Visual walkthrough — Quadratic forms — positive definite, negative definite, indefinite

4.5.39 · D2 · Maths › Linear Algebra (Full) › Quadratic forms — positive definite, negative definite, inde

Neeche sab kuch zero se build kiya gaya hai. Agar koi symbol aata hai, use pehle draw kiya gaya hai.

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Step 1 — "${\mathbf{x}}^{\top }A\mathbf{x}$" ka matlab asal mein kya hai

KYA. Hum ek arrow $\mathbf{x}$ machine mein daaalte hain aur wapas milta hai ek single number $Q(\mathbf{x})$.

EK NUMBER KYUN. Hum pooch rahe hain "is direction ke upar surface kitni uchi hai?" Height ek number hoti hai. Definiteness ki poori kahaani yeh hai: kya woh number positive hai, negative hai, ya iska sign badlta rehta hai jab arrow ghoomta hai?

PICTURE. Neeche, arrow $\mathbf{x}$ (amber) plane mein baith a hai. $A$ se multiply karne par pehle woh ek naye arrow $A\mathbf{x}$ (cyan) mein mod jaata hai. Phir ${\mathbf{x}}^{\top }(A\mathbf{x})$ measure karta hai $A\mathbf{x}$ original $\mathbf{x}$ ki taraf kitna point karta hai — woh overlap hi output number hai.

$$Q(\mathbf{x})=\underset{\text{arrow laid flat}}{\underbrace{{\mathbf{x}}^{\top }}}\underset{\text{bends the arrow}}{\underbrace{A}}\underset{\text{the arrow}}{\underbrace{\mathbf{x}}}=\text{one number}$$

Agar dono arrows roughly ek hi direction mein hain, toh number positive hoga. Agar $A\mathbf{x}$ ulti direction mein point kare $\mathbf{x}$ ke relative, toh number negative hoga. Wahi sign poora game hai.

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Step 2 — Problem: cross-terms sign padhna mushkil kar dete hain

KYA. Teen terms hain. $x^2$ aur $y^2$ terms ke honest signs hain ($x^2\ge 0$ hamesha). Lekin middle term $2bxy$ positive ya negative ho sakta hai depending on us quadrant par jahan arrow land karta hai.

YEH PROBLEM KYUN HAI. Maano $a>0$ aur $c>0$ — tum guess kar sakte ho "hamesha positive." Lekin agar $b$ bada aur negative ho, toh cross-term $2bxy$ kuch arrows ke liye total ko zero se neeche kheench sakta hai. Sign directly coefficients se nahi padha ja sakta. Cross-term $xy$ dono directions ko couple karta hai.

PICTURE. Neeche ki surface tilted aur twisted hai — $xy$ term woh shear hai jo bowl ko itna skew kar deta hai ki uski sabse nichli valleys $x$- ya $y$-axis ke saath align nahi karti. Hume ise un-twist karna hai.

Recall "

$2b$" kyun aur "$b$" kyun nahi? Off-diagonal $b$ $A$ ke dono top-right aur bottom-left mein aata hai. Jab tum ${\mathbf{x}}^{\top }A\mathbf{x}$ multiply karte ho, $xy$ contribution do baar aati hai — ek baar $A_{12}xy$ ke roop mein, ek baar $A_{21}yx$ ke roop mein — jo $2bxy$ deta hai. ::: Isliye exactly (parent note mein) off-diagonal cross-coefficient ka aadha hota hai.

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Step 3 — Rescue: symmetric matrices ke perpendicular eigen-directions hote hain

KYA. Ek symmetric $A$ ke liye (hamaara case, Symmetric Matrices se), Spectral Theorem ek saath do cheezein guarantee karta hai:

1. Eigenvalues ${\lambda }_1,{\lambda }_2$ real numbers hain, aur
2. Eigenvectors ko ek dusre ke perpendicular choose kiya ja sakta hai.

