Worked examples — Orthogonal sets and orthonormal basis

4.5.34 · D3 · Maths › Linear Algebra (Full) › Orthogonal sets and orthonormal basis

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Scenario matrix

Is topic ka har problem inhi cells mein se kisi ek mein fit hota hai. Neeche ke examples mein us cell ka label diya gaya hai. (Yahan $W$ wahi flat sheet hai jo basis span karta hai, jaise abhi define kiya.)

Cell	Situation	Kya tricky hai	Example
C1	Orthogonal (unit nahin) basis, $\mathbf{y}$ inside $W$	Denominator ${\mathbf{v}}_j\cdot {\mathbf{v}}_j$ rakhna zaroori	A
C2	Orthonormal basis	Denominator $1$ hota hai — use mat likho	B
C3	Size $p<n$ ka orthogonal set; $\mathbf{y}$ span ke bahar	Formula ek projection deta hai, $\mathbf{y}$ nahin	C
C4	Degenerate: set mein ek zero vector	Independence toot jaati hai; length-squared $=0$ se divide hoga	D
C5	Sign cases — negative coordinates, mixed-sign components	Dot products mein signs ka hisaab	E
C6	Orthogonal matrix length/angle preserve karta hai	$U^{\mathsf{T}}U=I$ verify karo; norm check karo	F
C7	Limiting case — near-orthogonal vectors, "almost" ki keemat	Small off-diagonal error	G
C8	Word problem (signals / forces) ko coordinates mein map karo	Words ko dot products mein translate karna	H
C9	Exam twist — "orthogonal columns" jo orthonormal nahin hain	Naming trap; pehle normalize karna hoga	I

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Example A — Cell C1: orthogonal basis, denominator rakhna zaroori hai

Step 1 — orthogonal check. ${\mathbf{v}}_1\cdot {\mathbf{v}}_2=(2)(-1)+(1)(2)=-2+2=0.$ ✓ Yeh step kyun? Tabhi shortcut formula $c_j=\frac{\mathbf{y}\cdot {\mathbf{v}}_j}{{\mathbf{v}}_j\cdot {\mathbf{v}}_j}$ apply hogi. (Yahan parent note ke generic ${\mathbf{u}}_j$ hi hamare ${\mathbf{v}}_j$ hain — same formula, matching names.)

Step 2 — numerator ke do dot products. $\mathbf{y}\cdot {\mathbf{v}}_1=(4)(2)+(3)(1)=11,\mathbf{y}\cdot {\mathbf{v}}_2=(4)(-1)+(3)(2)=2.$ Yeh step kyun? Numerator measure karta hai "kitna $\mathbf{y}$, ${\mathbf{v}}_j$ ki direction mein point kar raha hai."

Step 3 — denominators (length-squared). ${\mathbf{v}}_1\cdot {\mathbf{v}}_1=4+1=5,{\mathbf{v}}_2\cdot {\mathbf{v}}_2=1+4=5.$ Yeh step kyun? Kyunki ${\mathbf{v}}_j$ unit vectors nahin hain, hume unka apna size divide out karna hoga. Isse skip karna classic error hai.

Step 4 — assemble karo. $c_1=\frac{11}{5},c_2=\frac{2}{5},\mathbf{y}=\frac{11}{5}{\mathbf{v}}_1+\frac{2}{5}{\mathbf{v}}_2.$

Verify: $\frac{11}{5}\left[\begin{array}{c}2\\ 1\end{array}\right]+\frac{2}{5}\left[\begin{array}{c}-1\\ 2\end{array}\right]=\left[\begin{array}{c}(22-2)/5\\ (11+4)/5\end{array}\right]=\left[\begin{array}{c}4\\ 3\end{array}\right]=\mathbf{y}.$ ✓

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Example B — Cell C2: orthonormal, denominator nahin

Step 1 — unit length confirm karo. ${\mathbf{q}}_1\cdot {\mathbf{q}}_1=\frac{1}{5}(4+1)=1.$ ✓ Yeh step kyun? Jab ek baar har ${\mathbf{q}}_j\cdot {\mathbf{q}}_j=1$ ho, to coordinate bas $\mathbf{y}\cdot {\mathbf{q}}_j$ hoti hai.

