Exercises — Rank of a matrix — definition, row rank = column rank theorem

Kya notice karein: row 1 exactly $2\times$ row 2 hai. Toh dono rows ek hi direction mein point karti hain — sirf ek independent row hai. Isse rank kyun fix hoti hai: row rank = rows ke span ki dimension. Ek direction ⇒ dimension $1$. $\mathrm{rank}(A)=1.$ Columns se check karein: column 2 $=2\times$ column 1, toh column rank $=1$ bhi. theorem hold karta hai. ✔️

Kya notice karein: har column apne axis par already baitha hai — $(1,0,0),(0,1,0),(0,0,1)$ teen perpendicular directions mein point karte hain, toh koi bhi doosron ka mix nahi hai. Kyun: teen independent columns ek $3$-dimensional column space span karte hain. $\mathrm{rank}(I_3)=3.$ Yeh ek $3\times 3$ matrix ke liye full rank hai — dekho Invertible Matrix Theorem (full rank ⇔ invertible).

Kya notice karein: har row aur har column zero vector hai. Bunch of zero vectors ka span sirf ek point $\{0\}$ hota hai, jiski dimension $0$ hai. $\mathrm{rank}(0_{2\times 4})=0.$ Yeh bound $0\le \mathrm{rank}(A)$ ka degenerate floor hai.

Step 1 — column 1 clear karo. $R_2\to R_2-2R_1$, $R_3\to R_3-R_1$. Kyun? Row operations ek row ko rows ke combination se replace karte hain, jo row space kabhi nahi badalta (dekho Gaussian Elimination & Echelon Form), toh rank preserved rehti hai. $\to \left(\begin{array}{ccc}1& 2& 3\\ 0& 0& 0\\ 0& -1& -2\end{array}\right)$ Step 2 — staircase theek karo. $R_2\leftrightarrow R_3$ swap karo taaki zero row neeche aa jaye. $\to \left(\begin{array}{ccc}1& 2& 3\\ 0& -1& -2\\ 0& 0& 0\end{array}\right)$ Step 3 — pivots count karo: positions $(1,1)$ aur $(2,2)$. Do pivots. $\mathrm{rank}(A)=2.$

Step 1: $R_3\to R_3-R_1$. $\to \left(\begin{array}{ccc}1& 0& 2\\ 0& 1& 1\\ 0& 1& 1\end{array}\right)$ Step 2: $R_3\to R_3-R_2$ (duplicate row ko khatam karo). $\to \left(\begin{array}{ccc}1& 0& 2\\ 0& 1& 1\\ 0& 0& 0\end{array}\right)$ Step 3: $(1,1),(2,2)$ par do pivots. $\mathrm{rank}(B)=2.$ Notice karo column 3 $=2\cdot (\text{col }1)+1\cdot (\text{col }2)$ — missing pivot column exactly dependent column hai.

Idea: rank full se tab girta hai jab rows ek doosre ke multiples ban jaati hain. Row 2 $=3\times$ Row 1 ka matlab hai $(3,k)=(3,6)$, yaani $k=6$. Elimination check: $R_2\to R_2-3R_1$ se $(0,k-6)$ milta hai. Woh row zero hai ⇔ $k=6$. $\boxed{k=6}\Rightarrow \mathrm{rank}=1,\text{warna }\mathrm{rank}=2.$ Equivalently determinant $1\cdot k-2\cdot 3=k-6$ $k=6$ par vanish karta hai.

Tool — Rank–Nullity: $\mathrm{rank}(A)+\dim \ker (A)=n$, jahan $n$ columns ki sankhya hai. Dekho Rank–Nullity Theorem. $n$ kyun aur $m$ kyun nahi? Kyunki $x$ ${\mathbb{R}}^n$ mein rehta hai (uske paas multiply hone ke liye har column ke liye ek entry honi chahiye), toh input space jise hum split karte hain uski dimension $n=6$ hai. $\dim \ker (A)=6-3=3.$

Pehle shape: $AB$ $3\times 5$ hai, toh $\mathrm{rank}(AB)\le \min (3,5)=3$. Deeper bound: $AB$ ka har column, $A$ ke columns ka combination hai (dekho Matrix Multiplication as Linear Combination), toh $\mathrm{rank}(AB)\le \mathrm{rank}(A)\le 2$. $AB$ ki rows, $B$ ki rows ke combinations hain, toh bhi $\le \mathrm{rank}(B)\le 2$. $0\le \mathrm{rank}(AB)\le 2.$ Inner dimension $2$ product ko throttle karta hai — shape akele se jo cap mila tha usse genuinely tighter cap hai.

