Worked examples — Linear independence — formal definition, testing

4.5.15 · D3 · Maths › Linear Algebra (Full) › Linear independence — formal definition, testing

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Scenario matrix

Har linear-independence problem in cells mein se kisi ek mein hoti hai. Hum sab ko cover karenge.

Cell	Situation	Sahi tool	Answer pattern
A	Square, $n$ vectors in ${\mathbb{R}}^n$	determinant	$\det \ne 0$ → indep
B	Square but parallel/multiple (degenerate)	determinant $=0$ ya inspection	dependent
C	Non-square, fewer vectors than dimension	row reduce, count pivots	dono ho sakta hai
D	Non-square, more vectors than dimension	count vs dimension	hamesha dependent
E	Set mein zero vector hai	inspection	hamesha dependent
F	Ek vector doosron ka sum hai (hidden)	row reduce se pata chalta hai	dependent
G	Real-world word problem	translate → matrix → test	context-dependent
H	Exam twist (parameter $k$: kis $k$ pe dependent?)	$\det$ as function of $k$	solve $\det =0$

Vectors ki sankhya $n$ bas "kitne arrows list kiye" hai. Dimension woh hai ki har arrow mein kitne coordinates hain (jaise $(3,1)$ dimension $2$ mein rehta hai). Pivot row reduction ke baad ek leading nonzero entry hoti hai — ek genuinely naye direction ke liye.

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Example 1 — Cell A (square, determinant)

Forecast: Dono clearly alag disha mein point karte hain — na hi ek doosre ka stretch lagta hai. Andaza: independent. Aage padhne se pehle figure dekho.

1. Columns mein stack karo. $A=\left[\begin{array}{cc}3& 1\\ 1& 2\end{array}\right]$. Yeh step kyun? Equation $c_1{\mathbf{v}}_1+c_2{\mathbf{v}}_2=0$ exactly $A\mathbf{c}=0$ hai; $A$ ke columns hamare vectors hain.
2. ${\mathbb{R}}^2$ mein do vectors → square → determinant use karo. Yeh step kyun? $\det$ sirf square matrices ke liye exist karta hai, aur yahan $A$ $2\times 2$ hai. Geometrically $\det$ un do vectors se bane parallelogram ka signed area hai (figure mein blue). Zero area matlab wo ek line pe hain (dependent); nonzero area matlab wo genuinely 2D patch banate hain.
3. Compute karo. $\det A=3\cdot 2-1\cdot 1=6-1=5\ne 0$.

Answer: independent.

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Example 2 — Cell B (square, degenerate: parallel)

Forecast: Notice karo $-6=-3\cdot 2$ aur $3=-3\cdot (-1)$. Lagta hai multiple hai. Andaza: dependent.

1. Scalar multiple spot karo. ${\mathbf{v}}_2=-3{\mathbf{v}}_1$. Yeh step kyun? Agar ek vector doosre ka scalar multiple hai, to wo origin se guzarne wali same line pe hain (figure dekho — dono arrows aur unka extension ek dashed line share karte hain). Yeh dependence ki picture hai.
2. Nontrivial combo likho. $3{\mathbf{v}}_1+1\cdot {\mathbf{v}}_2=3(2,-1)+(-6,3)=(0,0)=0$, jahan $c_1=3\ne 0$. Yeh step kyun? Dependence ke liye kam se kam ek nonzero coefficient chahiye jo $0$ produce kare; humne wo exhibit kar diya.
3. Determinant se confirm karo. $\det \left[\begin{array}{cc}2& -6\\ -1& 3\end{array}\right]=2\cdot 3-(-6)(-1)=6-6=0$.

Answer: dependent.

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Example 3 — Cell C (non-square, dimension se kam)

Forecast: $3$-dimensional space mein sirf $2$ vectors, aur na hi ek doosre ka multiple hai. Andaza: independent (lekin $\det$ off-limits hai — matrix $3\times 2$ hai).

