Pseudoinverse: Worked examples — Hinglish deep dive

4.5.42 · D3 · Maths › Linear Algebra (Full) › Pseudoinverse

Ye page Pseudoinverse ka drill hall hai. Parent note ne teen formulas build kiye the; yahan hum har tarah ki matrix unpar throw karte hain — tall, fat, square-invertible, square-singular, rank-deficient (tall aur fat dono flavours), aur ek real-world data-fit — taaki exam mein koi bhi scenario tumhe surprise na kar sake.

Shuru karne se pehle, ek vaada: neeche har symbol parent note mein earn kiya gaya tha. Sirf teen tools ka lightning refresher — plus wo chaar rules jo define karte hain ki pseudoinverse hota kya hai:

Recall Teen formulas (parent se)

Tall & full column rank ( $n$ independent columns, $m \geq n$ ): $A^{+} = (A^{⊤} A)^{- 1} A^{⊤}$ . Ye ek left inverse hai: $A^{+} A = I_{n}$ .
Fat & full row rank ( $m$ independent rows, $n \geq m$ ): $A^{+} = A^{⊤} (A A^{⊤})^{- 1}$ . Ye ek right inverse hai: $A A^{+} = I_{m}$ .
Koi bhi shape, koi bhi rank: $A^{+} = V Σ^{+} U^{⊤}$ Singular Value Decomposition se, har nonzero $σ_{i}$ ko invert karke, zeros ko dead chhod ke.

Yahan $A^{⊤}$ ka matlab hai " $A$ ko uski diagonal ke across flip karo" (rows columns ban jaati hain), aur $∥ x ∥$ ka matlab hai ordinary length $x_{1}^{2} + x_{2}^{2} + \dots$ .

Recall Chaar Penrose conditions ("

$A^{+}$ " ka matlab kya hai) — aur do parent mistakes jinpar hum lean karte hain $A^{+}$ wo unique matrix hai jo chaaon satisfy karti hai:

$A A^{+} A = A$ 2. $A^{+} A A^{+} = A^{+}$ 3. $(A A^{+})^{⊤} = A A^{+}$ 4. $(A^{+} A)^{⊤} = A^{+} A$ . Condition 1 wahi hai jise hum neeche baar baar "plug-back" check ke taur par use karte hain.

Parent note ke do traps jo hum deliberately trigger karenge:

Mistake 1: " $A^{+} = (A^{⊤} A)^{- 1} A^{⊤}$ hamesha." Ise full column rank chahiye; warna $A^{⊤} A$ singular hoti hai (Examples 5, 8).
Mistake 2: " $A^{+} A = I$ hamesha." Sirf full column rank ke liye. Generally $A^{+} A$ ek projection hoti hai (uska eigenvalue 0 ho sakta hai), jo Examples 5, 8, 9 mein dikhta hai.

Scenario matrix

Har matrix jo tum pseudoinverse ko feed kar sakte ho, exactly inhi boxes mein se ek mein aati hai. $A^{+}$ ka pura point yahi hai ki ye kabhi crash nahi karta — lekin kaunsa formula safe hai ye box-to-box badalta hai. Neeche "rank" = genuinely independent rows/columns ki sankhya (wo directions jinhe matrix zero tak squash nahi karta).

Cell	Shape & rank	$A^{- 1}$ ke saath kya gadbad hai?	Kaunsa formula safe hai	Example
C-tall	tall, full column rank	square nahi — koi $A^{- 1}$ nahi; $A x = b$ ka koi exact solution nahi	$(A^{⊤} A)^{- 1} A^{⊤}$ (least squares)	Ex 1, Ex 2
C-fat	fat, full row rank	square nahi; infinitely many solutions	$A^{⊤} (A A^{⊤})^{- 1}$ (min-norm)	Ex 3
C-sqinv	square, invertible	kuch nahi! $A^{- 1}$ exist karta hai	koi bhi — sab agree karte hain, $A^{+} = A^{- 1}$	Ex 4
C-sqsing	square, singular (rank deficient)	$det = 0$ , koi $A^{- 1}$ nahi; $A^{⊤} A$ aur $A A^{⊤}$ dono singular	SVD use karna hi hoga	Ex 5
C-zero	zero matrix (degenerate limit)	sab kuch nullspace hai	SVD deta hai $A^{+} = 0$ (Ex 6 mein dekho kyun)	Ex 6
C-word	real data fit (tall in disguise)	overdetermined, noisy	least squares	Ex 7
C-twist-t	exam trap: rank-deficient tall	lagta hai C-tall jaisa lekin $A^{⊤} A$ singular hai!	SVD, tall formula nahi	Ex 8
C-twist-f	exam trap: rank-deficient fat	lagta hai C-fat jaisa lekin $A A^{⊤}$ singular hai!	SVD, fat formula nahi	Ex 9

