Geometric interpretation — signed volume: Exercises

4.5.23 · D4 · Maths › Linear Algebra (Full) › Geometric interpretation — signed volume

Neeche sab kuch usi ek formula par tika hai jo parent note ne axioms se derive kiya tha:

Figure — Geometric interpretation — signed volume

Level 1 — Recognition

Recall Solution 1.1

KYA: hum sign turn se judge karte hain aur size box se. KYUN: "right" ( $a$ ) se "up" ( $b$ ) ki taraf turn karna ek CCW quarter-turn hai → positive. KAISA DIKHTA HAI: ek seedha $2 \times 3$ rectangle — Figure 1 mein shaded (mint) box, jo right-pointing aur up-pointing arrows se bounded hai — jiska area $6$ hai. Toh answer hai $+ 6$ . Check karo: $2 \cdot 3 - 0 \cdot 0 = 6$ . ✓

Recall Solution 1.2

(a) Box ek line (2D) ya plane (3D) par flat ho gaya hai — zero volume. (b) Columns linearly dependent hain (ek doosre ka combination hai). (c) Map non-invertible hai — yeh information squash kar deta hai, isliye undo nahi ho sakta. Dekho Invertibility and the inverse matrix aur Linear independence and rank.

Recall Solution 1.3

False. Magnitude $∣ - 5 ∣ = 5$ physical volume hai; minus sign sirf yeh record karta hai ki orientation reverse ho gayi (ek mirror flip). Negative determinants bilkul ordinary aur meaningful hote hain.

Level 2 — Application

Recall Solution 2.1

KYA: $a_{1} b_{2} - a_{2} b_{1}$ apply karo jahan $a_{1} = 4, a_{2} = 2, b_{1} = 1, b_{2} = 3$ . $4 \cdot 3 - 2 \cdot 1 = 12 - 2 = 10.$ Positive → orientation preserved; box ka area $10$ hai.

Recall Solution 2.2

$1 \cdot 6 - 2 \cdot 3 = 6 - 6 = 0$ . Doosra column $(3, 6) = 3 \cdot (1, 2)$ pehle column ke parallel hai, isliye parallelogram ek line par collapse ho jaata hai — zero area, dependent columns.

Recall Solution 2.3

KYUN triple product: $b \times c$ ek aisa vector deta hai jiska length base-parallelogram ka area hai aur direction base ka normal hai; $a$ ke saath dot karna us normal ke saath height se multiply karta hai → base × height = volume (dekho Cross product and scalar triple product). Pehle $b \times c$ jahan $b = (0, 3, 0), c = (1, 1, 1)$ : $b \times c = (3 \cdot 1 - 0 \cdot 1, 0 \cdot 1 - 0 \cdot 1, 0 \cdot 1 - 3 \cdot 1) = (3, 0, - 3) .$ Phir $a = (1, 0, 2)$ ke saath dot karo: $1 \cdot 3 + 0 \cdot 0 + 2 \cdot (- 3) = 3 - 6 = - 3.$ Signed volume $- 3$ : magnitude $3$ , orientation reversed (ek left-handed frame). Dekho Orientation and handedness of bases.

Recall Solution 2.4

KYUN $c^{n}$ : poori matrix ko $2$ se multiply karna teeno columns ko scale karta hai, aur har scaling volume ko $2$ se multiply karti hai: $2 \times 2 \times 2 = 2^{3}$ . $det (2 A) = 2^{3} \cdot 4 = 8 \cdot 4 = 32.$