Step 2 — dot products. $c_1=\mathbf{y}\cdot {\mathbf{q}}_1=\frac{1}{\sqrt{5}}(8+3)=\frac{11}{\sqrt{5}},c_2=\mathbf{y}\cdot {\mathbf{q}}_2=\frac{1}{\sqrt{5}}(-4+6)=\frac{2}{\sqrt{5}}.$ Yeh step kyun? Denominator ki zaroorat nahin — orthonormality ne already woh division kar di hai.

Verify: $c_1{\mathbf{q}}_1+c_2{\mathbf{q}}_2=\frac{11}{\sqrt{5}}\cdot \frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]+\frac{2}{\sqrt{5}}\cdot \frac{1}{\sqrt{5}}\left[\begin{array}{c}-1\\ 2\end{array}\right]=\frac{11}{5}\left[\begin{array}{c}2\\ 1\end{array}\right]+\frac{2}{5}\left[\begin{array}{c}-1\\ 2\end{array}\right]=\left[\begin{array}{c}4\\ 3\end{array}\right].$ ✓ Same $\mathbf{y}$, alag coordinates — A ke saath consistent.

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Example C — Cell C3: $\mathbf{y}$ span ke bahar hai → projection

Step 1 — formula waise bhi apply karo. $c_1=\frac{\mathbf{y}\cdot {\mathbf{u}}_1}{{\mathbf{u}}_1\cdot {\mathbf{u}}_1}=\frac{2}{1}=2,c_2=\frac{\mathbf{y}\cdot {\mathbf{u}}_2}{{\mathbf{u}}_2\cdot {\mathbf{u}}_2}=\frac{3}{1}=3.$ Yeh step kyun? Formula ne kabhi nahin poocha ki $\mathbf{y}\in W$ hai ya nahin; woh bas project karta hai.

Step 2 — result banao. $\hat{\mathbf{y}}=2{\mathbf{u}}_1+3{\mathbf{u}}_2=\left[\begin{array}{c}2\\ 3\\ 0\end{array}\right].$ Yeh step kyun? Yeh $\hat{\mathbf{y}}$ woh point hai $W$ ka jo $\mathbf{y}$ ke sabse paas hai — plane par Orthogonal projection. $z=5$ wala part drop ho jaata hai kyunki koi bhi basis vector upar point nahin karta.

Step 3 — jo bacha woh $W$ se orthogonal hai. $\mathbf{y}-\hat{\mathbf{y}}=\left[\begin{array}{c}0\\ 0\\ 5\end{array}\right],(\mathbf{y}-\hat{\mathbf{y}})\cdot {\mathbf{u}}_1=0,(\mathbf{y}-\hat{\mathbf{y}})\cdot {\mathbf{u}}_2=0.$ Yeh step kyun? Yahi projection ki defining property hai — error subspace ke perpendicular hoti hai, jo Least squares ka beej hai.

Verify: $\hat{\mathbf{y}}\ne \mathbf{y}$ (teesre slot mein alag hain), aur residual $[0,0,5]$ dono basis vectors ke saath $0$ dot karta hai. ✓ C3 ka lesson: formula hamesha projection return karta hai; woh $\mathbf{y}$ ke barabar tabhi hoga jab $\mathbf{y}$ actually span mein ho.

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Example D — Cell C4: ek zero vector poora set kharaab kar deta hai

Step 1 — kya yeh "orthogonal" hai? ${\mathbf{w}}_1\cdot {\mathbf{w}}_2=0$. Technically haan — zero vector sab se orthogonal hota hai. Yeh step kyun? Definition sirf nonzero distinct pairs ko non-perpendicular hone se rokti hai; zero ke baare mein kuch nahin kehti, isliye zero ghus aata hai.