Neeche ke do figures do "no exact inverse" cells ke peeche ki geometry dikhate hain: tall case mein $b$ neeche project hoti hai (Ex 1), fat case mein sabse chhota arrow pick hota hai (Ex 3).

Cell C-tall — least squares

Worked example Example 1 (cell

C-tall) — projection picture $A = (11)$ , $b = (31)$ . $A x = b$ ko best possible solve karo ( $x$ ek single number hai).

Forecast: padhne se pehle ek best number $x$ guess karo. (Do equations $x = 3$ aur $x = 1$ agree nahi karti — fair compromise kya hai?)

Step 1 — cell identify karo. $A$ hai $2 \times 1$ : do rows, ek column. Ek column, aur wo nonzero hai, isliye full column rank → cell C-tall → use karo $A^{+} = (A^{⊤} A)^{- 1} A^{⊤}$ . Ye step kyun? Formula pick karne se pehle classify karna zaroori hai; tall formula sirf tabhi legal hai jab column(s) independent hon.

Step 2 — $A^{⊤} A$ build karo. $A^{⊤} A = (11) (11) = 2$ . Ye step kyun? $A^{⊤} A$ wo chhoti square matrix hai (yahan sirf number 2) jise hume invert karna hai; ye invertible hai exactly isliye kyunki column independent hai.

Step 3 — $A^{+}$ assemble karo. $A^{+} = (2)^{- 1} (11) = (\frac{1}{2} \frac{1}{2})$ . Ye step kyun? Ye row vector hi refund machine hai: koi bhi $b$ isko multiply karo aur best $x$ nikal aata hai.

Step 4 — $b$ par apply karo. $x = A^{+} b = \frac{1}{2} \cdot 3 + \frac{1}{2} \cdot 1 = 2$ . Ye step kyun? $x$ literally 3 aur 1 ka average hai — sabse fair compromise, exactly jaisa intuition demand karta hai.

Ye kaisa dikhta hai: figure s01 mein, $b = (3, 1)$ seedha $(1, 1)$ se span hone wali line par drop hoti hai; us perpendicular ka foot hai $A x = (2, 2)$ , aur bacha hua arrow $b - A x = (1, - 1)$ residual hai, line ke perpendicular.

Verify karo: residual $b - A x = (3, 1) - (2, 2) = (1, - 1)$ column $(1, 1)$ ke orthogonal hona chahiye: dot product $1 \cdot 1 + (- 1) \cdot 1 = 0$ ✓. Residual ki orthogonality hi least-squares condition hai (dekho Orthogonal Projection).

Worked example Example 2 (cell

C-tall) — parent ka line-fit, re-verified Parent ke Example 1 jaisa hi: $A = 111012$ , $b = 600$ , $y = c + d t$ fit karna.

Forecast: teen points $(0, 6), (1, 0), (2, 0)$ — koi straight line teeno ko hit nahi karti. Slope ka sign guess karo.

Step 1 — cell. $3 \times 2$ , columns $(1, 1, 1)$ aur $(0, 1, 2)$ independent hain → C-tall. Ye step kyun? Pehle jaisa hi test; yahan hum rank $2 =$ columns ki sankhya confirm karte hain.

Step 2 — $A^{⊤} A$ aur $A^{⊤} b$ . $A^{⊤} A = (3335)$ , $A^{⊤} b = (60)$ . Ye step kyun? Ye normal equations $A^{⊤} A x = A^{⊤} b$ hain.

Step 3 — $2 \times 2$ invert karo. $det = 3 \cdot 5 - 3 \cdot 3 = 6$ , isliye $(A^{⊤} A)^{- 1} = \frac{1}{6} (5 - 3 - 3 3)$ . Ye step kyun? $2 \times 2$ inverse hai $\frac{1}{d e t} (d - c - b a)$ .

