Basis — definition, uniqueness of representation: Exercises

4.5.17 · D4 · Maths › Linear Algebra (Full) › Basis — definition, uniqueness of representation

Reminders ek jagah (neeche sab kuch yahaan earn kiya gaya hai, kabhi assume nahi):

Ek basis woh set hai jo dono linearly independent ho (sirf all-zero combination $0$ deta hai) aur spanning ho (har vector banata hai).
Coordinates $[w]_{B} = (c_{1}, \dots, c_{n})$ woh unique scalars hain jahan $w = c_{1} v_{1} + \dots + c_{n} v_{n}$ .
$dim V$ kisi bhi basis ka size hai — space ke liye ek fixed number (dekho Dimension).

Level 1 — Recognition

L1.1

Kya $B = {(2, 0), (0, 3)}$ , $R^{2}$ ki basis hai? Batao tum kaun si do properties check karte ho.

Recall Solution

Hum independence aur spanning check karte hain.

Independence: $c_{1} (2, 0) + c_{2} (0, 3) = (2 c_{1}, 3 c_{2}) = (0, 0)$ se $2 c_{1} = 0$ aur $3 c_{2} = 0$ milta hai, yaani $c_{1} = c_{2} = 0$ . Sirf trivial combination $0$ deta hai. ✔
Spanning: koi bhi $(x, y)$ hai $\frac{x}{2} (2, 0) + \frac{y}{3} (0, 3)$ . ✔ Dono hold karte hain, isliye haan, $B$ ek basis hai. (2-dimensional space mein do independent vectors hamesha ek basis banate hain.)

L1.2

Inme se kaun sa $R^{2}$ ka basis hai? (a) ${(1, 2)}$ , (b) ${(1, 0), (0, 1), (1, 1)}$ , (c) ${(1, 2), (3, 5)}$ .

Recall Solution

(a) Ek vector $R^{2}$ ko span nahi kar sakta (uska span sirf ek line hai). Basis nahi.
(b) $R^{2}$ mein teen vectors dependent hain ( $(1, 1) = (1, 0) + (0, 1)$ ). Span karta hai par independent nahi. Basis nahi.
(c) Do vectors hain, aur woh independent hain kyunki $det (1235) = 1 \cdot 5 - 3 \cdot 2 = - 1 \neq = 0$ . Yeh basis hai.

L1.3

Sach ya jhooth: standard basis ${e_{1}, e_{2}, e_{3}}$ mein $w = (5, - 1, 4)$ ke coordinates $(5, - 1, 4)$ hain.

Recall Solution

Sach. Har $e_{i}$ exactly ek slot on karta hai: $w = 5 e_{1} - 1 e_{2} + 4 e_{3}$ , isliye $[w]_{B} = (5, - 1, 4)$ . Coordinates aur entries ki yeh equality standard basis ke liye special hai.

Level 2 — Application

L2.1

Basis $B = {(1, 1), (1, - 1)}$ mein $w = (7, 1)$ ke coordinates nikalo.

Recall Solution

$c_{1} (1, 1) + c_{2} (1, - 1) = (7, 1)$ solve karo: $c_{1} + c_{2} = 7, c_{1} - c_{2} = 1.$ System kyun banate hain: spanning guarantee karta hai ki solution exist karta hai; independence guarantee karti hai ki woh unique hai — isliye solve karna valid hai. Dono equations add karo: $2 c_{1} = 8 \Rightarrow c_{1} = 4$ . Phir $c_{2} = 7 - 4 = 3$ . $[w]_{B} = (4, 3) .$

L2.2

$P_{2}$ (degree $\leq 2$ ke polynomials) mein basis $B = {1, x, x^{2}}$ ke saath, $p (x) = 2 - 4 x + x^{2}$ ke coordinates do.

Recall Solution

Order $1, x, x^{2}$ mein coefficients padhlo: $[p]_{B} = (2, - 4, 1) .$ Kyun: ${1, x, x^{2}}$ ki independence ka matlab hai " $a + b x + c x^{2} = 0$ sabhi $x$ ke liye" se $a = b = c = 0$ force hota hai, isliye coefficient list unique hai.

L2.3

Woh vector reconstruct karo jiske coordinates $B = {(2, 1), (1, 3)}$ mein $[w]_{B} = (3, 1)$ hain.

Recall Solution

Coordinates tumhe recipe batate hain; bas execute karo: $w = 3 (2, 1) + 1 (1, 3) = (6, 3) + (1, 3) = (7, 6) .$

L2.4

$B = {(1, 1), (1, - 1)}$ mein zero vector $(0, 0)$ ke coordinates nikalo.

