Exercises — Linear independence — formal definition, testing

• ${\mathbf{v}}_1,\dots ,{\mathbf{v}}_n$ — hamare vectors (arrows / numbers ki lists).
• $c_1,\dots ,c_n$ — unknown scalars (plain numbers) jinhe hum vectors se multiply karte hain.
• $0$ — zero vector (length zero ka arrow; saare zeros ki list).
• $A\mathbf{c}=0$ — vectors ko matrix $A$ ke columns ki tarah stack karo; phir $A\mathbf{c}=c_1{\mathbf{v}}_1+\cdots +c_n{\mathbf{v}}_n$. "Yeh $0$ kab hai?" poochna hi independence test hai.
• $\mathrm{rank}(A)$ — row reduction ke baad pivots ki sankhya = genuinely independent directions ki sankhya.
• $\det A$ — ek single number, sirf square $A$ ke liye defined; zero matlab columns flat ho gaye (dependent).

Independent. $(1,0)$ purely East ki taraf point karta hai, $(0,1)$ purely North — na hi doosre ka scalar multiple hai, isliye koi redundant nahi hai. Algebraically $c_1(1,0)+c_2(0,1)=(c_1,c_2)=0$ se $c_1=c_2=0$ aata hai: sirf trivial (boring) solution.

Dependent. Dekho: $(2,4,6)=2\cdot (1,2,3)$. Yeh ek visible scalar multiple hai. Toh $1\cdot (2,4,6)-2\cdot (1,2,3)=0$ ek nontrivial combo hai (coefficient $1$ zero nahi hai). Dependent.

Dependent. $0$ wala koi bhi set dependent hota hai, kyunki $1\cdot (0,0)+0\cdot (1,5)+0\cdot (2,9)=0$ mein coefficient $c_1=1\ne 0$ hai — zero tak pahunchne ka ek nontrivial tarika.

$A=\left[\begin{array}{cc}1& 3\\ 2& 5\end{array}\right],\det A=1\cdot 5-3\cdot 2=5-6=-1.$ $\det A=-1\ne 0$, toh independent. Yahan determinant kyun? Plane mein do vectors ke liye, $\det A$ unke span karne wale parallelogram ka signed area hai (neeche Figure). Zero area matlab woh ek line par hain (parallel = dependent); nonzero area matlab woh genuinely 2D region kholte hain.

$\det \left[\begin{array}{cc}2& -4\\ 1& -2\end{array}\right]=2\cdot (-2)-(-4)\cdot 1=-4+4=0.$ $\det A=0\Rightarrow$ dependent. Waise bhi $(-4,-2)=-2\cdot (2,1)$: dono arrows ek hi line par hain, toh parallelogram ek line segment tak squash ho jaata hai — zero area (figure dekho).

$A=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right]\stackrel{R_3-R_1}{\to }\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 1\end{array}\right]\stackrel{R_3-R_2}{\to }\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right].$ Do pivots (columns 1 aur 2), do columns ke liye. Har column mein pivot hai $\Rightarrow$ koi free variable nahi $\Rightarrow$ sirf $\mathbf{c}=0$. Independent.

$A=\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 1\\ 2& 1& 5\end{array}\right]\stackrel{R_3-2R_1}{\to }\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 1\\ 0& 1& 1\end{array}\right]\stackrel{R_3-R_2}{\to }\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 1\\ 0& 0& 0\end{array}\right].$ Teen columns ke liye do pivots → ek free variable → dependent. Combo dhundho: $A\mathbf{c}=0$ solve karo. Reduced rows se, $c_1+2c_3=0$ aur $c_2+c_3=0$. $c_3=1$ lo: toh $c_1=-2,c_2=-1$. Check: $-2(1,0,2)-1(0,1,1)+1(2,1,5)=(-2+2,-1+1,-4-1+5)=(0,0,0).✓$ Toh ${\mathbf{v}}_3=2{\mathbf{v}}_1+{\mathbf{v}}_2$ — redundant passenger.

Dependent $⟺\det A=0$: $\det \left[\begin{array}{cc}1& 3\\ 2& k\end{array}\right]=k-6=0\Rightarrow k=6.$ $k=6$ par, ${\mathbf{v}}_2=(3,6)=3(1,2)=3{\mathbf{v}}_1$ — parallel, dependent. Har doosre $k$ ke liye, independent.

$4$ vectors ko columns ki tarah stack karo: $A$ $3\times 4$ hai. Row reduction zyada se zyada ek pivot per row deta hai, aur sirf $3$ rows hain, toh $\mathrm{rank}(A)\le 3$. Lekin $4$ columns ki independence ke liye har ek of the $4$ columns mein pivot chahiye, yaani $\mathrm{rank}=4$. Kyunki $3<4$, yeh impossible hai: ek free variable hamesha exist karta hai → dependent. Maximum independent count $=$ rows ki sankhya $=$ dimension $=3$. (Isliye $\dim {\mathbb{R}}^3=3$; dekho Basis and dimension.)

