Exercises — Invertible matrix theorem — 12+ equivalent conditions

Hum kya karenge: ek $2\times 2$ matrix ka determinant compute karenge. $\det A=(4)(3)-(2)(6)=12-12=0.$ Yeh tool kyun: IMT #13 kehta hai invertible $⟺\det \ne 0$, aur ek $2\times 2$ ke liye determinant sabse fast check hai. Kyunki $\det A=0$ hai, $A$ invertible NAHI hai. (Dhyan do row 2 $=1.5\times$ row 1 hai — rows dependent hain, jo $\det =0$ se consistent hai.)

Hum kya karenge: ek triangular matrix (diagonal ke neeche sab zeros) ka determinant uske diagonal entries ke product ke barabar hota hai. $\det A=1\cdot 2\cdot 3=6\ne 0.$ Yeh tool kyun: IMT #13 phir se — aur triangular matrices ke liye determinant sirf diagonal product hai, isliye koi row reduction ki zaroorat nahi. $\det A=6\ne 0\Rightarrow$ $A$ invertible HAI.

Hum kya karenge: kuch bhi karne se pehle IMT ki hypothesis check karo. IMT sirf square matrices par apply hota hai. Ek $3\times 5$ matrix square nahi hai, isliye "invertible" (two-sided inverse) word yahan lagta hi nahi. Aur baat yeh hai ki ${\mathbb{R}}^3$ mein 5 columns independent nahi ho sakte — tum 3-dimensional space mein 3 se zyada independent vectors nahi fit kar sakte (dekho Basis and Dimension). Toh yeh premise actually impossible hai. Answer: invertible NAHI — IMT non-square matrices par apply nahi hota.

Hum kya karenge: $3\times 3$ determinant ko first row ke along expand karenge. $\det A=2\det \left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)-1\det \left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)+0.$ Har $2\times 2$ compute karo: $\det \left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)=4-1=3$, aur $\det \left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)=2-0=2$. $\det A=2(3)-1(2)+0=6-2=4.$ Yeh tool kyun: cofactor expansion ek $3\times 3$ ko chote $2\times 2$'s mein tod deta hai; position $(1,3)$ par zero hone se last term zero ho jata hai, kaam bach jata hai. $\det A=4\ne 0\Rightarrow$ invertible.

Hum kya karenge: columns mein ek linear dependence dhundhenge. Pehle do columns $(1,2,3)$ aur $(2,4,6)=2\times$ column 1 hain. Toh column $2-2\times$ column $1=0$, yaani weights $(-2,1,0)$ columns ko zero mein combine karte hain. Yeh tool kyun: matrix–vector product $Ax$ columns ka linear combination hai jisme $x$ weights hain (IMT Link B). Columns mein dependence exactly ek null vector hai. $x=(-2,1,0{)}^{\mathsf{T}}$ se directly check karo: $Ax=\left(\begin{array}{c}-2+2+0\\ -4+4+0\\ -6+6+0\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right).✓$ Ek nonzero null vector exist karta hai $\Rightarrow$ IMT #4 fail $\Rightarrow$ invertible NAHI.

Hum kya karenge: elementary row operations apply karenge (dekho Elementary Matrices and Row Reduction). $\left(\begin{array}{cc}1& 2\\ 3& 5\end{array}\right)\stackrel{R_2\to R_2-3R_1}{\to }\left(\begin{array}{cc}1& 2\\ 0& -1\end{array}\right)\stackrel{R_2\to -R_2}{\to }\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\stackrel{R_1\to R_1-2R_2}{\to }\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I.$ Yeh tool kyun: IMT #2/#3 — har row aur column mein pivot ke saath $I$ tak pahunchna invertibility ka direct certificate hai. Do pivots, RREF $=I\Rightarrow$ invertible. Cross-check: $\det =1\cdot 5-2\cdot 3=-1\ne 0.✓$

Hum kya karenge: invertibility tab fail hoti hai jab $\det A=0$, isliye $\det A=0$ solve karo aur un $k$ ko exclude karo. $\det A=(1)(4)-(k)(k)=4-k^2.$ Yeh tool kyun: IMT #13. Determinant ko $k$ ka function maanne par "invertible?" ek simple equation ban jaata hai. $4-k^2=0$ set karo $\Rightarrow k^2=4\Rightarrow k=\pm 2$. Toh $A$, $k=2$ ya $k=-2$ ke liye invertible NAHI hai, aur baaki sabhi $k$ ke liye invertible hai, yaani $k\in \mathbb{R}\setminus \{-2,2\}$.

