Exercises — Subspaces — four fundamental subspaces of a matrix

KYA/KYU: dimensions seedha dimension formulas $r,r,n-r,m-r$ se aati hain jahan $m=3$, $n=5$, $r=2$.

• $\dim C(A)=r=2$, output space ${\mathbb{R}}^3$ mein rehta hai.
• $\dim C(A^T)=r=2$, input space ${\mathbb{R}}^5$ mein rehta hai.
• $\dim N(A)=n-r=5-2=3$, input space ${\mathbb{R}}^5$ mein rehta hai.
• $\dim N(A^T)=m-r=3-2=1$, output space ${\mathbb{R}}^3$ mein rehta hai. Sums check karo: input pieces $2+3=5=n$ ✓; output pieces $2+1=3=m$ ✓.

False. $\dim N(A)=n-r$ aur $\dim C(A)=r$; ye tab barabar hote hain jab $n-r=r$, yaani $n=2r$. Generally ye differ karte hain (ye to alag-alag spaces ${\mathbb{R}}^n$ vs ${\mathbb{R}}^m$ mein rehte hain). Sahi "same dimension" pairing hai row space aur column space, dono $=r$.

Step 1 — Row reduce karo (KYU: pivots rank aur free variables reveal karte hain). $R_2\to R_2-2R_1$: $\left[\begin{array}{ccc}1& 2& 2\\ 0& 0& 1\end{array}\right]$. Pivot 2 ke upar clear karo: $R_1\to R_1-2R_2$: $\left[\begin{array}{ccc}1& 2& 0\\ 0& 0& 1\end{array}\right]$. Pivots columns $1$ aur $3$ mein hain, toh $r=2$. Step 2 — Column space (KYU: original $A$ ke pivot columns $C(A)$ span karte hain). $C(A)=\text{span}\{(1,2),(2,5)\}$. ${\mathbb{R}}^2$ mein do independent vectors, toh $C(A)={\mathbb{R}}^2$, $\dim =2$. Step 3 — Null space (KYU: free variable = column 2). RREF se: $x_1+2x_2=0$ aur $x_3=0$. Free $x_2=1$: $x_1=-2,x_3=0$. Special solution $(-2,1,0)$. $N(A)=\text{span}\{(-2,1,0)\}$, $\dim =n-r=3-2=1$. Sanity: $2+1=3=n$ ✓.

Step 1 — Row reduce karo. $R_2\to R_2-2R_1$, $R_3\to R_3+R_1$: $\left[\begin{array}{cc}1& 3\\ 0& 0\\ 0& 0\end{array}\right]$. Ek pivot, $r=1$ ($m=3,n=2$). Column space: pivot column 1 hai $\Rightarrow C(B)=\text{span}\{(1,2,-1)\}$, ${\mathbb{R}}^3$ mein ek line, $\dim =1$. Row space: nonzero RREF row $\Rightarrow C(B^T)=\text{span}\{(1,3)\}$, ${\mathbb{R}}^2$ mein ek line, $\dim =1$. Null space: $x_1+3x_2=0$, free $x_2=1\Rightarrow x_1=-3$. $N(B)=\text{span}\{(-3,1)\}$, $\dim =n-r=2-1=1$. Orthogonality check (input space ${\mathbb{R}}^2$): row $(1,3)$ ko null $(-3,1)$ se dot karo $=1(-3)+3(1)=0$ ✓ — bilkul wahi perpendicularity jaise Orthogonal Complements mein hai.