Step 4 — solve karo. $x = \frac{1}{6} (5 - 3 - 3 3) (60) = (5 - 3)$ . Best line $y = 5 - 3 t$ . Slope negative hai. Ye step kyun? Normal equations solve karne se least-squares coefficients $(c, d) = (5, - 3)$ milte hain.

Verify karo: predictions $(5, 2, - 1)$ at $t = 0, 1, 2$ ; residual $r = b - A x = (6, 0, 0) - (5, 2, - 1) = (1, - 2, 1)$ . Check karo $A^{⊤} r = 0$ : column $(1, 1, 1) \cdot (1, - 2, 1) = 0$ ✓, column $(0, 1, 2) \cdot (1, - 2, 1) = 0 - 2 + 2 = 0$ ✓.

Cell C-fat — minimum norm

Worked example Example 3 (cell

C-fat) — sabse chhota arrow $A = (12)$ , $b = (4)$ . $x_{1} + 2 x_{2} = 4$ ko sabse chhote $∥ x ∥$ ke saath solve karo.

Forecast: solutions ek poori line form karte hain plane mein. Us line par origin ke sabse close kaun sa point hai?

Step 1 — cell. $1 \times 2$ : ek row, aur wo nonzero hai → full row rank → cell C-fat → use karo $A^{+} = A^{⊤} (A A^{⊤})^{- 1}$ . Ye step kyun? Ek equation, do unknowns: infinitely many solutions. Hume sabse chhota chahiye, jo fat formula ka kaam hai.

Step 2 — $x$ ko row space mein force karo. Parent ki derivation likhti hai $x = A^{⊤} w$ , jahan $w$ ek chhota helper vector hai (yahan ek single unknown number) jo isliye choose kiya gaya hai taaki sahi equation $A x = b$ satisfy ho. $x = A^{⊤} w$ substitute karne se $A x = b$ ban jaata hai $A A^{⊤} w = b$ . Ye step kyun? Har solution ek row-space part plus nullspace junk mein split hoti hai; $x = A^{⊤} w$ likhna $x$ ko poori tarah row direction $(1, 2)$ se build karta hai, junk ki guarantee zero karti hai — aur exactly yahi "smallest norm" demand karta hai. Letter $w$ sirf wo coefficient hai jo batata hai ki row direction ka kitna lena hai.

Step 3 — $w$ ke liye solve karo, phir $x$ . $A A^{⊤} = (12) (12) = 1 + 4 = 5$ , isliye $w = (A A^{⊤})^{- 1} b = \frac{1}{5} \cdot 4 = \frac{4}{5}$ . Phir $x = A^{⊤} w = (12) \cdot \frac{4}{5} = (4/5 8/5)$ . Ye step kyun? Jab $w$ pata chal jaata hai, $x = A^{⊤} w$ seedha answer deta hai. Do steps ko bundle karne se closed form milta hai $A^{+} = A^{⊤} (A A^{⊤})^{- 1} = (1/5 2/5)$ , aur indeed $A^{+} b = (4/5 8/5)$ .

Ye kaisa dikhta hai: figure s02 mein dashed line $x_{1} + 2 x_{2} = 4$ ke sare solutions hain; plum arrow $(4/5, 8/5)$ tak us line ke perpendicular hai — origin ka closest point.

Verify karo (do tarike): (i) ye equation solve karta hai: $\frac{4}{5} + 2 \cdot \frac{8}{5} = \frac{4}{5} + \frac{16}{5} = \frac{20}{5} = 4$ ✓. (ii) $x$ nullspace direction $(2, - 1)$ ke perpendicular hai: $\frac{4}{5} \cdot 2 + \frac{8}{5} \cdot (- 1) = \frac{8}{5} - \frac{8}{5} = 0$ ✓ (dekho Four Fundamental Subspaces).

Cell C-sqinv — honest inverse

Worked example Example 4 (cell

C-sqinv) — $A^{+}$ collapse hokar $A^{- 1}$ ban jaata hai $A = (2111)$ . Kyunki $det A = 2 - 1 = 1 \neq = 0$ , $A$ square aur invertible hai.

Forecast: pseudoinverse ko exactly ordinary inverse dena chahiye. $A^{+}$ predict karo.