Recall Solution

$c_{1} (1, 1) + c_{2} (1, - 1) = (0, 0)$ solve karo: $c_{1} + c_{2} = 0$ aur $c_{1} - c_{2} = 0$ milta hai, isliye $c_{1} = c_{2} = 0$ . $[0]_{B} = (0, 0) .$ Kyun important hai: zero vector ke har basis mein all-zero coordinates hote hain — yeh exactly independence condition hai disguise mein.

Level 3 — Analysis

L3.1

Dikhao ki ${(1, 2), (2, 4)}$ $R^{2}$ ka basis nahi hai, aur batao kaun si property fail hoti hai.

Recall Solution

Note karo $(2, 4) = 2 \cdot (1, 2)$ , isliye $2 (1, 2) + (- 1) (2, 4) = (0, 0)$ — ek nontrivial combination $0$ ke barabar hai. Yeh independence tod deta hai. Is wajah se span sirf $(1, 2)$ se guzarne wali line hai, isliye spanning bhi fail hoti hai. Basis nahi.

L3.2

Set ${(1, 0), (0, 1), (1, 1)}$ $R^{2}$ ko span karta hai. $(2, 3)$ ki do alag representations dikhao, proving karo ki coordinates yahaan unique nahi hain.

Recall Solution

Sirf pehle do use karke: $(2, 3) = 2 (1, 0) + 3 (0, 1) + 0 (1, 1)$ .
Teesra ek baar use karke: kyunki $(1, 1) = (1, 0) + (0, 1)$ , $(2, 3) = 1 (1, 0) + 2 (0, 1) + 1 (1, 1) .$ Check: $1 (1, 0) + 2 (0, 1) + 1 (1, 1) = (1 + 0 + 1, 0 + 2 + 1) = (2, 3)$ . ✔ Ek hi point ke liye do alag addresses $(2, 3, 0)$ aur $(1, 2, 1)$ → unique nahi → dependence culprit hai, isliye basis nahi.

L3.3

Kya $B = {(1, 0, 1), (0, 1, 1), (1, 1, 0)}$ $R^{3}$ ka basis hai? Determinant se justify karo.

Recall Solution

Vectors ko columns ke roop mein stack karo aur determinant nikalo:

=1(1\cdot0-1\cdot1)-0+1(0\cdot1-1\cdot1)=1(-1)+1(-1)=-2.$$ **Determinant kyun:** nonzero determinant ka matlab hai columns independent hain *aur* $\mathbb R^3$ ko span karte hain (matrix invertible hai). Kyunki $-2\ne 0$, **haan, yeh basis hai**.

L3.4

L3.3 ki basis mein $w = (1, 2, 3)$ ke coordinates nikalo.

Recall Solution

$c_{1} (1, 0, 1) + c_{2} (0, 1, 1) + c_{3} (1, 1, 0) = (1, 2, 3)$ solve karo: $c_{1} + c_{3} = 1, c_{2} + c_{3} = 2, c_{1} + c_{2} = 3.$ Teenon equations kyun add karte hain: har equation ek sach statement hai (uska left side literally right side ke barabar hai), aur sach equations add karne se ek aur sach equation milti hai — bilkul waise jaise dono pair apne weight jaanne ke liye scale par khadhe ho. Left side par, har unknown $c_{1}, c_{2}, c_{3}$ teen equations mein se exactly do mein appear karta hai, isliye sum karne par $2 c_{1} + 2 c_{2} + 2 c_{3}$ milta hai; right side par $1 + 2 + 3 = 6$ milta hai. Isliye: $2 (c_{1} + c_{2} + c_{3}) = 6 \Rightarrow c_{1} + c_{2} + c_{3} = 3.$ Ab har original equation is total se subtract karne par ek-ek unknown isolate hota hai (yeh bhi valid move hai — barabar cheez se barabar cheez subtract karne par equality bani rehti hai):

$(c_{1} + c_{2} + c_{3}) - (c_{1} + c_{3}) = c_{2} = 3 - 1 = 2.$
$(c_{1} + c_{2} + c_{3}) - (c_{2} + c_{3}) = c_{1} = 3 - 2 = 1.$
$(c_{1} + c_{2} + c_{3}) - (c_{1} + c_{2}) = c_{3} = 3 - 3 = 0.$ $[w]_{B} = (1, 2, 0) .$ Check: $1 (1, 0, 1) + 2 (0, 1, 1) + 0 (1, 1, 0) = (1, 2, 3)$ . ✔

Level 4 — Synthesis

L4.1

Prove karo: agar $B = {v_{1}, v_{2}, v_{3}}$ , $V$ ka basis hai, toh $B^{'} = {v_{1}, v_{1} + v_{2}, v_{1} + v_{2} + v_{3}}$ bhi basis hai.