Hum kya karenge: $\det A$ ko $t$ ka function compute karo, phir uske roots dhundho. First row ke along expand karo: $\det A=1\det \left(\begin{array}{cc}2& 3\\ 4& t\end{array}\right)-1\det \left(\begin{array}{cc}1& 3\\ 1& t\end{array}\right)+1\det \left(\begin{array}{cc}1& 2\\ 1& 4\end{array}\right).$ Har minor: $\det \left(\begin{array}{cc}2& 3\\ 4& t\end{array}\right)=2t-12$, $\det \left(\begin{array}{cc}1& 3\\ 1& t\end{array}\right)=t-3$, $\det \left(\begin{array}{cc}1& 2\\ 1& 4\end{array}\right)=4-2=2$. $\det A=(2t-12)-(t-3)+2=2t-12-t+3+2=t-7.$ Yeh tool kyun: IMT #13; determinant ko linear expression $t-7$ ki tarah likhne se singular case obvious ho jaata hai. $\det A=0⟺t=7$. Toh $A$ sirf $t=7$ par singular hai, aur baaki har $t$ ke liye invertible hai.

Hum kya karenge: eigenvalues aur invertibility ke beech link use karenge (dekho Eigenvalues and Eigenvectors). Yeh tool kyun: IMT #14 — $A$ invertible hai $⟺0$ eigenvalue nahi hai. Yahan $0$ listed eigenvalue hai. Isliye $A$ invertible NAHI hai. $\det =$ eigenvalues ka product se confirm karo: $\det A=2\cdot (-1)\cdot 0=0,$ IMT #13 se consistent. Zero eigenvalue $\lambda =0$ ek nonzero $x$ deta hai jiske liye $Ax=0x=0$ — ek nontrivial null vector, toh IMT #4 bhi fail hota hai.

Hum kya karenge: dikhao ki map $x\mapsto Ax$ onto hai, phir squareness use karke full invertibility tak jump karo. Step 1 (surjective). Kisi bhi $b\in {\mathbb{R}}^n$ ke liye, $x=Db$ set karo. Tab $Ax=A(Db)=(AD)b=Ib=b$. Toh har $b$ ka ek preimage hai — map onto hai (IMT #9), equivalently $Ax=b$ sabhi $b$ ke liye solvable hai (IMT #7). Yeh step kyun: $AD=I$ literally humein ek formula $x=Db$ deta hai jo $Ax=b$ solve karta hai. Step 2 (IMT climb). Ek square matrix ke liye, onto (#9) equivalence loop mein hai, toh sab conditions hold hoti hain — particularly $A$ invertible hai (#1). Invertibility ek genuine $A^{-1}$ deta hai jisme $A^{-1}A=A{A}^{-1}=I$. Step 3 ($D$ identify karo). $AD=I$ ko left mein $A^{-1}$ se multiply karo: $A^{-1}AD=A^{-1}\Rightarrow D=A^{-1}$. Isliye $DA=A^{-1}A=I$. $■$ Key point: squareness hi woh cheez hai jo ek one-sided inverse ko two-sided banata hai (IMT #10, #11).

Hum kya karenge: $A$ aur $B$ ke liye one-sided inverses dikhao, phir Exercise 4.1 ka principle apply karo. Maano $M=(AB{)}^{-1}$, toh $(AB)M=I$ aur $M(AB)=I$. $B$ invertible hai: $M(AB)=I$ se hum $(MA)B=I$ padhte hain, $B$ ke liye ek left inverse. Square $B$ ke liye, IMT #10 $\Rightarrow B$ invertible. $A$ invertible hai: $(AB)M=I$ se hum $A(BM)=I$ padhte hain, $A$ ke liye ek right inverse. Square $A$ ke liye, IMT #11 $\Rightarrow A$ invertible. Yeh tool kyun: humne kabhi inverses compute nahi kiye; bas one-sided inverses dhundhe aur IMT (Ex. 4.1 ke zariye) ko unhe upgrade karne diya. $■$

Hum kya karenge: invertibility ko determinant ke do facts use karke transpose ke across transfer karenge. Fact: $\det (A^{\mathsf{T}})=\det A$ (transpose karne se determinant nahi badalta). Kyunki $A$ invertible $\Rightarrow \det A\ne 0$ (IMT #13), hum paate hain $\det (A^{\mathsf{T}})=\det A\ne 0$. IMT #13 ko $A^{\mathsf{T}}$ par apply karne se, $A^{\mathsf{T}}$ invertible hai (yeh IMT #12 hai). $■$ Yeh tool kyun: determinant ek single number hai jo transpose ke "through" dekh sakta hai, isliye poora IMT machinery instantly carry over ho jaata hai.

