Basis — definition, uniqueness of representation: Worked examples

4.5.17 · D3 · Maths › Linear Algebra (Full) › Basis — definition, uniqueness of representation

Shuru karne se pehle, do baatein plain words mein yaad kar lo.

Woh scalars dhundhne ka matlab almost hamesha hota hai ek chota system of equations solve karna — yahi ek skill hai jo yeh page drill karta hai. Yaad rakho ki basis ko ek saath do kaam karne hote hain, aur har kaam ek property guarantee karta hai: spanning guarantee karta hai ki solution exist karta hai (har vector ka kam se kam ek address hota hai), jabki independence guarantee karta hai ki solution unique hai (kisi bhi vector ke do addresses nahi hote). Hum dono ka sahaara lete hain throughout.

Scenario matrix

Har basis problem jo kabhi milega woh inhi cells mein se kisi mein hoga. Figure 1 usi landscape ko visually encode karta hai — teen coloured regimes (kam, bilkul sahi, zyada) hi sab kuch ka backbone hain, toh table padhne se pehle ek nazar maar lo.

Figure 1 — Poora landscape ek nazar mein. Vectors ka ek set teen regimes mein se kisi ek mein hota hai, depending on uska size aur $dim V$ ka comparison. Red (too few): independent hai lekin har vector tak nahi pahunch sakta → kuch addresses missing hain (Cell E). Green (just right): exactly $dim V$ independent vectors → genuine basis, har vector ka exactly ek address (Cells A, B, F, G, I). Orange (too many): sab kuch reach kar sakta hai lekin redundant hai → same vector ke kai addresses milte hain (Cell D). Green box zero-vector case (Cell C) ko bhi cover karta hai, jiska address hamesha all-zeros hota hai. Grey arrows follow karo: vector add karne se "just right" se "too many" ki taraf shift ho jaate ho.

#	Case class	Tricky kyun hai	Regime (Fig 1)	Worked example
A	Standard basis, plain vector	coordinates equal components hote hain — "invisible" case	green	Ex 1
B	Non-standard basis, mixed signs	$2 \times 2$ system solve karna padta hai; signs matter karte hain	green	Ex 2
C	Degenerate INPUT: zero vector	coordinates sab $0$ hone chahiye — lekin sirf tab jab $B$ basis ho	green	Ex 3
D	Degenerate SET: dependent "basis"	coordinates unique nahi hain — failure detect karni padti hai	orange	Ex 4
E	Degenerate SET: too few (doesn't span)	kuch vectors ke koi coordinates nahi hote — failure detect karo	red	Ex 5
F	Non-arrow space (polynomials)	"vectors" functions hain; same machinery kaam karti hai	green	Ex 6
G	Non-arrow space (matrices)	$2 \times 2$ matrices ek $4$ -dimensional space ke roop mein	green	Ex 7
H	Real-world word problem	ek mixing/recipe story ko coordinates mein translate karo	green	Ex 8
I	Exam twist: verify karo ki set $R^{3}$ mein basis HAI	independence + spanning ek saath determinant se check karo	green	Ex 9

Ab har cell walk karte hain.

Worked examples

Cell A — invisible standard basis

Worked example

Ex 1. $R^{3}$ mein $B = {e_{1}, e_{2}, e_{3}} = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}$ ke saath, $w = (- 5, 8, 0)$ ke coordinates dhundho. Forecast: aage padhne se pehle teen numbers guess karo.

Defining equation likho: $c_{1} e_{1} + c_{2} e_{2} + c_{3} e_{3} = (- 5, 8, 0)$ . Yeh step kyun? Coordinates defined hote hain as the scalars in this combination — hum hamesha definition se shuru karte hain, guess nahi karte.
Har $e_{i}$ exactly ek slot on karta hai, toh $(c_{1}, c_{2}, c_{3}) = (- 5, 8, 0)$ seedha mil jaata hai. Yeh step kyun? Left side literally $(c_{1}, c_{2}, c_{3})$ padh raha hai; entries match karna instant hai.

Answer: $[w]_{B} = (- 5, 8, 0)$ .