Observe karo (pehle KYU): rows hain $(1,2,3),(1,2,3),(2,4,6)$ — row 2, row 1 ke barabar hai, row 3, row 1 ka $2\times$ hai. Toh $r=1$. (a) Left null space. $N(A^T)=\{y:A^{T}y=0\}$, equivalently rows ke woh combinations jo zero dete hain. Chahiye $y_1{r}_1+y_2{r}_2+y_3{r}_3=0$ jahan saari rows $(1,2,3)$ ke multiples hain: $(y_1+y_2+2y_3)(1,2,3)=0\Rightarrow y_1+y_2+2y_3=0.$ Ek equation, $3$ unknowns $\Rightarrow$ do free directions, $\dim =m-r=3-1=2$ ✓. Free $(y_2,y_3)$: $y_2=1,y_3=0\Rightarrow (-1,1,0)$; $y_2=0,y_3=1\Rightarrow (-2,0,1)$. $N(A^T)=\text{span}\{(-1,1,0),(-2,0,1)\}$. (b) Solvability. Solving Ax=b ke hisaab se, $Ax=b$ solvable hai $⟺b\in C(A)⟺b\perp N(A^T)$. Toh $b$ ko $N(A^T)$ ke har basis vector se dot karo aur zero set karo: $-b_1+b_2=0$ aur $-2b_1+b_3=0$, yaani $\boxed{b_2=b_1,b_3=2b_1}$. Read-off: $b$ ek $(1,1,2)$ ka multiple hona chahiye — jo exactly woh single column direction hai, jaise rank $1$ ke liye expected hai.

Rank: $\dim N(A)=n-r=3-r=0\Rightarrow r=3$ (full column rank). Left null space: $\dim N(A^T)=m-r=5-3=2$. Yeh nonzero hai, toh forbidden output directions exist karte hain. Har $b$ ke liye solvable? Nahi — $\dim C(A)=r=3<5=m$, toh column space ${\mathbb{R}}^5$ ke andar ek 3-plane hai. Sirf woh $b$ jo us 3-plane mein hai (equivalently $b\perp N(A^T)$) reachable hai. Yahi woh setting hai jahan Least Squares & Projections best approximate answer dhundhne ke liye aata hai.

Step 1 — Reduce karo. $R_2\to R_2-R_1$, $R_3\to R_3-R_1$: $\left[\begin{array}{cc}1& 1\\ 0& 1\\ 0& 2\end{array}\right]$; phir $R_3\to R_3-2R_2$: $\left[\begin{array}{cc}1& 1\\ 0& 1\\ 0& 0\end{array}\right]$; upar clear karo: $\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$. Do pivots, $r=2$ ($m=3,n=2$). Column space: original $A$ ke pivot columns $1,2$: $C(A)=\text{span}\{(1,1,1),(1,2,3)\}$, ${\mathbb{R}}^3$ mein ek plane, $\dim 2$. Row space: $C(A^T)=\text{span}\{(1,0),(0,1)\}={\mathbb{R}}^2$, $\dim 2$. Null space: $r=n=2\Rightarrow N(A)=\{0\}$, $\dim 0$. Left null space: $\dim =m-r=3-2=1$. $y_1(1,1,1)+y_2(1,2,3)=0$ solve karo jo $A^T$ ke columns across padha jaata hai: $y_1+y_2+y_3=0$ aur $y_1+2y_2+3y_3=0$. Subtract karo: $y_2+2y_3=0\Rightarrow y_2=-2y_3$; phir $y_1=-y_2-y_3=2y_3-y_3=y_3$. $y_3=1$ lo: $y=(1,-2,1)$. $N(A^T)=\text{span}\{(1,-2,1)\}$. (b) Perpendicular check. Left-null vector ko har column se dot karo: $(1,-2,1)\cdot (1,1,1)=1-2+1=0$ ✓; $(1,-2,1)\cdot (1,2,3)=1-4+3=0$ ✓. Toh line $N(A^T)$, plane $C(A)$ ka normal hai — neeche figure dekho. (c) $b=(1,0,0)$ ka direction split. Kyunki $C(A)\oplus N(A^T)={\mathbb{R}}^3$ aur woh perpendicular hain, $b$ uniquely split hota hai $b=b_{\parallel }+b_{\perp }$ jahan $b_{\parallel }\in C(A)$ ($(1,1,1),(1,2,3)$ ka combination) aur $b_{\perp }$, $(1,-2,1)$ ka multiple hai. Is clean split ka sirf exist karna hi Fundamental Theorem, Part 2 hai — woh machinery jo Least Squares & Projections ke peeche hai.