Step 1 — cell. Square, $det \neq = 0$ → C-sqinv → sare formulas agree karte hain aur $A^{- 1}$ ke barabar hote hain. Ye step kyun? Jab $A^{- 1}$ exist karta hai toh wo uniquely chaaon Penrose conditions satisfy karta hai, isliye $A^{+} = A^{- 1}$ .

Step 2 — ordinary inverse. $A^{- 1} = \frac{1}{1} (1 - 1 - 1 2) = (1 - 1 - 1 2)$ . Ye step kyun? $2 \times 2$ inverse formula with $det = 1$ .

Step 3 — tall formula se sanity check. Full column rank ke saath tall formula same cheez deni chahiye: $A^{+} = (A^{⊤} A)^{- 1} A^{⊤}$ . Kyunki $A^{⊤} A$ invertible hai aur $A$ square hai, $(A^{⊤} A)^{- 1} A^{⊤} = A^{- 1} (A^{⊤})^{- 1} A^{⊤} = A^{- 1}$ . Ye step kyun? Ye algebraic collapse hi pura reason hai ki $A^{+}$ inverse ko "generalize" karta hai (dekho Matrix Inverse).

Verify karo: $A A^{- 1} = (2111) (1 - 1 - 1 2) = (1001) = I$ ✓.

Cell C-sqsing — square lekin singular

Worked example Example 5 (cell

C-sqsing) — jab $det = 0$ , sirf SVD bachti hai $A = (1111)$ . Yahan $det = 1 - 1 = 0$ : nahi invertible, aur dono $A^{⊤} A$ aur $A A^{⊤}$ singular hain, isliye na tall na fat formula legal hai.

Forecast: $A$ sab kuch line $(1, 1)$ par crush kar deta hai. Refund sirf us ek surviving direction ke saath kaam kar sakta hai. Guess karo $A^{+}$ kitna "spread out" dikhega.

Step 1 — cell. Square, $det = 0$ → C-sqsing → zaroori hai SVD use karo $A^{+} = V Σ^{+} U^{⊤}$ . Ye step kyun? Parent ki Mistake 1 ne warn kiya tha: $A^{⊤} A$ singular ⇒ tall formula explode karta hai.

Step 2 — surviving direction dhundho. $A$ ka rank 1 hai: ye direction $u = \frac{1}{2} (1, 1)$ ko 2 se scale karta hai aur perpendicular direction $u_{⊥} = \frac{1}{2} (1, - 1)$ ko kill karta hai. Isliye single nonzero singular value hai $σ_{1} = 2$ . Ye step kyun? $A u = \frac{1}{2} (2, 2) = 2 u$ aur $A u_{⊥} = \frac{1}{2} (0, 0) = 0$ : ye Eigenvalues and Eigenvectors $2$ aur $0$ hain, jo yahan singular directions se coincide karte hain kyunki $A$ symmetric hai.

Step 3 — full $U$ , $Σ$ , $V$ explicitly likhte hain. Kyunki $A$ symmetric hai aur uske eigenvalues $2, 0$ non-negative hain, hum $U = V$ le sakte hain jo column-by-column $u$ se (for $σ_{1} = 2$ ) phir $u_{⊥}$ se (for $σ_{2} = 0$ ) bana ho: $U = V = (\frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{1}{2}), Σ = (2000) .$ Ye step kyun? $U$ aur $V$ genuine $2 \times 2$ orthonormal matrices honi chahiye; unke columns singular-value order mein unit singular vectors hain. Tum check kar sakte ho $U^{⊤} U = I$ (columns perpendicular unit vectors hain) aur $U Σ V^{⊤} = A$ .

Step 4 — sirf nonzeros invert karo. $Σ^{+} = (1/2 0 00)$ (2 ko 1/2 karo, 0 ko dead chhodo). Ye step kyun? Zero se divide karna impossible hai; dead direction mein restore karne ke liye koi information nahi hai.

Step 5 — reassemble karo. $A^{+} = V Σ^{+} U^{⊤} = \frac{1}{2} u u^{⊤} = \frac{1}{2} \cdot \frac{1}{2} (1111) = (1/4 1/4 1/4 1/4)$ . Ye step kyun? Sirf surviving direction $u$ rebuild hota hai, $1/ σ_{1} = 1/2$ se scale karke.