Verify: $- 5 e_{1} + 8 e_{2} + 0 e_{3} = (- 5, 8, 0) = w$ . ✓ Notice karo ki teesri coordinate $0$ hai even though vector "real" hai — ek component zero ho sakta hai. Coordinates components ke equal sirf yahan hote hain.

Cell B — non-standard basis, mixed signs

Worked example

Ex 2. $R^{2}$ mein $B = {(2, 1), (- 1, 3)}$ lo. $w = (3, - 4)$ ke liye $[w]_{B}$ dhundho. Forecast: kya coordinates entries $(3, - 4)$ se bade honge ya chote? Har ek ka sign guess karo.

Set up karo $c_{1} (2, 1) + c_{2} (- 1, 3) = (3, - 4)$ , jo do scalar equations deta hai: $2 c_{1} - c_{2} = 3, c_{1} + 3 c_{2} = - 4.$ Yeh step kyun? Spanning promise karta hai ki solution exist karta hai; equations sirf "har coordinate match karo" hai.
Pehli equation se $c_{2} = 2 c_{1} - 3$ . Dusri mein substitute karo: $c_{1} + 3 (2 c_{1} - 3) = - 4 \Rightarrow 7 c_{1} - 9 = - 4 \Rightarrow c_{1} = \frac{5}{7} .$ Yeh step kyun? Substitution do unknowns ko ek mein collapse kar deta hai — $2 \times 2$ ke liye sabse clean route.
Phir $c_{2} = 2 \cdot \frac{5}{7} - 3 = \frac{10}{7} - \frac{21}{7} = - \frac{11}{7}$ . Yeh step kyun? Back-substitute karke doosra coordinate recover karo.

Answer: $[w]_{B} = (\frac{5}{7}, - \frac{11}{7})$ .

Verify: $\frac{5}{7} \cdot (2, 1) + (- \frac{11}{7}) \cdot (- 1, 3) = (\frac{10 + 11}{7}, \frac{5 - 33}{7}) = (\frac{21}{7}, \frac{- 28}{7}) = (3, - 4) = w$ . ✓ Coordinates $\neq =$ components — yahi norm hai, exception nahi. Figure 2 yahi sum geometrically dikhata hai.

Figure 2 — Coordinates as a recipe of scaled basis arrows. Blue arrow $u = (2, 1)$ aur green arrow $v = (- 1, 3)$ basis $B$ ki do directions hain. Orange arrow target $w = (3, - 4)$ hai. $w$ banane ke liye hum $u$ ke saath $\frac{5}{7}$ walk karte hain (dashed blue, blue dot par land karte hain), phir $v$ ke saath $- \frac{11}{7}$ (dashed green — negative coefficient ka matlab hum $v$ ke opposite travel karte hain). Dashed path exactly orange arrow ke tip par land karta hai: picture ke roop mein " $[w]_{B} = (\frac{5}{7}, - \frac{11}{7})$ " iska yahi matlab hai.

Cell C — degenerate input: zero vector

Cell D — degenerate SET: dependent, coordinates not unique

Worked example

Ex 4. Kya $S = {(1, 0), (0, 1), (1, 1)}$ $R^{2}$ ki basis hai? $w = (2, 3)$ ko coordinate karne ki koshish karo. Forecast: 2-D space mein teen vectors — kuch problem smell ho raha hai?

Try karo $c_{1} (1, 0) + c_{2} (0, 1) + c_{3} (1, 1) = (2, 3)$ , yaani $c_{1} + c_{3} = 2, c_{2} + c_{3} = 3.$ Yeh step kyun? Coordinate system honestly set up karo aur dekho kya hota hai.
Do equations, teen unknowns → $c_{3}$ free hai. $c_{3} = 0$ rakho: $(c_{1}, c_{2}, c_{3}) = (2, 3, 0)$ . $c_{3} = 1$ rakho: $(1, 2, 1)$ . Yeh step kyun? Free variable ka matlab infinitely many solutions hain — coordinates unique nahi hain.
Failure seedha zero vector se check karo: $(1, 0) + (0, 1) - (1, 1) = (0, 0)$ with coefficients $(1, 1, - 1) \neq = (0, 0, 0)$ . Yeh step kyun? $0$ tak pahunchne wala non-trivial combination exactly dependence ki definition hai — yahi uniqueness todta hai.

