$: $(2,3)=a(1,0)+b(1,1)=(a+b,b)$. Toh $b=3$, $a+b=2\Rightarrow a=-1$. Isliye $v_1=-1\cdot w_1+3\cdot w_2.$ $w_1$ par coefficient $-1\ne 0$ hai, isliye hum $w_1$ ke liye solve kar sakte hain: $w_1=\frac{1}{-1}(v_1-3w_2)=-v_1+3w_2.$ Isliye $w_1$, $\{v_1,w_2\}$ se expressible hai, toh $\{v_1,w_2\}=\{(2,3),(1,1)\}$ abhi bhi spans karta hai ${\mathbb{R}}^2$ ko (independence check: $\det \begin{array}{cc}2& 1\\ 3& 1\end{array}=2-3=-1\ne 0$). Ek $v$ swap in hua, ek $w$ bahar gaya, spanning size preserve hui. Figure dekho.

Maano har $c_j=0$ hai. Toh $v_{k+1}=b_1{v}_1+\cdots +b_k{v}_k$ — yaani $v_{k+1}$ pehle ke $v$'s ka combination hai. Lekin $L=\{v_1,\dots ,v_m\}$ independent hai, matlab koi bhi $v$ doosron ka combination nahi hai. Contradiction. Isliye koi $c_j\ne 0$ hai, jo guarantee karta hai ki ek remaining $w$ present hai jise bahar nikala ja sake. Agar ye fail ho jaata, toh hum $w$'s khatam kar lete swap karne ke liye aur size-$n$ spanning invariant collapse ho jaata — poora bound $m\le n$ unproved reh jaata. Yahi ek jagah hai jahan $L$ ki independence enter hoti hai (dekho Linear Independence).

• Direction 1: $B_1$ independent hai, $B_2$ spans karta hai ⇒ lemma deta hai $p\le q$.
• Direction 2: $B_2$ independent hai, $B_1$ spans karta hai ⇒ lemma deta hai $q\le p$.

Dono hold karte hain kyunki ek basis simultaneously independent aur spanning hota hai, isliye har ek dono roles play kar sakta hai. $p\le q$ aur $q\le p$ se milta hai $p=q$. Woh common value $\dim V$ hai.

Constraint solve karo: $y,z$ free lo, phir $x=-y-z$. Toh $(x,y,z)=(-y-z,y,z)=y(-1,1,0)+z(-1,0,1).$ Do vectors $(-1,1,0),(-1,0,1)$ $W$ ko span karte hain aur independent hain (koi bhi dusre ka scalar multiple nahi hai). Isliye woh ek basis banate hain aur $\dim W=2$. Intuition: ek linear equation ek 3-dimensional space se ek degree of freedom hata deti hai, chhodke $3-1=2$. Ye Rank-Nullity Theorem ka chhota roop hai: map $T(x,y,z)=x+y+z$ ka image $1$-dimensional hai aur kernel $=W$ $2$-dimensional hai.

Humein total size $=\dim {\mathbb{R}}^3=3$ chahiye, isliye exactly ek aur vector — bas wo do ka combination na ho. Try karo $(0,0,1)$: kya $\{(1,1,0),(1,0,1),(0,0,1)\}$ independent hai? Determinant compute karo $\det \left(\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 0& 1\end{array}\right).$ Bottom row ke along expand karo: $=1\cdot \det \left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)=1\cdot (1\cdot 0-1\cdot 1)=-1\ne 0.$ Independent ⇒ basis. Exactly ek kyun? Har basis ki cardinality $3$ hoti hai; hamare paas pehle se $2$ independent vectors hain, aur exchange lemma ke by hum hamesha exchange kar sakte hain full dimension tak pahunchne ke liye bina overshoot kiye.

Rank-Nullity Theorem: $\dim (\ker T)+\dim (\text{im }T)=\dim (\text{domain})=5$. $\dim \ker =2$ ke saath, $\dim (\text{im }T)=5-2=3$. Kyunki codomain ${\mathbb{R}}^3$ ka dimension $3$ hai aur image uska ek $3$-dimensional subspace hai, image barabar poore ${\mathbb{R}}^3$ ke hai (ek subspace jiska dimension poore space ke barabar ho woh poora space hai — yahi basis-cardinality phir se). Toh haan, $T$ surjective hai.

Uska basis empty set $\varnothing$ hai. Empty set vacuously independent hai (koi bhi coefficients constrain nahi hote, isliye ek hi combination — empty wala — $\vec{0}$ deta hai) aur ye $\{\vec{0}\}$ ko spans karta hai (empty combination $\vec{0}$ ke barabar hai, jo ek hi vector present hai). Uski cardinality $0$ hai, isliye $\dim \{\vec{0}\}=0$. Note karo ki $\{\vec{0}\}$ khud ek basis nahi hai: koi bhi set jo $\vec{0}$ contain kare dependent hota hai kyunki $1\cdot \vec{0}=\vec{0}$ ek nontrivial relation hai.

Lo $V={\mathbb{R}}^n$ jo standard basis (size $n$) se span hota hai. Wohi standard basis ek independent set hai size $n$ ka. Isliye ek independent set spanning-set size tak pahunch sakta hai exactly; bound $m\le n$ equality ke saath achieve hota hai. Equality tab hoti hai jab independent set bhi spanning ho — yaani woh ek basis hai.

Lo $P$ sab polynomials ka space (koi bhi degree). Set $\{1,x,x^2,x^3,\dots \}$ independent aur spanning hai lekin infinite — koi finite set $P$ ko span nahi kar sakta (ek finite set ki ek maximum degree $d$ hoti hai, isliye woh $x^{d+1}$ tak nahi pahunch sakta). Yahan $\dim P=\infty$. Exchange lemma abhi bhi relatively apply hota hai: kisi bhi finite spanning set ke liye kisi bhi subspace $W\subseteq P$ ke — $W$ mein independent sets us spanning size se bounded hain. Lemma ka engine kabhi nahi toot‍ta; sirf assumption "koi finite spanning set exist karta hai" tooti hai.

Change-of-basis matrix (dekho Coordinates and Change of Basis) mein utne columns hote hain jitni old basis size hai aur utne rows jitni new basis coordinate length — dono kisi basis mein vectors ki sankhya ke barabar hote hain. Kyunki Dimension Theorem dono bases ko same count $\dim V=3$ force karta hai, matrix $3\times 3$ hai, isliye square hai, isliye potentially invertible (aur woh hai invertible, kyunki woh ek basis ko doosre se map karta hai). Non-square impossible hota — matlab do bases ka size alag hoga, jo basis cardinality ke against hai.

Recognition: dimension = basis ki size. Application: independent ≤ spanning. Analysis: ek exchange spanning preserve karta hai kyunki nonzero coefficient se hum solve kar sakte hain. Synthesis: rank-nullity aur extension ke saath combine karo. Mastery: $\{\vec{0}\}$ ka empty-set basis, tightness, aur infinite case.