YEH HUME KYUN BACHATA HAI. Perpendicular special directions matlab hum unhe ek bilkul nayi pair of axes ki tarah use kar sakte hain. In axes ke along koi twist nahi hoti — cross-term gayab ho jaata hai. Step 2 ka twisted bowl un-twisted bowl ban jaata hai.

PICTURE. Dono eigen-directions (cyan) twisted surface par draw kiye gaye hain. Dhyaan do ki woh shape ke valley aur ridge ke saath exactly align hain — woh natural axes jo surface "chahti" hai.

$$A\underset{\text{special arrow}}{\underbrace{{\mathbf{v}}_i}}=\underset{\text{stretch factor}}{\underbrace{{\lambda }_i}}{\mathbf{v}}_i{\mathbf{v}}_1\perp {\mathbf{v}}_2$$

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Step 4 — Eigen-axes par rotate karo (change of coordinates)

KYA. Hum nayi coordinates define karte hain $\mathbf{y}=Q_{\text{rot}}^{\top }\mathbf{x}$. Yeh bas "wahi arrow hai, eigen-axes use karke describe kiya gaya, standard axes ki jagah."

YEH HARMLESS KYUN HAI. Rotation kabhi arrow ki length nahi badalta, aur kabhi non-zero arrow ko zero arrow mein nahi badalta. Isliye $\mathbf{x}\ne 0$ tab aur sirf tab jab $\mathbf{y}\ne 0$. $\mathbf{y}$ ke signs ke baare mein hum jo bhi conclude karein woh $\mathbf{x}$ ke liye exactly sach hoga.

PICTURE. Wahi arrow, do rulers. Left par, standard grid (white). Right par, tilted eigen-grid (cyan) components $y_1,y_2$ read karta hai.

$$\underset{\text{arrow in eigen-coords}}{\underbrace{\mathbf{y}}}=\underset{\text{rotate the ruler}}{\underbrace{Q_{\text{rot}}^{\top }}}\underset{\text{same arrow}}{\underbrace{\mathbf{x}}}$$

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Step 5 — Twist gayab hoti hai: pure squares

KYA. Kyunki $\Lambda$ diagonal hai, ${\mathbf{y}}^{\top }\Lambda \mathbf{y}$ mein koi cross-term nahi hai. Yeh collapse ho jaata hai:

$$Q=\underset{\text{stretch 1}}{\underbrace{{\lambda }_1}}\underset{\ge 0}{\underbrace{y_1^2}}+\underset{\text{stretch 2}}{\underbrace{{\lambda }_2}}\underset{\ge 0}{\underbrace{y_2^2}}.$$

YEH POORA POINT KYUN HAI. Dekho har piece kya kar sakta hai:

• $y_1^2$ aur $y_2^2$ squares hain — yeh kabhi bhi negative nahi ho sakte.
• Isliye har term ka sign entirely uske eigenvalue ${\lambda }_i$ ke sign se decide hota hai.

Twist (Step 2) gayi. Humne machine ko do independent one-dimensional springs mein separate kar diya hai.

PICTURE. Un-twisted surface, ab axes ke saath aligned. $y_1$-axis ke along yeh ${\lambda }_1$ steepness se curve karta hai; $y_2$ ke along ${\lambda }_2$ steepness se. Do independent parabolas.

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Step 6 — Signs se chaar cases padho

KYA. Har row bas "do signed squares add karo" hai.

HUM YEH CONCLUSION KYUN NIKALTE HAIN.

• Dono positive: $0$ paane ka ek hi tarika hai $y_1=y_2=0$, matlab zero arrow. Har doosra arrow $Q>0$ deta hai → bowl.
• Mixed signs: arrow ko ${\mathbf{v}}_1$ ke along point karo (toh $y_2=0$) aur ${\lambda }_1{y}_1^2$ milega; ${\mathbf{v}}_2$ ke along ${\lambda }_2{y}_2^2$ milega. Ek $+$ hai, doosra $-$. Dono signs ek hi machine se aate hain → saddle.
• Ek zero: us flat eigen-direction ke along $Q$ exactly $0$ rehta hai chahe arrow non-zero ho → boundary case, semidefinite not definite.