Level 3 — Analysis

Recall Solution 3.1

Notation: $det (a, b)$ likhte hain edge columns $a, b$ wale box ke signed area ke liye — yeh exactly $det (a_{1} a_{2} b_{1} b_{2})$ hai, bas apne do columns ki function ke roop mein dekha gaya. Multilinearity axiom (recap): parent note ne establish kiya tha ki signed area multilinear hai — yeh har column mein separately additive hai, $det (u + w, b) = det (u, b) + det (w, b)$ (aur isi tarah doosre slot mein bhi). Yahi cheez hume ek slot ke andar ki sum ko do alag determinants mein todne deti hai — neeche ke expansion ka yeh concrete engine hai. Alternating axiom (recap): parent note ne yeh bhi establish kiya tha ki signed area alternating hai — do equal edges wala box flat hota hai, isliye kisi bhi $v$ ke liye $det (v, v) = 0$ . Algebra: alternating axiom ko vector $a + b$ par dono slots mein apply karo, toh $det (a + b, a + b) = 0$ . Ab multilinearity se expand karo — pehle slot ki sum split karo, phir har resulting doosre slot ki sum split karo: $det (a, a) + det (a, b) + det (b, a) + det (b, b) = 0.$ Do equal-argument terms $det (a, a)$ aur $det (b, b)$ alternating axiom se vanish ho jaate hain, aur bachhta hai $det (a, b) = - det (b, a)$ . Geometry: edges ko swap karna turn ka sense reverse kar deta hai (CCW ↔ CW), isliye same box ko opposite sign milta hai — ek mirror flip. Numeric check: $det (4213) = 10$ jabki $det (1342) = 1 \cdot 2 - 3 \cdot 4 = - 10$ . ✓

Recall Solution 3.2

KYUN yeh $det$ formula: $a_{1} b_{2} - a_{2} b_{1}$ jahan $a_{1} = cos θ, a_{2} = sin θ, b_{1} = - sin θ, b_{2} = cos θ$ : $cos θ \cdot cos θ - sin θ \cdot (- sin θ) = cos^{2} θ + sin^{2} θ = 1.$ Interpretation: rotation kabhi kuch stretch ya flip nahi karta — ek spun square apna area ( $∣ det ∣ = 1$ ) aur apni handedness (sign $+$ ) rakhta hai. Pythagorean identity exactly "rigid turn" ki algebra hai.

Recall Solution 3.3

$det S = 1 \cdot 1 - 0 \cdot k = 1$ har $k$ ke liye. Geometry: sheared unit square ek aisa parallelogram ban jaata hai jiska base $1$ aur height $1$ same rehti hai — sirf top horizontally slide hoti hai. Area = base × height shear se untouched rehta hai, isliye determinant $1$ hi rehta hai chahe shear kitna bhi extreme ho.

Level 4 — Synthesis

Recall Solution 4.1

KYUN multiplicativity: pehle map $B$ karna volume ko $det B$ se scale karta hai phir map $A$ karna $det A$ se, isliye composite product se scale karta hai. Kyunki $A A^{- 1} = I$ aur $det I = 1$ : $det A \cdot det (A^{- 1}) = 1 \Rightarrow det (A^{- 1}) = \frac{1}{d e t A} .$ Yeh exactly tab fail hota hai jab $det A = 0$ — ek flattened map undo nahi ho sakta, isliye inverse exist nahi karta (Invertibility and the inverse matrix). Example: agar $det A = 5$ , toh $det (A^{- 1}) = \frac{1}{5}$ .

Recall Solution 4.2

KYUN Jacobian determinant: locally har smooth map linear lagta hai, aur uska linear part $J$ tiny areas ko $∣ det J ∣$ se scale karta hai — yahi factor change-of-variables integral mein appear hota hai (Change of variables and the Jacobian). $det J = 2 \cdot 3 - 0 \cdot 1 = 6, new area = 4 \cdot 6 = 24.$ Orientation preserved hai (sign $+$ ).

Recall Solution 4.3

KYUN product: determinant eigenvalues ka product hota hai, kyunki har eigendirection mein map us eigenvalue se scale karta hai, aur volume per-direction stretches ka product hai (Eigenvalues — product equals determinant). $det A = λ_{1} λ_{2} = 3 \cdot (- 2) = - 6.$ Magnitude $6$ : areas chhe guna stretch hote hain. Sign $-$ : orientation reverse ho jaati hai — single negative eigenvalue plane ko apni eigendirection ke saath reflect karta hai.