Square case hai, toh determinant use karo. Columns ${\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3$ ke saath: $A=\left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 1& t\end{array}\right].$ Pehli row ke along expand karo: $\det A=1\cdot (0\cdot t-1\cdot 1)-1\cdot (1\cdot t-1\cdot 0)+0=(-1)-(t)=-1-t.$ $\det A\ne 0⟺-1-t\ne 0⟺t\ne -1$. Toh ye saare $t\ne -1$ ke liye basis banate hain. $t=-1$ par, $\det A=0$: dependent, basis nahi. Yeh pura answer kyun hai: ${\mathbb{R}}^n$ mein $n$ vectors ke liye, invertible $⟺$ independent $⟺$ spanning $⟺$ basis. Ek determinant sab kuch settle kar deta hai.

(a) Nahi. $5$ columns ki independence ke liye $5$ pivots chahiye, lekin rank $=3$ sirf $3$ pivots deta hai → $2$ free variables → nontrivial solutions → dependent. (b) Ek maximal independent subset ka size $=\mathrm{rank}=3$ columns (pivot columns). (c) Rank–nullity idea se, $\dim (\text{null space})=(\#\text{columns})-\mathrm{rank}=5-3=2$. Do free variables ⇔ nontrivial combinations ka $2$-dimensional space. (Dekho Homogeneous systems and null space.)

Row reduce karo. $\stackrel{R_2-2R_1,R_3-R_1}{\to }\left[\begin{array}{cccc}1& 2& 3& 1\\ 0& 0& 0& -2\\ 0& 0& 0& 4\end{array}\right]\stackrel{R_3+2R_2}{\to }\left[\begin{array}{cccc}1& 2& 3& 1\\ 0& 0& 0& -2\\ 0& 0& 0& 0\end{array}\right].$ Pivots column 1 aur column 4 mein hain. (a) $\mathrm{rank}(A)=2$. (b) Dependent ($2$ pivots $<4$ columns). (c) Pivot columns maximal independent subset dete hain: $\{{\mathbf{v}}_1=(1,2,1),{\mathbf{v}}_4=(1,0,5)\}$. (d) Non-pivot columns ${\mathbf{v}}_2,{\mathbf{v}}_3$ ${\mathbf{v}}_1$ ke combinations hain. Directly: ${\mathbf{v}}_2=(2,4,2)=2{\mathbf{v}}_1$ aur ${\mathbf{v}}_3=(3,6,3)=3{\mathbf{v}}_1$ (har ek $0\cdot {\mathbf{v}}_4$ use karta hai). Check: $2\cdot (1,2,1)=(2,4,2)={\mathbf{v}}_2.✓$

Maano $a({\mathbf{v}}_1+{\mathbf{v}}_2)+b({\mathbf{v}}_2+{\mathbf{v}}_3)+c({\mathbf{v}}_1+{\mathbf{v}}_3)=0$. Original vector ke hisaab se group karo: $(a+c){\mathbf{v}}_1+(a+b){\mathbf{v}}_2+(b+c){\mathbf{v}}_3=0.$ Kyunki $\{{\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3\}$ independent hai, har coefficient $0$ hona chahiye: $a+c=0,a+b=0,b+c=0.$ Pehle do se, $c=-a$ aur $b=-a$. Teesre mein daalo: $(-a)+(-a)=-2a=0\Rightarrow a=0$, isliye $b=c=0$. Sirf trivial solution → naya set independent hai. $■$ (System $a+c=a+b=b+c=0$ ka coefficient matrix determinant $\det \left[\begin{array}{ccc}1& 0& 1\\ 1& 1& 0\\ 0& 1& 1\end{array}\right]=2\ne 0$ hai, jo sirf trivial solution confirm karta hai.)

Coordinate vectors (order $[\text{const},x,x^2]$ mein): $p_1=(1,1,0),p_2=(0,1,1),p_3=(1,0,1)$. Yahi setup hai jo L5.2 ke implicit matrix mein tha. Determinant: $\det \left[\begin{array}{ccc}1& 0& 1\\ 1& 1& 0\\ 0& 1& 1\end{array}\right]=1(1\cdot 1-0\cdot 1)-0+1(1\cdot 1-1\cdot 0)=1+1=2\ne 0.$ $\det \ne 0\Rightarrow$ independent. Linear independence ek coordinate statement hai — yeh polynomials ke liye bhi exactly same kaam karta hai jab ek baar tumne coefficients vector ki tarah list kar liye.