Verify karo (Penrose condition 1): $A A^{+} A = A$ . Compute karo $A^{+} A = (1/2 1/2 1/2 1/2)$ (row space par projection, $I$ nahi — parent ki Mistake 2!). Phir $A \cdot (A^{+} A) = (1111) = A$ ✓. Ye bhi confirm karo $U Σ V^{⊤} = A$ : $U Σ = (2200)$ , $V^{⊤}$ se multiply karne par $(1111)$ milta hai ✓.

Cell C-zero — degenerate limit

Worked example Example 6 (cell

C-zero) — all-zero matrix $A = (0000)$ . Har direction nullspace hai. $A^{+}$ kya hai?

Forecast: agar $A$ sab kuch throw away kar deta hai, toh fairest refund deta hai... kya?

Step 1 — cell. Zero matrix → C-zero, C-sqsing ka extreme (rank 0) → SVD. Ye step kyun? Koi bhi formula jisme $(A^{⊤} A)^{- 1}$ ho kaam nahi kar sakta; invert karne ke liye kuch nahi hai.

Step 2 — SVD. Sare singular values $0$ hain: $Σ = 0$ , isliye $Σ^{+} = 0$ (invert karo kuch nahi, kyunki koi $σ_{i}$ nonzero nahi hai). Ye step kyun? "Nonzeros invert karo, zeros chhodo" ka rule sab kuch zero chhod deta hai.

Step 3 — assemble karo. $A^{+} = V \cdot 0 \cdot U^{⊤} = (0000)$ . Ye step kyun? Koi invertible direction nahi, toh best refund hai "kuch wapas mat do" — jo exactly $x = 0$ hai, minimum-norm least-squares answer.

Verify karo (kyun $A^{+} = 0$ sahi answer hai, cop-out nahi): pseudoinverse ko $A x = b$ ka least-squares, minimum-norm solution dena chahiye. $A = 0$ ke liye, har $x$ error deta hai $∥0 \cdot x - b ∥ = ∥ b ∥$ , isliye error reduce nahi ho sakti; un sare $x$ mein jo us error ke liye tied hain, sabse chhote-norm wala $x = 0$ hai. Kyunki $x = A^{+} b = 0$ har $b$ ke liye, hume chahiye $A^{+} = 0$ — wo unique matrix jo fairest refund deliver karta hai. Chaaon Penrose conditions phir trivially hold karti hain ( $A A^{+} A = 0 = A$ ✓).

Cell C-word — ek real-world fit

Worked example Example 7 (cell

C-word) — noisy readings se resistance Ek physicist voltage $V$ measure karta hai ek resistor par currents $I = 1, 2, 3$ amps par, milta hai $V = 2.1, 3.9, 6.1$ volts. Ohm's law kehta hai $V = R I$ (origin se hoti ek line). Best resistance $R$ ohms mein dhundho.

Forecast: readings slope-2 line ke paas hain. Guess karo $R \approx 2 Ω$ .

Step 1 — model karo $A x = b$ ki tarah. $A = 123$ (currents), $x = (R)$ , $b = 2.1 3.9 6.1$ (voltages). Ye step kyun? Ek unknown $R$ , teen noisy equations → overdetermined → cell C-word = C-tall in disguise.

Step 2 — cell & formula. $3 \times 1$ , nonzero column → full column rank → $R = (A^{⊤} A)^{- 1} A^{⊤} b$ . Ye step kyun? Noise par least squares, exactly Example 1 jaisa.

Step 3 — $A^{⊤} A$ aur $A^{⊤} b$ compute karo. $A^{⊤} A = 1^{2} + 2^{2} + 3^{2} = 1 + 4 + 9 = 14$ . $A^{⊤} b = 1 (2.1) + 2 (3.9) + 3 (6.1) = 2.1 + 7.8 + 18.3 = 28.2$ . Ye step kyun? Ye do numbers hi sab kuch hain jo normal equation $A^{⊤} A R = A^{⊤} b$ ko chahiye.

Step 4 — $R$ ke liye solve karo. $R = \frac{A ^{⊤} b}{A ^{⊤} A} = \frac{28.2}{14} = 2.0142857 \dots \approx 2.01 Ω$ . Ye step kyun? Formula higher-current readings ko zyada weight deta hai (larger $I^{2}$ ), jo sahi hai: unme zyada signal hota hai.