Recall Solution

$V$ 3-dimensional hai (uske paas size 3 ka basis hai), isliye counting theorem se sirf $B^{'}$ independent prove karna kaafi hai — 3-dim space mein teen independent vectors automatically span karte hain. Maano $a v_{1} + b (v_{1} + v_{2}) + c (v_{1} + v_{2} + v_{3}) = 0 .$ Har original basis vector ke coefficients collect karo: $(a + b + c) v_{1} + (b + c) v_{2} + c v_{3} = 0 .$ Kyun collect karte hain: $B$ independent hai, isliye har coefficient zero hona chahiye: $a + b + c = 0, b + c = 0, c = 0.$ Back-substitute karo: $c = 0 \Rightarrow b = 0 \Rightarrow a = 0$ . Sirf trivial combination kaam karta hai, isliye $B^{'}$ independent hai, yani basis hai. ∎

L4.2

Coordinate change. L4.1 wala $B^{'}$ lekar aur $w = 2 v_{1} - v_{2} + 4 v_{3}$ ke saath, $[w]_{B^{'}}$ nikalo.

Recall Solution

Hum $d_{1}, d_{2}, d_{3}$ chahte hain jahan $d_{1} v_{1} + d_{2} (v_{1} + v_{2}) + d_{3} (v_{1} + v_{2} + v_{3}) = 2 v_{1} - v_{2} + 4 v_{3} .$ $v_{1}, v_{2}, v_{3}$ ke hisaab se group karo aur match karo (valid kyunki ${v_{1}, v_{2}, v_{3}}$ independent hai): $d_{1} + d_{2} + d_{3} = 2, d_{2} + d_{3} = - 1, d_{3} = 4.$ Neeche se upar: $d_{3} = 4$ ; phir $d_{2} = - 1 - 4 = - 5$ ; phir $d_{1} = 2 - (- 5) - 4 = 3$ . $[w]_{B^{'}} = (3, - 5, 4) .$ Check: $3 v_{1} - 5 (v_{1} + v_{2}) + 4 (v_{1} + v_{2} + v_{3}) = (3 - 5 + 4) v_{1} + (- 5 + 4) v_{2} + 4 v_{3} = 2 v_{1} - v_{2} + 4 v_{3} .$ ✔

L4.3

Structural claim. Maano $V$ mein $n$ vectors ka set $B$ hai jisme yeh property hai ki har $w \in V$ ke liye zyaada se zyaada ek representation $\sum c_{i} v_{i}$ ke roop mein hai (uniqueness), par tumhe nahi bataya ki woh span karta hai. Kaun si basis property guaranteed hai, aur kaun si fail ho sakti hai? Failure ki ek example do.

Recall Solution

"Sabhi $w$ ke liye zyaada se zyaada ek representation" ko $w = 0$ par apply karne se pata chalta hai ki $0$ likhne ka ek hi tarika hai, all-zero coefficients — yahi independence hai ✔ (guaranteed). Spanning fail ho sakti hai: $R^{3}$ mein $B = {(1, 0, 0), (0, 1, 0)}$ har jagah at-most-one representation deta hai (woh independent hai), phir bhi $(0, 0, 1)$ ki koi representation nahi — spanning fail hoti hai. Isliye uniqueness alone $\Rightarrow$ independence, basis nahi.

Level 5 — Mastery

L5.1

Full uniqueness proof, tumhara version. Maano $B = {v_{1}, \dots, v_{n}}$ ek basis hai. Prove karo ki coordinate map $w \mapsto [w]_{B}$ well-defined (single-valued) hai, independence se derive karke — koi hand-waving nahi.

