Question bank — QR decomposition

True — yeh defining property hai, columns construction se orthonormal hain, isliye $Q$ ki Gram matrix identity hai.

False thin QR ke liye — $Q{Q}^{\top }$ ek $m\times m$ projection hai $A$ ke column space par, identity nahi, jab tak $m=n$ na ho (full/square case).

Weak sense mein True: $A=QR$ kisi bhi $A$ ke liye exist karta hai, lekin positive diagonal ke saath uniqueness ke liye linearly independent columns chahiye; dependent columns ke saath kuch $r_{kk}=0$ ho jaata hai aur $Q$ fully determined nahi hota.

False — yeh forced hai: ${\mathbf{a}}_k$ sirf ${\mathbf{q}}_1,\dots ,{\mathbf{q}}_k$ se bana hota hai, isliye koi bhi coefficient $r_{ik}$ jahan $i>k$ ho, zero hona hi chahiye, yahi upper-triangularity hai.

False — independence ka matlab hai ${\mathbf{u}}_k\ne 0$ har step par, isliye $r_{kk}=\Vert {\mathbf{u}}_k\Vert >0$; zero hona linear dependence signal karega.

True — Gram–Schmidt paata hai ki har column pehle se unit-length aur perpendicular hai, isliye har $r_{kk}=1$ aur saare off-diagonals vanish ho jaate hain, giving $Q=A$, $R=I$.

True — ${\mathbf{a}}_1$ ko scale karne se sirf ${\mathbf{u}}_1$ scale hota hai; normalize karne ke baad, ${\mathbf{q}}_1$ same rahta hai, halaanki $r_{11}$ (aur $R$ ki pehli row) double ho jaati hai.

False — LU mein do triangular factors hote hain; QR ek orthogonal factor ko ek triangular ke saath pair karta hai, isliye yeh sirf trivial cases mein coincide karte hain, generically kabhi nahi.

True — yeh triangular hai aur saare diagonal entries $>0$ hain, aur triangular matrix ka determinant diagonal ka product hota hai, jo yahan nonzero hai.

False — columns reorder karne se yeh badal jaata hai ki kaun kis ke against orthogonalize hota hai, isliye $Q$ aur $R$ dono poori tarah se change ho sakte hain; QR order-dependent hai.

Error — $Q^{-1}$ ke liye $Q$ ka square hona zaroori hai; ek tall $m\times n$ $Q$ ka koi two-sided inverse nahi hota. Sirf $Q^{\top }$ ek left inverse ki tarah kaam karta hai ($Q^{\top }Q=I$), $Q{Q}^{\top }$ nahi.

Spirit mein Error — QR $A^{\top }A$ banane se bachata hai, jiska condition number squared ho jaata hai; $Rx=Q^{\top }b$ solve karna original conditioning maintain karta hai, yahi poora numerical point hai.

Practice mein Error — tum $R$ kabhi invert nahi karte; triangular hone ki wajah se tum bottom row se upar ki taraf back-substitute karte ho, jo $R^{-1}$ banane se zyada fast aur stable hai.

Error — sign conventions tak yeh same $Q,R$ produce karte hain; woh sirf numerical accuracy mein differ karte hain, exact answer mein nahi.

Error — negative $r_{kk}$ ka matlab ${\mathbf{q}}_k$ ka sign flip karna hai; yeh ek valid QR hai, lekin tum uniqueness kho dete ho, isliye positive-diagonal convention fix kiya jaata hai.

Error — tumhe $Q^{\top }$ se multiply karna hoga, giving $Rx=Q^{\top }b$, na ki $Rx=b$; $Q^{\top }b$ generally $b$ se alag hota hai (aur $b$ shayad $Q$ ke column space mein bhi na ho).

Error — projection vector $({\mathbf{q}}_i^{\top }{\mathbf{a}}_k){\mathbf{q}}_i$ hai; tum unit direction ${\mathbf{q}}_i$ ko scale karte ho, na ki ${\mathbf{a}}_k$ ko. Yeh scalar ${\mathbf{q}}_i^{\top }{\mathbf{a}}_k$ exactly entry $r_{ik}$ hai.