Verify karo (units + sanity): $[A^{⊤} b] =$ amp·volt, $[A^{⊤} A] =$ amp $^{2}$ , ratio ki units hain volt/amp $= Ω$ ✓. Value $\approx 2.01 Ω$ forecast $\approx 2 Ω$ se match karti hai ✓ (Least Squares Regression ka application).

Cell C-twist-t — tall exam trap

Worked example Example 8 (cell

C-twist-t) — ek tall matrix jahan tall formula phir bhi fail karta hai $A = 123246$ . Ye $3 \times 2$ (tall) hai, toh ek jaldi student $(A^{⊤} A)^{- 1} A^{⊤}$ ke liye haath badhata hai. Trap: column 2 exactly $2 \times$ column 1 hai, isliye rank $= 1$ , aur $A^{⊤} A$ singular hai. $A^{+}$ sahi tarike se dhundho.

Forecast: kya tall formula kaam karega? (Nahi — aur yahi exam ka pura point hai.)

Step 1 — trap detect karo. Columns $(1, 2, 3)$ aur $(2, 4, 6)$ proportional hain → rank $1 < 2$ → full column rank nahi → cell C-twist-t → SVD required. Ye step kyun? Tall shape zaroori hai lekin sufficient nahi (parent ki Mistake 1); hume independent columns chahiye.

Step 2 — $A$ ko ek outer product ki tarah likho (rank-1). $A = 123 (12)$ , yani $A = c r^{⊤}$ jahan column-direction $c = (1, 2, 3)$ aur row-direction $r = (1, 2)$ . Ye step kyun? Har rank-1 matrix ek single column times single row hoti hai; ye SVD ko trivially read off karna easy bana deta hai.

Step 3 — unit singular vectors aur $σ_{1}$ nikalo. Normalise karo: left singular vector hai $u = \frac{c}{∥ c ∥} = \frac{1}{14} (1, 2, 3)$ , aur right singular vector hai $v = \frac{r}{∥ r ∥} = \frac{1}{5} (1, 2)$ . Akela singular value hai $σ_{1} = ∥ c ∥ ∥ r ∥ = 14 \cdot 5 = 70$ . Ye step kyun? Kyunki $A = c r^{⊤} = (∥ c ∥ u) (∥ r ∥ v)^{⊤} = (∥ c ∥∥ r ∥) u v^{⊤}$ , $A = σ_{1} u v^{⊤}$ se compare karne par force hota hai $σ_{1} = ∥ c ∥∥ r ∥$ . Yahan $u$ ab explicitly define hai — normalised column direction.

Step 4 — rank-1 matrix ka pseudoinverse. $A = σ_{1} u v^{⊤}$ ke liye, SVD rule deta hai $A^{+} = \frac{1}{σ _{1}} v u^{⊤}$ . Substitute karke: $A^{+} = \frac{1}{70} \cdot \frac{1}{5} (12) \cdot \frac{1}{14} (123) = \frac{1}{70} (122436) .$ Ye step kyun? $70 \cdot 5 \cdot 14 = 70 \cdot 70 = 70$ ; sirf surviving direction invert hoti hai.

Verify karo (Penrose condition 1): $A A^{+} A = A$ . Kyunki $A^{+} A = v v^{⊤} = \frac{1}{5} (1224)$ (row-space projection, $I$ nahi — Mistake 2 phir se), $A (A^{+} A)$ : pehla column $\frac{1}{5} [(1, 2, 3) \cdot 1 + (2, 4, 6) \cdot 2] = \frac{1}{5} (5, 10, 15) = (1, 2, 3)$ ✓, $A$ ka column 1 recover hota hai; column 2 similarly $(2, 4, 6)$ deta hai ✓.

Cell C-twist-f — fat exam trap

Worked example Example 9 (cell

C-twist-f) — ek fat matrix jahan fat formula phir bhi fail karta hai $A = (122436)$ . Ye $2 \times 3$ (fat) hai, toh ek jaldi student $A^{⊤} (A A^{⊤})^{- 1}$ ke liye haath badhata hai. Trap: row 2 exactly $2 \times$ row 1 hai, isliye rank $= 1$ , aur $A A^{⊤}$ singular hai — Example 8 ke trap ka direct analogue, ek dimension upar.

Forecast: kya fat formula kaam karega? (Nahi — aur ye Ex 8 ka mirror image hai.)