Level 5 — Mastery

Recall Solution 5.1

KYA: edge vectors $u = Q - P$ aur $v = R - P$ use karke $P$ ko origin par shift karo. KYUN: translation area change nahi karta, aur do edges $u, v$ ek parallelogram span karte hain jiska signed area $det [u v]$ hai. Triangle exactly us parallelogram ka aadha hota hai (diagonal use do congruent triangles mein kaata hai). Absolute value orientation discard karta hai kyunki ek physical triangle ka area positive hota hai. Compute: $u = Q - P = (3, 0)$ , $v = R - P = (1, 4)$ . $det (3014) = 3 \cdot 4 - 0 \cdot 1 = 12, Area = \frac{1}{2} ∣12∣ = 6.$

Neeche wali figure Solution 5.1 ko picture mein dikhati hai. Isme corner $P$ share karne wale do nested shapes hain: bada parallelogram (pale butter-yellow region, area $12$ ) jo do edge vectors $u, v$ se spanned hai, aur uske andar triangle $P QR$ (lavender-purple region, area $6$ ) — exactly woh aadha jo tum rakhte ho. $Q$ se $R$ tak jaane wala diagonal parallelogram ko do congruent triangles mein split karta hai; yahi split hai kyun factor $\frac{1}{2}$ aata hai, aur kyun absolute value sirf ek physical area se orientation sign haatne ke liye chahiye. Color ke bina bhi, ise aise padho: ek tircha four-sided box jisme triangle uska aadha occupy kar raha hai, corner $P$ bottom-left par.

Recall Solution 5.2

Logic: teen points tab collinear hote hain jab unse bana triangle ka area $0$ ho, yaani edge vectors $u = Q - P$ aur $v = R - P$ parallel hon → unka determinant $0$ ho. $u = (2, 4)$ , $v = (4, 8)$ : $det (2448) = 2 \cdot 8 - 4 \cdot 4 = 16 - 16 = 0.$ Determinant $0$ hai, isliye points collinear hain — "triangle" flat hai.

Recall Solution 5.3

Hume $det = 1$ chahiye jabki dono columns unit length se lambe hon. Try karo $A = (2111), det A = 2 \cdot 1 - 1 \cdot 1 = 1.$ Column lengths hain $2^{2} + 1^{2} = 5 \approx 2.24$ aur $1^{2} + 1^{2} = 2 \approx 1.41$ — dono $1$ se zyaada. Determinant $1$ matlab area- aur orientation-preserving, phir bhi yeh rotation nahi hai (rotations har length keep karti hain; yahan columns $5$ aur $2$ tak stretch hote hain, toh lengths change hoti hain — rotation impossible hai). Toh yeh kaunsa map hai? Yeh ek general area-preserving shearing map hai. Concretely, ise elementary shears mein factor karo: $(2111) = (1011) (1101),$ ek horizontal shear phir ek vertical shear (har ek ka determinant $1$ hai, aur $1 \cdot 1 = 1$ ). Toh $A$ space ko sideways slide karta hai, phir upward slide karta hai, unit square ko ek aisi slanted parallelogram mein distort karta hai jiska area $1$ same hai aur handedness same hai — ek pure shear-composite, kabhi rigid rotation nahi. Koi bhi integer matrix jiska $det = 1$ ho (ek element of $S L_{2} (Z)$ ) aisa map deta hai.

Active Recall

Recall Answers cover karo aur khud test karo

2D mein edges $a, b$ ka signed area ::: $a_{1} b_{2} - a_{2} b_{1}$ $n \times n$ ke liye $det (c A)$ ::: $c^{n} det A$ (sign flip hoti hai agar $c < 0$ aur $n$ odd ho) $det (A^{- 1})$ ::: $1/ det A$ (zaroorat hai $det A \neq = 0$ ) $det (A B)$ ::: $det A \cdot det B$ Vertices se triangle area ::: uske do edge vectors ke determinant ke absolute value ka aadha Teen points ka collinearity test ::: edge-vector determinant $0$ ke barabar ho

Connections

Geometric interpretation — signed volume
Determinant — definition and cofactor expansion
Cross product and scalar triple product
Linear independence and rank
Invertibility and the inverse matrix
Change of variables and the Jacobian
Eigenvalues — product equals determinant
Orientation and handedness of bases

Exercises — Geometric interpretation — signed volume