Step 2 — independence fail hoti hai. Relation $0\cdot {\mathbf{w}}_1+5\cdot {\mathbf{w}}_2=0$ ek nonzero coefficient ke saath hold karti hai. To set dependent hai — yeh basis nahin hai. Yeh step kyun? Parent ke Section 2 mein nonzero orthogonal vectors chahiye the. Proof $c_j\Vert {\mathbf{w}}_j{\Vert }^2=0$ isolate karta hai; jab $\Vert {\mathbf{w}}_2\Vert =0$ ho to $c_2=0$ conclude nahin ho sakta.

Step 3 — formula zero se divide karta hai. $c_2=\frac{\mathbf{y}\cdot {\mathbf{w}}_2}{{\mathbf{w}}_2\cdot {\mathbf{w}}_2}=\frac{0}{0}\text{— undefined.}$ Yeh step kyun? Denominator ${\mathbf{w}}_2\cdot {\mathbf{w}}_2=0$ exactly length-squared hai, jo sirf zero vector ke liye $0$ hoti hai.

Verify: ${\mathbf{w}}_2\cdot {\mathbf{w}}_2=0$, jisse confirm hota hai ki division illegal hai. ✓ Rule: set ko orthogonal kehne se pehle hamesha zero vectors nikal do.

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Example E — Cell C5: saare signs

Step 1 — orthogonality (teen checks). ${\mathbf{a}}_1\cdot {\mathbf{a}}_2=1-1+0=0,{\mathbf{a}}_1\cdot {\mathbf{a}}_3=1+1-2=0,{\mathbf{a}}_2\cdot {\mathbf{a}}_3=1-1+0=0.$ Yeh step kyun? Har pair vanish honi chahiye warna shortcut invalid hai.

Step 2 — numerators (signs dhyan se). $\mathbf{y}\cdot {\mathbf{a}}_1=3-1-2=0,\mathbf{y}\cdot {\mathbf{a}}_2=3+1+0=4,\mathbf{y}\cdot {\mathbf{a}}_3=3-1+4=6.$ Yeh step kyun? Zero numerator ($c_1$) legal aur meaningful hai — $\mathbf{y}$ ka ${\mathbf{a}}_1$ ki direction mein koi component nahin.

Step 3 — denominators, phir divide karo. ${\mathbf{a}}_1\cdot {\mathbf{a}}_1=3,{\mathbf{a}}_2\cdot {\mathbf{a}}_2=2,{\mathbf{a}}_3\cdot {\mathbf{a}}_3=6.$ $c_1=\frac{0}{3}=0,c_2=\frac{4}{2}=2,c_3=\frac{6}{6}=1.$ Yeh step kyun? ${\mathbf{a}}_j$ orthogonal hain lekin unit length nahin, isliye hume har numerator ko apne length-squared se divide karna hoga — warna reconstruct kiya vector galat scale mein aayega.

Verify: $0{\mathbf{a}}_1+2{\mathbf{a}}_2+1{\mathbf{a}}_3=\left[\begin{array}{c}0+2+1\\ 0-2+1\\ 0+0-2\end{array}\right]=\left[\begin{array}{c}3\\ -1\\ -2\end{array}\right]=\mathbf{y}.$ ✓ Sign bookkeeping sahi rahi.

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Example F — Cell C6: rotation length aur angle preserve karta hai

Step 1 — orthonormal columns confirm karo. Column 1 length${}^2=\frac{1}{4}+\frac{3}{4}=1$; columns dot karne par $\frac{1}{2}(-\frac{\sqrt{3}}{2})+\frac{\sqrt{3}}{2}\cdot \frac{1}{2}=0$ aata hai. To $U^{\mathsf{T}}U=I$ (wahi clean $1$-diagonal grid). Yeh step kyun? Length-preservation $U^{\mathsf{T}}U=I$ se follow hoti hai, na ki is baat se ki $U$ "koi" matrix hai — QR decomposition dekho jahan yeh $U$-with-orthonormal-columns wala $Q$ hota hai.