Step 1 — trap detect karo. Rows $(1, 2, 3)$ aur $(2, 4, 6)$ proportional hain → rank $1 < 2$ → full row rank nahi → cell C-twist-f → SVD required. Note karo $A = A_{8}^{⊤}$ (Example 8 ki matrix ka transpose), isliye sab kuch reuse kar sakte hain. Ye step kyun? Fat shape zaroori hai lekin sufficient nahi; hume independent rows chahiye.

Step 2 — $A$ ko ek outer product ki tarah likho. $A = (12) (123)$ , yani $A = c r^{⊤}$ jahan column-direction $c = (1, 2)$ aur row-direction $r = (1, 2, 3)$ . Ye step kyun? Pehle jaisi rank-1 reading, ab do directions ke roles swap hain.

Step 3 — unit singular vectors aur $σ_{1}$ . $u = \frac{c}{∥ c ∥} = \frac{1}{5} (1, 2)$ (left, size 2), $v = \frac{r}{∥ r ∥} = \frac{1}{14} (1, 2, 3)$ (right, size 3), aur $σ_{1} = ∥ c ∥∥ r ∥ = 5 \cdot 14 = 70$ . Ye step kyun? Ex 8 jaisi hi logic; yahan $u$ $R^{2}$ mein rehta hai aur $v$ $R^{3}$ mein.

Step 4 — $A^{+}$ assemble karo. $A^{+} = \frac{1}{σ _{1}} v u^{⊤} = \frac{1}{70} \cdot \frac{1}{14} 123 \cdot \frac{1}{5} (12) = \frac{1}{70} 123246$ . Ye step kyun? $A^{+}$ hai $3 \times 2$ as required, aur ye $(A_{8}^{+})^{⊤}$ ke barabar hai — ek accha sanity check, kyunki $(A^{⊤})^{+} = (A^{+})^{⊤}$ hamesha.

Verify karo (Penrose condition 1): $A A^{+} A = A$ . Yahan $A A^{+} = u u^{⊤} = \frac{1}{5} (1224)$ (column-space projection line $(1, 2)$ par). Phir $(A A^{+}) A$ : row 1 ke liye, $\frac{1}{5} [1 \cdot (1, 2, 3) + 2 \cdot (2, 4, 6)] = \frac{1}{5} (5, 10, 15) = (1, 2, 3)$ ✓; row 2 deta hai $(2, 4, 6)$ ✓.

Flashcards

x = 3

aur

x = 1

fit karne wala best single number (Ex 1)?

x = 2

, average; residual

(1, - 1)

⟂ column

(1, 1)

x_{1} + 2 x_{2} = 4

ka min-norm solution (Ex 3)?

x = (4/5, 8/5)

, nullspace direction

(2, - 1)

ke perpendicular.

Tall matrix ko kabhi kabhi SVD kyun chahiye (Ex 8)?

Agar uske columns dependent hain (rank deficient), toh

A^{⊤} A

singular hai isliye tall formula fail karta hai.

Fat matrix ko kabhi kabhi SVD kyun chahiye (Ex 9)?

Agar uske rows dependent hain (rank deficient), toh

A A^{⊤}

singular hai isliye fat formula fail karta hai.

(1111)

ka pseudoinverse (Ex 5)?

\frac{1}{4} (1111)

; note karo

A^{+} A

ek projection hai,

I

nahi.

Zero matrix ka pseudoinverse (Ex 6)?

Zero matrix khud hi.

V = 2.1, 3.9, 6.1

I = 1, 2, 3

se best-fit resistance (Ex 7)?

R = 28.2/14 \approx 2.01 Ω

Transpose aur pseudoinverse ko jodne wali handy identity (Ex 9)?

(A^{⊤})^{+} = (A^{+})^{⊤}

Connections

Pseudoinverse — parent; ye page uska scenario drill hai.
Singular Value Decomposition — cells C-sqsing, C-zero, C-twist-t, C-twist-f ke liye ek safe tool.
Least Squares Regression — cells C-tall aur C-word.
Orthogonal Projection — Ex 1 mein residual-⟂-column fact.
Four Fundamental Subspaces — Ex 3 mein min-norm ⟂ nullspace, Ex 8–9 mein row/column-space projections.
Matrix Inverse — Ex 4 mein collapse $A^{+} = A^{- 1}$ .
Eigenvalues and Eigenvectors — Ex 5 mein singular values $2, 0$ .

Worked examples — Pseudoinverse