Taaki $Q$ ke columns unit length ke hon, making $Q^{\top }Q=I$; iske bina $R$ ka diagonal $Q$ mein tangle ho jaata aur lengths preserve nahi hote.

Kyunki ${\mathbf{q}}_i$ ek unit vector hai, dot product ${\mathbf{q}}_i^{\top }{\mathbf{a}}_k$ exactly ${\mathbf{a}}_k$ ke shadow ki signed length hai ${\mathbf{q}}_i$ ki direction mein — ${\mathbf{q}}_i$ ki woh amount jo ${\mathbf{a}}_k$ ko rebuild karne mein use hoti hai.

Normal equations $A^{\top }A$ use karti hain, condition number ko square karke rounding amplify karti hain; QR $A$ ke saath directly kaam karta hai orthogonal $Q$ ke through, jo kabhi errors inflate nahi karta.

Ek dependent column ${\mathbf{u}}_k=0$ produce karta hai, isliye ${\mathbf{q}}_k$ undefined ho jaata hai (zero se divide karna) aur $r_{kk}=0$ — recipe toot jaata hai aur direction arbitrary ho jaati hai.

Repeatedly $A_{k+1}=R_k{Q}_k$ set karna ek orthogonal similarity transform hai jo eigenvalues preserve karta hai while matrix ko triangular form ki taraf push karta hai, diagonal par eigenvalues reveal karta hai.

${\mathbf{q}}_k$ ko negate karna aur saath mein $R$ ki row $k$ ko negate karna $QR$ ko unchanged rakhta hai, kyunki product mein do sign flips cancel ho jaate hain — so validity survive karti hai lekin uniqueness ko sign rule chahiye.

Kyunki $\Vert Q\mathbf{x}{\Vert }^2={\mathbf{x}}^{\top }{Q}^{\top }Q\mathbf{x}={\mathbf{x}}^{\top }\mathbf{x}=\Vert \mathbf{x}{\Vert }^2$; orthonormality middle ko identity mein collapse kar deti hai, isliye $Q$ rotation/reflection ki tarah act karta hai.

Gram–Schmidt step-by-step procedure hai; har ${\mathbf{a}}_k={\sum }_i{r}_{ik}{\mathbf{q}}_i$ likhna aur un relations ko column-by-column stack karna literally matrix equation $A=QR$ hai.

$Q=[{\mathbf{a}}_1/\Vert {\mathbf{a}}_1\Vert ]$ aur $R=[\Vert {\mathbf{a}}_1\Vert ]$ — ek $1\times 1$ triangular matrix; orthogonalize karne ke liye koi baad ke columns nahi hain.

Doosre ka projection ke baad zero leftover hoga (${\mathbf{u}}_2=0$), isliye $r_{22}=0$ aur ${\mathbf{q}}_2$ undefined hai — positive diagonal ke saath QR exist nahi karta kyunki columns dependent hain.

$Q=I$ aur $R=I$ — columns pehle se orthonormal hain, isliye Gram–Schmidt kuch nahi karta, "orthogonal in ⇒ do nothing" rule ke saath match karta hai.

Woh column dependent hai (yeh baaki ka trivial combination hai), ${\mathbf{u}}_k=0$, aur koi positive-diagonal QR exist nahi karta; matrix mein independent columns nahi hain.

Haan — jab $m=n$ ho to pad karne ke liye koi extra rows nahi hain, isliye $Q$ pehle se $n\times n$ orthogonal hai aur $Q{Q}^{\top }=I$ bhi hold karta hai.

Zero — projection exact hai, $Ax=b$ ka ek true solution hai, aur $Rx=Q^{\top }b$ solve karna use bina kisi leftover error ke recover karta hai.

Yeh $\Vert {\mathbf{a}}_1\Vert$ hai, pehle column ki length, kyunki ${\mathbf{u}}_1={\mathbf{a}}_1$ hai jab kuch subtract nahi hua — $R$ ka pehla diagonal bas us column ki length hai.