Answer: $S$ basis nahi hai. Yeh $R^{2}$ ko span karta hai lekin dependent hai, toh addresses unique nahi hain. Dimension ke hisaab se, $R^{2}$ ki koi bhi basis mein exactly $2$ vectors hote hain, kabhi $3$ nahi. (Yeh Figure 1 ka orange "too many" regime hai.)

Verify: dono $(2, 3, 0)$ aur $(1, 2, 1)$ $w$ rebuild karte hain: $(2, 3, 0) \to (2 + 0, 3 + 0) = (2, 3)$ ✓ aur $(1, 2, 1) \to (1 + 1, 2 + 1) = (2, 3)$ ✓. Do valid addresses = failure.

Cell E — degenerate SET: too few, doesn't span

Worked example

Ex 5. Kya $S = {(1, 0, 0), (0, 1, 0)}$ $R^{3}$ ki basis hai? $w = (0, 0, 1)$ coordinate karo. Forecast: $S$ mein kaun si direction bilkul missing hai?

Try karo $c_{1} (1, 0, 0) + c_{2} (0, 1, 0) = (0, 0, 1)$ . Teesri coordinate: $0 = 1$ . Yeh step kyun? Equation set up karne se contradiction turant samne aa jaati hai.
$0 = 1$ impossible hai, toh koi solution exist nahi karta. Yeh step kyun? No solution ka matlab $w$ ke $S$ mein koi coordinates nahi hain — spanning fail ho gayi.

Answer: $S$ basis nahi hai: yeh independent hai lekin $R^{3}$ span nahi karta ( $z$ -direction unreachable hai). Yeh Figure 1 ka red "too few" regime hai; $R^{3}$ ki basis mein exactly $3$ vectors chahiye.

Verify: koi bhi combination $c_{1} (1, 0, 0) + c_{2} (0, 1, 0) = (c_{1}, c_{2}, 0)$ ki teesri entry hamesha $0 \neq = 1$ hogi. ✓ contradiction confirmed. Yeh Ex 4 ka exact mirror hai: too few $\Rightarrow$ existence fail; too many $\Rightarrow$ uniqueness fail.

Cell F — arrows ke bina space: polynomials

Worked example

Ex 6. $P_{2}$ mein (degree $\leq 2$ ke polynomials), $B = {1, 1 + x, 1 + x + x^{2}}$ lo. $p (x) = 2 + 3 x + 4 x^{2}$ ke liye $[p]_{B}$ dhundho. Forecast: yeh basis "triangular" hai (Figure 3 batata hai kyun) — guess karo ki coefficients $x^{2}$ (top) se padhni chahiye ya $1$ (bottom) se pehle.

Likho $p = a \cdot 1 + b \cdot (1 + x) + c \cdot (1 + x + x^{2})$ aur expand karo: $p = (a + b + c) + (b + c) x + c x^{2} .$ Yeh step kyun? Do polynomials equal hote hain sirf tab jab har coefficient match kare (yahi $1, x, x^{2}$ ki independence hai), toh hum power ke hisaab se group karte hain.
Sabse highest power se match karo: $c = 4$ . Phir $b + c = 3 \Rightarrow b = - 1$ . Phir $a + b + c = 2 \Rightarrow a = 2 - (- 1) - 4 = - 1$ . Yeh step kyun? "Triangular" shape — har basis polynomial exactly ek naya power add karta hai (Figure 3) — top-down solve karne deta hai bina algebra tangle ke; yeh ek common exam shortcut hai.

Answer: $[p]_{B} = (a, b, c) = (- 1, - 1, 4)$ .

Verify: $- 1 \cdot 1 - 1 \cdot (1 + x) + 4 \cdot (1 + x + x^{2}) = (- 1 - 1 + 4) + (- 1 + 4) x + 4 x^{2} = 2 + 3 x + 4 x^{2} = p$ . ✓ $R^{3}$ jaisi hi machinery — yahi basis ka poora point hai: abstract spaces ke concrete addresses milte hain. Dekho Coordinate Vectors and Change of Basis.