Step 2 — $U$ apply karo. $U\mathbf{x}=\left[\begin{array}{c}\frac{1}{2}\cdot 2\\ \frac{\sqrt{3}}{2}\cdot 2\end{array}\right]=\left[\begin{array}{c}1\\ \sqrt{3}\end{array}\right].$ Yeh step kyun? Hum $2$ units lambe vector ko $60^{\circ }$ se rotate kar rahe hain; figure us arc ko dikhata hai jo woh sweep karta hai.

Step 3 — norms compare karo. $\Vert U\mathbf{x}\Vert =\sqrt{1+3}=2=\Vert \mathbf{x}\Vert .$ Yeh step kyun? Yahi is example ka poora point hai — dono lengths ko side by side rakhna geometric claim "rotation doesn't stretch" ko confirm karta hai, abstract identity $U^{\mathsf{T}}U=I$ ko ek concrete number match mein badalta hai.

Verify: $\Vert U\mathbf{x}{\Vert }^2=1^2+(\sqrt{3}{)}^2=4$ aur $\Vert \mathbf{x}{\Vert }^2=2^2=4$. ✓ Length untouched, bilkul jaisa Dot product and norms orthonormal maps ke liye predict karta hai.

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Example G — Cell C7: limiting / "almost orthogonal" case

Step 1 — true off-diagonal. ${\mathbf{b}}_1\cdot {\mathbf{b}}_2=ϵ=0.01\ne 0$. Yeh step kyun? Yahi nonzero entry hai jo shortcut zero assume karta hai; yahi saari error ka source hai.

Step 2 — naive coordinates. $c_1^{\text{naive}}=\mathbf{y}\cdot {\mathbf{b}}_1=1,c_2^{\text{naive}}=\mathbf{y}\cdot {\mathbf{b}}_2=ϵ+1=\mathrm{1.01.}$ Yeh step kyun? Yahi ek careless solver likhta hai, non-orthogonality ko ignore karke.

Step 3 — reconstruct karo aur error measure karo. $\hat{\mathbf{y}}=1\cdot {\mathbf{b}}_1+1.01\cdot {\mathbf{b}}_2=\left[\begin{array}{c}1+1.01(0.01)\\ 1.01\end{array}\right]=\left[\begin{array}{c}1.0101\\ 1.01\end{array}\right].$ Error vector $\hat{\mathbf{y}}-\mathbf{y}=\left[\begin{array}{c}0.0101\\ 0.01\end{array}\right]$. Yeh step kyun? Yeh "almost orthogonal" ko "orthogonal" samajhne ki keemat quantify karta hai.

Verify: $ϵ=0.01$ ke saath, pehla error component $=ϵ+{ϵ}^2=0.0101$ aur doosra $=ϵ=0.01$. Dono $\to 0$ jaise $ϵ\to 0$. ✓ Moral: near-orthogonality, orthogonality nahin hai — exactness ke liye pehle Gram-Schmidt process run karo.

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Example H — Cell C8: word problem (do microphones)

Step 1 — orthonormal confirm karo. ${\mathbf{q}}_1\cdot {\mathbf{q}}_1=0.36+0.64=1$; ${\mathbf{q}}_1\cdot {\mathbf{q}}_2=-0.48+0.48=0$. ✓ Yeh step kyun? Sirf orthonormal channels mein har reading bina crosstalk ke ek single dot product hoti hai.

Step 2 — channel readings = dot products. $c_1=\mathbf{s}\cdot {\mathbf{q}}_1=3+8=11,c_2=\mathbf{s}\cdot {\mathbf{q}}_2=-4+6=2.$ Yeh step kyun? Har channel apni direction mein $\mathbf{s}$ ka component measure karta hai.