Plan (KYU): column space $=$ ek vector ka span matlab har column $(1,2,2)$ ka multiple hai, toh $A=(1,2,2{)}^T{v}^T$ kisi row vector $v=(v_1,v_2,v_3)$ ke liye (ek rank-1 matrix). Null space phir woh saare $x$ hain jahan $v\cdot x=0$. Hume $v$ aisa choose karna hai taaki $N=\{x:v\cdot x=0\}$ given plane ke barabar ho — yaani $v$ us plane ka normal ho. $v$ dhundho (KYU: normal ⟂ dono plane directions): chahiye $v\cdot (1,-1,0)=0$ aur $v\cdot (1,0,-1)=0$: $v_1-v_2=0$ aur $v_1-v_3=0\Rightarrow v=(1,1,1)$. Assemble karo: $A=\left[\begin{array}{c}1\\ 2\\ 2\end{array}\right]\left[\begin{array}{ccc}1& 1& 1\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 2& 2& 2\\ 2& 2& 2\end{array}\right].$ Verify karo. Columns saari $=(1,2,2)$: $C(A)=\text{span}\{(1,2,2)\}$, $\dim 1=r$ ✓. $Ax=0⟺x_1+x_2+x_3=0$, jiska solution space exactly $\text{span}\{(1,-1,0),(1,0,-1)\}$ hai, $\dim 2=n-r$ ✓. $\dim C(A^T)=r=1$; $\dim N(A^T)=m-r=3-1=2$. Sums $1+2=3$ dono sides ✓.

Strategy: null spaces match karo, $N(A^{T}A)=N(A)$, phir orthogonal complements ke zariye convert karo. Step 1 — $N(A)\subseteq N(A^{T}A)$: agar $Ax=0$ toh $A^{T}Ax=A^{T}0=0$. Seedha. Step 2 — $N(A^{T}A)\subseteq N(A)$ (clever half): maano $A^{T}Ax=0$. $x$ se dot karo: $x^T{A}^{T}Ax=0\Rightarrow (Ax{)}^T(Ax)=0\Rightarrow \Vert Ax{\Vert }^2=0\Rightarrow Ax=0.$ KYU yeh kaam karta hai: $\Vert Ax{\Vert }^2$ squares ka sum hai; yeh zero tab hi hota hai jab vector khud zero ho — yahi woh ek fact hai jo $Ax$ ko pin down karta hai. Toh $N(A^{T}A)=N(A)$, dono ${\mathbb{R}}^n$ ke subspaces. Step 3 — row spaces ki taraf flip karo. ${\mathbb{R}}^n$ mein, row space null space ka orthogonal complement hai: $C(M^T)=N(M{)}^{\perp }$. Kyunki $A^{T}A$ symmetric hai, iska row space iske column space ke barabar hai, aur: $C(A^{T}A)=N(A^{T}A{)}^{\perp }=N(A{)}^{\perp }=C(A^T).$ Consequence: $\text{rank}(A^{T}A)=\dim C(A^T)=\text{rank}(A)$. ∎ Yahi woh exact fact hai jo normal equations $A^{T}A\hat{x}=A^{T}b$ ko Least Squares & Projections mein solvable banata hai.

Answers chhupaao. $m\times n$ rank $r$ ke liye $N(A)$ ki dimension? ::: $n-r$. Left null space kis space mein rehta hai aur kis ke perpendicular hota hai? ::: ${\mathbb{R}}^m$, $C(A)$ ke perpendicular. $N(A^T)$ ke terms mein $Ax=b$ solvability condition? ::: $b\perp N(A^T)$ (equivalently $b\in C(A)$). $A^{T}A$ ka rank versus $A$ ka rank? ::: Equal.