Haan. Ek square $n\times n$ matrix ke liye, $\mathrm{rank}=n$ ka matlab hai saare $n$ columns independent hain, toh woh ${\mathbb{R}}^5$ ka basis form karte hain; Invertible Matrix Theorem ke anusar full rank $\iff$ invertible. $\mathrm{rank}(A)=5=n\Rightarrow A^{-1}\text{ exist karta hai.}$

Step 1 — pivot columns dhundho. Reduce karne par (parent note) pivots columns $1$ aur $3$ mein milte hain. Toh $C$ ke liye woh original columns lo: $C=\left(\begin{array}{cc}1& 1\\ 2& 3\\ 3& 4\end{array}\right)(3\times 2).$ Step 2 — $A$ ke har column ko basis $\{c_1,c_3\}$ mein express karo. Yeh coefficient vectors $R$ ke columns bante hain:

• col $1=1\cdot c_1+0\cdot c_3\Rightarrow (1,0{)}^T$
• col $2=2\cdot c_1+0\cdot c_3\Rightarrow (2,0{)}^T$ (kyunki col $2=2\times$col $1$)
• col $3=0\cdot c_1+1\cdot c_3\Rightarrow (0,1{)}^T$ $R=\left(\begin{array}{ccc}1& 2& 0\\ 0& 0& 1\end{array}\right)(2\times 3).$ Step 3 — verify karo $CR=A$: $CR=\left(\begin{array}{cc}1& 1\\ 2& 3\\ 3& 4\end{array}\right)\left(\begin{array}{ccc}1& 2& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}1& 2& 1\\ 2& 4& 3\\ 3& 6& 4\end{array}\right)=A.✔️$ Shared inner dimension $2$ hi $\mathrm{rank}(A)$ hai.

Strategy: row $3$ ko rows $1$ aur $2$ ka combination banao, rows $1,2$ ko independent rakhte hue, aur koi zero entry avoid karo. Lo $M=\left(\begin{array}{ccc}1& 1& 2\\ 1& 2& 1\\ 2& 3& 3\end{array}\right),R_3=R_1+R_2.$ Rank $2$ kyun: rows $1,2$ independent hain (multiples nahi — pehli do entries mein $1:1$ vs $1:2$ compare karo), toh rank $\ge 2$; row $3$ kuch add nahi karta kyunki $R_3=R_1+R_2$, toh rank $\le 2$. Isliye exactly $2$. Saari nau entries nonzero hain. ✔️

Setup — transpose kya karta hai: $A^T$, $A$ ki rows ko columns mein aur columns ko rows mein badalta hai. Isliye $A^T$ ke columns literally $A$ ki rows hain, toh $\text{column rank}(A^T)=\text{row rank}(A).$ Ab central theorem $A^T$ par apply karo: kisi bhi matrix ke liye, row rank = column rank, isliye $\mathrm{rank}(A^T)=\text{column rank}(A^T)=\text{row rank}(A).$ Lekin theorem $A$ par applied kehta hai $\text{row rank}(A)=\mathrm{rank}(A)$. Chain karke: $\mathrm{rank}(A^T)=\mathrm{rank}(A).■$

Rank $0$: lo $B=-A$ jahan $A=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$. Tab $A+B=0$, rank $0$. (Overlap perfectly cancel ho jaata hai.) Rank $1$: lo $A=B=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$. Tab $A+B=2A$, phir bhi rank $1$. Rank $2$: lo $A=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$, $B=\left(\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)$. Tab $A+B=\mathrm{diag}(1,1,0)$, rank $2$. Conclusion: teeno $0,1,2$ occur hote hain, toh $\mathrm{rank}(A+B)=\mathrm{rank}A+\mathrm{rank}B$ generally false hai. Sirf inequality $\mathrm{rank}(A+B)\le \mathrm{rank}A+\mathrm{rank}B=2$ survive karta hai, aur woh tight hai (rank $2$ case). ✔️

Column argument: column space, $A$ ke $n$ columns se span hota hai, har ek ${\mathbb{R}}^m$ mein rehta hai. $n$ vectors ke span ki dimension $\le n$ hoti hai; ${\mathbb{R}}^m$ ke subspace ki dimension $\le m$ hoti hai. Toh $r\le n$ aur $r\le m$, jisse $r\le \min (m,n)$ milta hai (uses Linear Independence and Basis). Limiting cases:

• Agar $m\le n$ aur $r=m$: rows independent hain, $A$ ka full row rank hai — map $A$ ${\mathbb{R}}^m$ par onto hai.
• Agar $n\le m$ aur $r=n$: columns independent hain, $A$ ka full column rank hai — map one-to-one hai, $\ker (A)=\{0\}$ (Rank–Nullity se nullity $0$).
• Degenerate floor $r=0$ ⇔ $A=0$. $0\le \mathrm{rank}(A)\le \min (m,n).■$