Figure 3 — Basis "triangular" kyun hai. Har basis polynomial ek row hai; har power $1, x, x^{2}$ ek column hai. Filled cell ka matlab hai "yeh power appear karti hai." Filled cells lower-triangular staircase banate hain: $1$ sirf $1$ -column use karta hai; $1 + x$ $x$ -column add karta hai; $1 + x + x^{2}$ $x^{2}$ -column add karta hai. Kyunki sirf last basis polynomial $x^{2}$ column tak pahunchta hai, uska coefficient $c$ top power se pehle pin hota hai; phir $x$ pin hoti hai, phir $1$ — exactly step 2 ka top-down solve.

Cell G — matrices ka space

Worked example

Ex 7. $2 \times 2$ real matrices ka space $M_{2 \times 2}$ ka standard basis $B = {E_{11}, E_{12}, E_{21}, E_{22}}$ hai, jahan $E_{ij}$ mein slot $(i, j)$ mein $1$ aur baaki $0$ hai. $A = (70 - 2 5)$ ke coordinates dhundho. Forecast: ek $2 \times 2$ matrix ke kitne coordinates hote hain? (Hint: kitne slots hote hain?)

Likho $A = c_{1} E_{11} + c_{2} E_{12} + c_{3} E_{21} + c_{4} E_{22}$ . Yeh step kyun? Har $E_{ij}$ exactly ek slot switch on karta hai — standard basis ka matrix version.
Slot by slot padho: $c_{1} = 7, c_{2} = - 2, c_{3} = 0, c_{4} = 5$ . Yeh step kyun? Bilkul $R^{3}$ mein $e_{i}$ ki tarah, is standard basis ke liye coordinates entries ke equal hote hain.

Answer: $[A]_{B} = (7, - 2, 0, 5)$ , ek vector in $R^{4}$ .

Verify: rebuild karo: $7 E_{11} - 2 E_{12} + 0 E_{21} + 5 E_{22} = (70 - 2 5) = A$ . ✓ Toh $dim M_{2 \times 2} = 4$ : ek matrix "ek 4-vector hai" jab ek baar basis fix ho jaaye (yahi Matrix of a Linear Map ko power deta hai).

Cell H — real-world word problem

Worked example

Ex 8. Ek juice bar do base syrups mix karta hai. Syrup $u = (3, 1)$ mein (grams sugar, grams vitamin-C) per scoop hai; syrup $v = (1, 2)$ bhi isi tarah. Ek customer $(11, 12)$ grams ka drink chahta hai. Har ek ke kitne scoops chahiye? Forecast: sugary syrup zyada ya vitamin syrup? Solve karne se pehle guess karo.

Yeh poochh raha hai: basis $B = {u, v} = {(3, 1), (1, 2)}$ mein target $(11, 12)$ ke coordinates dhundho. Set up karo $c_{1} (3, 1) + c_{2} (1, 2) = (11, 12)$ : $3 c_{1} + c_{2} = 11, c_{1} + 2 c_{2} = 12.$ Yeh step kyun? "Do ingredients mix karke target hit karo" literally ek coordinate problem hai disguise mein.
Pehli se, $c_{2} = 11 - 3 c_{1}$ . Substitute karo: $c_{1} + 2 (11 - 3 c_{1}) = 12 \Rightarrow c_{1} - 6 c_{1} + 22 = 12 \Rightarrow - 5 c_{1} = - 10 \Rightarrow c_{1} = 2$ . Yeh step kyun? Substitution phir se — $2 \times 2$ ke liye reliable.
Phir $c_{2} = 11 - 3 (2) = 5$ . Yeh step kyun? Back-substitute karke dusra scoop count nikalo.

Answer: $u$ ke $2$ scoops aur $v$ ke $5$ scoops. (Dono non-negative — physically sensible.)