1. Determinant kyun nahi. $A=\left[\begin{array}{cc}1& 0\\ 0& 3\\ 2& 1\end{array}\right]$ $3\times 2$ hai — square nahi, isliye $\det$ exist nahi karta. Yeh step kyun? Determinant sirf square matrices ke liye defined hai; yahan iske liye reach karna ek classic error hai. Iski jagah rank via row reduction use karo.
2. Row reduce karo. $\left[\begin{array}{cc}1& 0\\ 0& 3\\ 2& 1\end{array}\right]\stackrel{R_3-2R_1}{\to }\left[\begin{array}{cc}1& 0\\ 0& 3\\ 0& 1\end{array}\right]\stackrel{R_3-\frac{1}{3}{R}_2}{\to }\left[\begin{array}{cc}1& 0\\ 0& 3\\ 0& 0\end{array}\right].$ Yeh step kyun? Row operations columns ke beech dependence relations nahi badlate; hum bas pivots count karna chahte hain (leading entries).
3. Pivots count karo. Column 1 mein pivot (woh $1$) aur column 2 mein (woh $3$): 2 columns ke liye 2 pivots. Yeh step kyun? Pivots $=$ columns ki sankhya ⇒ $\mathrm{rank}=2=n$ ⇒ koi free variable nahi ⇒ sirf trivial solution.

Answer: independent.

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Example 4 — Cell D (dimension se zyada vectors)

Forecast: $3$ arrows ek flat $2$-dimensional plane mein thuse hue. Andaza: dependent — $3$ independent directions ke liye simply jagah nahi hai.

1. Vectors vs dimension count karo. $n=3$ vectors, dimension $=2$. Matrix $A$ $2\times 3$ hai. Yeh step kyun? $\mathrm{rank}(A)\le \min (\text{rows},\text{cols})=2<3$, isliye kam se kam ek column mein pivot nahi ⇒ ek free variable guaranteed hai ⇒ nontrivial solution guaranteed. Compute kiye bina hi answer pata hai.
2. Ek explicit relation dhundho (satisfaction ke liye). Notice: ${\mathbf{v}}_3={\mathbf{v}}_1+{\mathbf{v}}_2$? Check karo: $(1,1)+(2,3)=(3,4)={\mathbf{v}}_3$. ✓ Yeh step kyun? Actual combo produce karna dependence concretely confirm karta hai: ${\mathbf{v}}_1+{\mathbf{v}}_2-{\mathbf{v}}_3=0$, coefficients $(1,1,-1)\ne 0$ ke saath.

Answer: dependent.

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Example 5 — Cell E (zero vector hai)

Forecast: Unmen se ek $0$ hai. Andaza: dependent — aur yeh chahe baaki vectors kuch bhi hon, hamesha hold karta hai.

1. Zero vector isolate karo. Unhe ${\mathbf{v}}_1,{\mathbf{v}}_2=0,{\mathbf{v}}_3$ kaho. Yeh step kyun? Koi bhi set jisme $0$ ho woh instantly dependent hai — humen baaki entries ki zaroorat bhi nahi.
2. Nontrivial combo banao. $0\cdot {\mathbf{v}}_1+1\cdot {\mathbf{v}}_2+0\cdot {\mathbf{v}}_3=1\cdot 0=0$. Yeh step kyun? Yahan $c_2=1\ne 0$ hai, phir bhi sum $0$ hai. Woh single nonzero coefficient dependence ke liye kaafi hai.

Answer: dependent.

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Example 6 — Cell F (hidden sum, row reduce karna zaroori)

Forecast: Koi bhi do visible multiples nahi hain. "Independent" bolne ka man karta hai, lekin trap yeh hai ki ek hidden combination ho sakti hai. Safe rehne ke liye row reduce karo.

1. Columns stack karo aur reduce karo. $A=\left[\begin{array}{ccc}1& 2& 4\\ 2& 1& 5\\ 1& 0& 2\end{array}\right]\stackrel{\begin{array}{c}{R}_2-2R_1\\ R_3-R_1\end{array}}{\to }\left[\begin{array}{ccc}1& 2& 4\\ 0& -3& -3\\ 0& -2& -2\end{array}\right]\stackrel{R_3-\frac{2}{3}{R}_2}{\to }\left[\begin{array}{ccc}1& 2& 4\\ 0& -3& -3\\ 0& 0& 0\end{array}\right].$ Yeh step kyun? Ek zero row appear hua — 3 columns ke liye sirf 2 pivots, isliye $\mathrm{rank}=2<3$ ⇒ free variable ⇒ dependent.
2. Relation recover karo. Guess karo ${\mathbf{v}}_3=2{\mathbf{v}}_1+{\mathbf{v}}_2$: $2(1,2,1)+(2,1,0)=(4,5,2)={\mathbf{v}}_3$. ✓ Yeh step kyun? Dependence ${\mathbf{v}}_3=2{\mathbf{v}}_1+{\mathbf{v}}_2$ thi, pehli nazar mein invisible — exactly isliye row reduction eyeballing se behtar hai.