Hum kya karenge: null space, pivots aur rank ko connect karenge (dekho Rank and Nullity). ($\Rightarrow$) Maano $Ax=0$ mein $x=0$ force hota hai. Har non-pivot (free) column ek free variable aur isliye ek nonzero solution create karta; kyunki koi exist nahi karta, koi free column nahi hai. Toh sab $n$ columns pivot columns hain $\Rightarrow$ rank $=$ pivots ki sankhya $=n$. ($\Leftarrow$) Maano rank $=n$. Tab $n$ pivots hain, har column mein ek, toh koi free variable nahi hai. Bina free variables wala homogeneous system ka unique solution $x=0$ hota hai. Rank–Nullity check: $\dim \mathrm{Nul}A=n-\mathrm{rank}A=n-n=0$, "trivial null space" se match karta hai (IMT #17). Geometry ($n=2$): $A$ ke columns plane mein do vectors hain.

Hum kya karenge: pivots count karenge, yeh use karte hue ki ek square matrix mein exactly $n$ rows aur $n$ columns hote hain. Spanning $\Rightarrow$ har row mein pivot. Agar columns ${\mathbb{R}}^n$ span karte hain, toh map $x\mapsto Ax$ onto hai (IMT #9), jiske liye har row mein pivot chahiye — warna koi $b$ unreachable hota. Woh $n$ pivots hain (har row mein ek). $n$ pivots $\Rightarrow$ har column mein pivot. $n$ columns mein $n$ pivots spread hone se, har column mein exactly ek pivot hota hai; koi free column nahi hoti. Free column nahi $\Rightarrow$ independent (IMT #5, via Link B). $■$ Yeh sirf squares ke liye kyun kaam karta hai: argument yeh use karta hai ki "$n$ pivots dono sabhi rows AND sabhi columns fill karte hain" — possible tab hi jab rows $=$ columns. Ek $2\times 3$ matrix ke liye columns ${\mathbb{R}}^2$ span kar sakte hain sirf 2 pivots ke saath, ek free column chod ke, toh woh independent nahi hote.

(a) Singularity ke teen certificates:

• Determinant (IMT #13): $\det A=(1)(4)-(2)(2)=4-4=0.$
• Null vector (IMT #4): column $2=2\times$ column $1$, toh $x=(2,-1{)}^{\mathsf{T}}$ se $Ax=(1\cdot 2+2\cdot (-1),2\cdot 2+4\cdot (-1){)}^{\mathsf{T}}=(0,0{)}^{\mathsf{T}}$ milta hai.
• Rank (IMT #15): row 2 $=2\times$ row 1, toh $R_2\to R_2-2R_1$ ke baad hum $\left(\begin{array}{cc}1& 2\\ 0& 0\end{array}\right)$ paate hain — ek pivot, toh rank $=1<2$. Teen tareekon se kyun: IMT ka poora point hi yeh hai ki yeh sab ek hi fact hain; yahan hum ek hi failure ko teen alag-alag costumes mein dekhte hain. (b) Geometry:

  

  Dono columns $(1,2)$ aur $(2,4)$ ek hi line $y=2x$ par lie karte hain. Har input vector us ek line par map ho jaata hai — map 2D plane ko 1D line par collapse karta hai. Poori perpendicular direction $(2,-1)$ origin par squash ho jaati hai (yahi null vector hai). Collapse information destroy karta hai, toh transformation ko undo nahi kiya ja sakta: invertible nahi. $■$

Yahan har exercise ek pivot count par reduce hoti hai — determinant, rank, null space, eigenvalues, spanning aur independence sirf alag-alag lenses hain is sawaal par ki "kya $A$ ke $n$ pivots hain?"