Verify (with units): sugar $= 2 (3) + 5 (1) = 11$ g ✓; vitamin-C $= 2 (1) + 5 (2) = 12$ g ✓. Kyunki $u, v$ independent hain, yeh recipe unique hai — customer ko koi second valid mix nahi milega.

Cell I — exam twist: prove karo ki ek set $R^{3}$ ki basis HAI

Worked example

Ex 9. Dikhao ki $B = {(1, 2, 0), (0, 1, 1), (1, 0, 1)}$ $R^{3}$ ki basis hai, phir $w = (2, 3, 1)$ coordinate karo. Forecast: $R^{3}$ mein, exactly $3$ vectors ka set basis hai iff ek single number non-zero ho — kaun sa number?

Vectors ko columns ki tarah stack karo aur determinant compute karo $det 120011101 .$ Yeh step kyun? Exactly $dim R^{3} = 3$ vectors ke saath, "independent" aur "spanning" same condition ban jaate hain, aur dono hold karte hain iff determinant $\neq = 0$ ho. Ek check do kaam karta hai. (Dekho Dimension.)
Pehli row along expand karo: $1 \cdot (1 \cdot 1 - 0 \cdot 1) - 0 + 1 \cdot (2 \cdot 1 - 1 \cdot 0) = 1 + 2 = 3 \neq = 0.$ Yeh step kyun? Non-zero determinant $\Rightarrow$ columns independent hain $\Rightarrow$ (sahi count ke saath) basis.
Ab $w$ coordinate karo: solve karo $c_{1} (1, 2, 0) + c_{2} (0, 1, 1) + c_{3} (1, 0, 1) = (2, 3, 1)$ , yaani $c_{1} + c_{3} = 2, 2 c_{1} + c_{2} = 3, c_{2} + c_{3} = 1.$ Pehli se, $c_{3} = 2 - c_{1}$ . Teesri se, $c_{2} = 1 - c_{3} = 1 - (2 - c_{1}) = c_{1} - 1$ . Dusri mein substitute karo: $2 c_{1} + (c_{1} - 1) = 3 \Rightarrow 3 c_{1} = 4 \Rightarrow c_{1} = \frac{4}{3}$ . Yeh step kyun? Easy equations chain karo ek unknown isolate karne ke liye.
Phir $c_{2} = \frac{4}{3} - 1 = \frac{1}{3}$ aur $c_{3} = 2 - \frac{4}{3} = \frac{2}{3}$ . Yeh step kyun? Baaki ke liye back-substitute karo.

Answer: $B$ basis hai (det $= 3$ ), aur $[w]_{B} = (\frac{4}{3}, \frac{1}{3}, \frac{2}{3})$ .

Verify: $\frac{4}{3} (1, 2, 0) + \frac{1}{3} (0, 1, 1) + \frac{2}{3} (1, 0, 1) = (\frac{4 + 2}{3}, \frac{8 + 1}{3}, \frac{1 + 2}{3}) = (2, 3, 1) = w$ . ✓

Recall Yeh problem kis cell ka hai? (self-test)

Problem diye jaane par, is order mein poochho: Kitne vectors hain $dim V$ ke comparison mein? ::: $dim$ se zyada → dependent (Cell D); kam → span nahi kar sakta (Cell E); equal → possible basis, det check karo (Cell I). Kya input zero vector hai? ::: Kisi bhi genuine basis mein coordinates sab $0$ honge (Cell C). Kya "vector" polynomial ya matrix hai? ::: Same machinery — coefficients / slots match karo (Cells F, G). Kya yeh mixing/recipe wali story hai? ::: Yeh coordinates in disguise hai — system set up karo (Cell H).

Connections

Linear Independence — yahi reason hai ki coordinates unique hote hain (Cells C, D, I).
Span and Spanning Sets — yahi reason hai ki coordinates exist karte hain (Cell E).
Dimension — "sahi count" jo Cells D, E, I decide karta hai.
Coordinate Vectors and Change of Basis — yahan har answer ek coordinate vector $[w]_{B}$ hai.
Subspaces — polynomial aur matrix examples bade spaces mein rehte hain.
Matrix of a Linear Map — Cell G ka matrix-as-4-vector is idea ka seed hai.