Recall Solution

Maano $w$ ke do coordinate lists hain: $w = \sum_{i} c_{i} v_{i} aur w = \sum_{i} d_{i} v_{i} .$ Subtract karo (same vector minus itself $0$ hai): $0 = w - w = \sum_{i} (c_{i} - d_{i}) v_{i} .$ Kyun subtract karte hain: yeh "do barabar sums" ko " $0$ ke barabar ek combination" mein convert karta hai, jo exactly woh situation hai jise independence control karta hai. Independence se, $\sum_{i} (c_{i} - d_{i}) v_{i} = 0$ har coefficient $c_{i} - d_{i} = 0$ force karta hai, yaani sabhi $i$ ke liye $c_{i} = d_{i}$ . Dono lists same hain, isliye $[w]_{B}$ single-valued hai. ∎

L5.2

Degenerate space. Zero space $V = {0}$ ka basis aur dimension kya hai? Explain karo ki empty set kyun kaam karta hai.

Recall Solution

Basis empty set $\emptyset$ hai, aur $dim V = 0$ .

Spanning: $\emptyset$ ka span (convention se) ${0}$ hai — ek empty sum $0$ ke barabar hota hai. Yahi poora space hai. ✔
Independence: combine karne ke liye koi vectors nahi hain, isliye koi nontrivial combination $0$ ke barabar nahi — vacuously independent. ✔ Isliye $\emptyset$ size $0$ ka basis hai, jisse $dim {0} = 0$ milta hai. Kisi space ke sabhi bases yeh size share karte hain (dekho Dimension).

L5.3

Linear maps meet bases. Maano $B = {v_{1}, v_{2}}$ , $V$ ka basis hai aur $T : V \to W$ linear hai. Prove karo ki $T$ poori tarah $T (v_{1}), T (v_{2})$ ke do values se determine hota hai.

Recall Solution

Koi bhi $w \in V$ lo. Kyunki $B$ ek basis hai, $w$ ka ek unique coordinate expression $w = c_{1} v_{1} + c_{2} v_{2}$ hai (uniqueness, L5.1 se). $T$ ki linearity apply karo: $T (w) = T (c_{1} v_{1} + c_{2} v_{2}) = c_{1} T (v_{1}) + c_{2} T (v_{2}) .$ Kyun yeh claim prove karta hai: right side sirf fixed data $T (v_{1}), T (v_{2})$ aur (unique) scalars $c_{1}, c_{2}$ use karta hai. Isliye jab woh do images choose ho jaate hain, $T (w)$ har $w$ ke liye force ho jaata hai. Yahi reason hai ki ek linear map ko ek matrix ke roop mein store kiya jaata hai jiske columns basis vectors ki images hain — dekho Matrix of a Linear Map. ∎

L5.4

Ek subtle counterexample. $R^{2}$ mein, kya $B = {(1, 1), (2, 2)}$ basis hai? Phir precisely explain karo ki uske span mein kisi vector ke liye uniqueness ke saath kya galat hota hai.

Recall Solution

$(2, 2) = 2 (1, 1)$ , isliye $B$ dependent hai — basis nahi. Uska span sirf line $y = x$ hai. Us line par kisi vector ke liye, maano $(3, 3)$ , uniqueness collapse ho jaati hai: $(3, 3) = 3 (1, 1) + 0 (2, 2) = 1 (1, 1) + 1 (2, 2) = (- 1) (1, 1) + 2 (2, 2), \dots$ infinitely many representations, kyunki redundancy ek vector ko doosre ke multiple se trade karne deti hai. Yeh L3 trap ka geometric chehra hai: dependence $\Rightarrow$ non-unique addresses.

Neeche ki figure yeh draw karti hai: dono dashed vectors $v_{1} = (1, 1)$ (blue) aur $v_{2} = (2, 2)$ (pink) ek hi dashed line $y = x$ par hain, aur yellow point $w = (3, 3)$ us line par hai. Kyunki $v_{2}$ sirf $v_{1}$ ka ek stretched copy hai, tum $w$ tak $v_{1}$ ke along aage jaake aur $v_{2}$ ke along peeche kheench ke (ya ulta) endlessly many tareekon se pahunch sakte ho — axes ke neeche print ki gayi teenon recipes sab ek hi yellow dot par land karti hain. Isko ek real basis se compare karo, jahan do arrows genuinely alag disha mein point karte hain aur ek single recipe pin karte hain.

Figure — Basis — definition, uniqueness of representation

Connections

Basis — definition, uniqueness of representation — woh parent theorem jinhe yeh drill karte hain.
Linear Independence — har uniqueness step ke peeche ka engine.
Span and Spanning Sets — L1–L2 mein test ki gayi existence half.
Dimension — L4 mein use kiya gaya counting shortcut.
Coordinate Vectors and Change of Basis — L2, L4.2 exactly yahi compute karte hain.
Matrix of a Linear Map — L5.3 hi reason hai ki matrices exist karti hain.