Answer: dependent.

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Example 7 — Cell G (real-world word problem)

Forecast: "Doosron ko mix karke reproduce karna" literally redundant vector ki definition hai. Scan karo: ${\mathbf{s}}_3$ lagta hai ${\mathbf{s}}_1+{\mathbf{s}}_2$ jaisa. Andaza: dependent (ek stock wasteful hai).

1. Story translate karo. "Solution X ko doosron se mix karo" = "X doosron ki linear combination hai" = dependence. Yeh step kyun? $a,b$ amounts se mix karna matlab $a{\mathbf{s}}_1+b{\mathbf{s}}_2$; yeh poochna ki kya yeh ${\mathbf{s}}_3$ ke equal hai, dependency ke baare mein poochna hai.
2. Guess test karo. ${\mathbf{s}}_1+{\mathbf{s}}_2=(2+0,0+4,1+2)=(2,4,3)={\mathbf{s}}_3$. ✓ Yeh step kyun? Ek explicit nontrivial combo (${\mathbf{s}}_1+{\mathbf{s}}_2-{\mathbf{s}}_3=0$) dependence prove karta hai — reduce karne ki zaroorat nahi.

Answer: dependent — ${\mathbf{s}}_3$ exactly $1$ part ${\mathbf{s}}_1$ + $1$ part ${\mathbf{s}}_2$ hai, isliye yeh ek redundant stock hai.

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Example 8 — Cell H (exam twist: ek parameter)

Forecast: Square $3\times 3$ case hai, isliye dependence ⇔ $\det =0$. Ek special value of $k$ hoga. Andaza: $\det (k)=0$ solve karo.

1. Determinant ko $k$ ka function set up karo. $A=\left[\begin{array}{ccc}1& 0& 1\\ 2& 1& 3\\ 3& 4& k\end{array}\right].$ Yeh step kyun? Square matrix ⇒ determinant test valid hai, aur $\det$ ek formula ban jaata hai $k$ mein; dependent values uske roots hain.
2. Row 1 ke saath expand karo. $\det A=1\cdot (1\cdot k-3\cdot 4)-0+1\cdot (2\cdot 4-1\cdot 3)=(k-12)+(8-3)=k-7.$ Yeh step kyun? Cofactor expansion ek $3\times 3$ determinant ko chhote $2\times 2$ waalon mein tod deta hai jo hum evaluate kar sakte hain.
3. $\det =0$ solve karo. $k-7=0\Rightarrow k=7$. Yeh step kyun? $\det =0$ exactly dependence condition hai (columns $<3$ directions mein collapse ho jaate hain).

Answer: set sirf $k=7$ pe dependent hai; har doosre $k$ ke liye independent.

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Case-coverage checklist

Recall Kya humne har cell cover ki?

A (square/det) → Ex 1 ::: ✓ B (parallel degenerate) → Ex 2 ::: ✓ C (non-square, fewer) → Ex 3 ::: ✓ D (more than dimension) → Ex 4 ::: ✓ E (contains zero vector) → Ex 5 ::: ✓ F (hidden sum) → Ex 6 ::: ✓ G (word problem) → Ex 7 ::: ✓ H (parameter twist) → Ex 8 ::: ✓

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Connections

• Determinant — $\det =0$ test jo Cells A, B, H ko power karta hai.
• Rank of a matrix — non-square Cells C, D, F ke liye pivot-counting.
• Homogeneous systems and null space — har test actually yahi poochh raha hai: "kya $A\mathbf{c}=0$ ka nontrivial solution hai?".
• Basis and dimension — Cell D "dimension ke liye bahut zyada" fact hai.
• Span and spanning sets — word problem (Cell G) poochh raha hai ki kya ek stock doosron ke span mein hai.
• Invertible matrix theorem — square cells ke liye, independent ⇔ invertible.