PICTURE. Teen surfaces side by side — bowl (dono upar), dome (dono neeche), saddle (ek taraf upar, doosri taraf neeche) — har ek apne eigen-axes ke saath aur $Q$ ka sign amber mein marked.

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Step 7 — Degenerate cases, carefully kiye gaye

KYA / KYUN — har corner covered:

• Dono eigenvalues zero ($A=0$): $Q\equiv 0$ sab arrows ke liye. Surface flat plane hai. Technically yeh PSD aur NSD dono hai.
• Ek zero, ek positive (PSD): ek eigen-line ke along flat (seedha tala trough), baaki jagah upar uthta hua.
• Ek zero, ek negative (NSD): ek eigen-line ke along flat, baaki jagah neeche girta hua.
• Dono non-zero, same sign: strictly definite (bowl ya dome) — origin ke alawa kabhi $0$ nahi chhoota.
• Dono non-zero, opposite sign: indefinite — ek genuine saddle.

PICTURE. PSD trough: ek valley jiska flat seedha bottom zero-eigenvalue direction ke along run karta hai (amber line), ek single lowest point nahi.

Recall Doosre test se cross-check

Yeh eigenvalue verdict Sylvester's leading-minor test (parent §4) aur Second Derivative Test via Hessian Matrix ke saath agree karna chahiye. Karta hai: bowl/PD Hessian ⇒ local minimum; dome/ND ⇒ maximum; saddle/indefinite ⇒ saddle point. Wahi algebra, do languages. Sylvester akela PSD kyun nahi pakad sakta? ::: Uska clean form sirf strict definiteness certify karta hai; boundary $D_k=0$ ko full principal-minor analysis chahiye (ya bas eigenvalues).

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Ek-picture summary

Upar sab ek hi idea hai: eigen-axes par rotate karo, aur twisted machine do independent signed springs ban jaati hai; springs ke signs eigenvalues hain.

$${\mathbf{x}}^{\top }A\mathbf{x}\underset{\text{Spectral Thm}}{\overset{\text{rotate }\mathbf{y}=Q_{\text{rot}}^{\top }\mathbf{x}}{\to }}{\lambda }_1{y}_1^2+{\lambda }_2{y}_2^2\stackrel{\text{read signs}}{\to }\{\text{bowl / dome / saddle / trough}\}$$

Recall Feynman retelling — ek 12-saal ke bachche ko batao

Tumhare paas ek machine hai: arrow daalo, number nikalo. Pehle machine messy lagti hai — ek "twist" term hai jo dono directions ko mix karta hai, toh tum seedha nahi bata sakte ki number positive hoga ya negative, sirf dekh ke.

Lekin ek symmetric machine ke do magic directions hote hain (eigenvectors) jo right angles par baithe hain. Agar tum apna sar ghuma lo taaki woh directions tumhara naya left-right aur up-down ban jaayein, twist gayab ho jaati hai. Ab number bas yeh hai: (pehla stretch) times (pehli distance squared) plus (doosra stretch) times (doosri distance squared).

Distances squared kabhi negative nahi hote — woh innocent part hain. Isliye answer ka sign decide karne wali ek hi cheez hai: kya stretches (eigenvalues) positive hain ya negative.

• Dono stretches upar → har arrow ek positive number deta hai → tum ek bowl ke bottom par khade ho.
• Dono neeche → dome, tum pahadi ki choti par ho.
• Ek upar, ek neeche → saddle, ek taraf uphill aur doosri taraf downhill.
• Exactly zero ka stretch → ek flat groove: us par chalo aur number kabhi nahi badlega.

Yahi poora theorem hai. Apna sar magic directions ki taraf ghuma lo, phir bas plus aur minus signs count karo.