Step 3 — energy check (Parseval). $c_1^2+c_2^2=121+4=125,\Vert \mathbf{s}{\Vert }^2=25+100=125.$ Yeh step kyun? Orthonormal bases total energy conserve karte hain — ek physical sanity check jis par engineer rely karta hai.

Verify: $c_1^2+c_2^2=125=\Vert \mathbf{s}{\Vert }^2$, aur $11{\mathbf{q}}_1+2{\mathbf{q}}_2=\left[\begin{array}{c}6.6-1.6\\ 8.8+1.2\end{array}\right]=\left[\begin{array}{c}5\\ 10\end{array}\right]=\mathbf{s}.$ ✓

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Example I — Cell C9: exam naming trap

Step 1 — columns orthogonal hain? $\left[\begin{array}{c}3\\ 1\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 3\end{array}\right]=-3+3=0.$ ✓ Perpendicular, haan. Yeh step kyun? Claim ka premise genuinely sach hai — yahi trap ko seductive banata hai.

Step 2 — lekin kya woh unit vectors hain? Har column ki length${}^2=9+1=10\ne 1$. Yeh step kyun? "Orthogonal matrix" (ek historical misnomer) ko orthonormal columns chahiye. Length matter karti hai, isliye hume usse test karna hoga.

Step 3 — $M^{\mathsf{T}}M$ test karo. $M^{\mathsf{T}}M=\left[\begin{array}{cc}3& 1\\ -1& 3\end{array}\right]\left[\begin{array}{cc}3& -1\\ 1& 3\end{array}\right]=\left[\begin{array}{cc}10& 0\\ 0& 10\end{array}\right]=10I\ne I.$ Yeh step kyun? Ek orthogonal matrix ko $M^{\mathsf{T}}M=I$ (wahi clean $1$-diagonal grid) satisfy karna hoga; yahan $I$ ki jagah $10I$ aata hai, jo dikhata hai ki $M^{\mathsf{T}}\ne M^{-1}$. Actually $M^{-1}=\frac{1}{10}{M}^{\mathsf{T}}$.

Step 4 — fix: normalize karo. Har column ko uski length $\sqrt{10}$ se divide karo. Tab $Q=\frac{1}{\sqrt{10}}M$ ke orthonormal columns honge, aur ab $Q^{-1}=Q^{\mathsf{T}}$. Yeh step kyun? Normalize karne se har column ki length-squared $1$ ho jaati hai, jo exactly woh missing condition hai, jo merely-orthogonal $M$ ko ek true orthogonal matrix $Q$ mein convert karti hai.

Verify: $M^{\mathsf{T}}M=10I$ (to $M$ nahin hai orthogonal), jabki $\left(\frac{1}{\sqrt{10}}M{)}^{\mathsf{T}}\right(\frac{1}{\sqrt{10}}M)=\frac{1}{10}(10I)=I$. ✓ "Orthogonal matrix" word secretly orthonormal + square ka matlab rakhta hai — wahi distinction jo Eigenvalues and eigenvectors of symmetric matrices ke liye bhi matter karti hai.

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Recall Which cell is this?

Ek basis ke perpendicular columns hain lekin lengths $\sqrt{7}$ hain. Coordinates pane ke liye tum... ::: denominator ${\mathbf{v}}_j\cdot {\mathbf{v}}_j=7$ rakhte ho (Cell C1 — orthogonal, orthonormal nahin). Coordinate formula theek se chalta hai lekin ek vector $\ne \mathbf{y}$ return karta hai. Kya hua? ::: $\mathbf{y}$ span ke bahar tha; tumhe uska projection mila (Cell C3). Tumhara denominator $0$ hai. Diagnosis? ::: Set mein ek zero vector hai — use discard karo (Cell C4). "Orthogonal columns ⇒ orthogonal matrix." Sach? ::: Nahin — unit length aur square dono chahiye (Cell C9); pehle normalize karo.