Worked examples — Diagonalization — conditions, procedure

4.5.31 · D3 · Maths › Linear Algebra (Full) › Diagonalization — conditions, procedure

Tumne procedure dekh li hai. Ab hum ise har tarah ki matrix ke against drill karte hain jo yeh topic tumhare saamne rakh sakta hai — sirf friendly wali nahi. Goal yeh hai: is page ke baad, koi bhi exam matrix tumhe surprise nahi karegi, kyunki tum uska type pehle se dekh chuke hoge.

Shuru karne se pehle, ek reminder un vocabulary ki jo hum non-stop use karenge (sab parent note aur Eigenvalues and Eigenvectors mein build ki gayi hain):

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Scenario matrix

Har diagonalization problem in cells mein se exactly ek mein fit hoti hai. Hum sab ko cover karenge.

#	Cell class	Kya special hai	Diagonalizable?
A	$n$ distinct real eigenvalues	sabse aasaan case, free win	✅ hamesha
B	Repeated eigenvalue, $g=a$	repeat ke bawajood kaafi arrows	✅
C	Repeated eigenvalue, $g<a$ (defective)	bahut kam arrows	❌
D	Complex eigenvalues (real matrix)	andar chhupa rotation	✅ $\mathbb{C}$ ke upar, ❌ $\mathbb{R}$ ke upar
E	Zero eigenvalue (singular $A$)	$\det A=0$, ek squashed direction	depends karta hai ($g$ vs $a$)
F	Symmetric matrix	Spectral Theorem guarantee	✅ hamesha, orthogonally
G	Real-world / word problem	population / [[Matrix Powers and Exponentials	$A^n$]]
H	Exam twist: diya $D,P$, $A$ reconstruct karo	machine ko ulta chalao	—

Jo signs hum cover karte hain unka legend: eigenvalues jo positive, negative, zero, aur complex ($\pm i$) hain sab neeche aate hain, aur symmetric aur non-symmetric dono matrices.

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Example A — distinct real eigenvalues (positive & negative)

Forecast: diagonal ke bahar ke dono entries equal hain, toh yeh matrix symmetric hai. Guess: do real eigenvalues, aur (spoiler) perpendicular eigenvectors.

1. Characteristic equation. $\det (A-\lambda I)=(1-\lambda {)}^2-4={\lambda }^2-2\lambda -3=(\lambda -3)(\lambda +1).$ Toh ${\lambda }_1=3,{\lambda }_2=-1$. Yeh step kyun? Eigenvalues woh $\lambda$ hain jo $A-\lambda I$ ko singular banate hain; is polynomial ke roots exactly wahi values hain. Notice karo ek positive hai, ek negative — ek valid, common case.

2. Eigenvector for $\lambda =3$. $A-3I=\left(\begin{array}{cc}-2& 2\\ 2& -2\end{array}\right)$, row $-2x_1+2x_2=0\Rightarrow x_1=x_2$. Lo ${\mathbf{p}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$. Yeh step kyun? Humein woh direction chahiye jo sirf stretch ho. $A-3I$ ka null space wahi direction hai.

3. Eigenvector for $\lambda =-1$. $A+I=\left(\begin{array}{cc}2& 2\\ 2& 2\end{array}\right)\Rightarrow x_1=-x_2$. Lo ${\mathbf{p}}_2=\left(\begin{array}{c}1\\ -1\end{array}\right)$. Yeh step kyun? Doosre eigenvalue ke liye same reasoning. Check karo: ${\mathbf{p}}_1\cdot {\mathbf{p}}_2=1-1=0$ — perpendicular, jaise forecast kiya tha (symmetric matrices hamesha yahi deti hain, cell F).

4. Assemble karo (order match hona chahiye!): $P=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right),D=\left(\begin{array}{cc}3& 0\\ 0& -1\end{array}\right).$

Verify karo: $\det P=-2\ne 0$ (invertible ✓). Column 1 check karo: $A{\mathbf{p}}_1=\left(\begin{array}{c}3\\ 3\end{array}\right)=3{\mathbf{p}}_1$ ✓; column 2: $A{\mathbf{p}}_2=\left(\begin{array}{c}-1\\ 1\end{array}\right)=-1\cdot {\mathbf{p}}_2$ ✓.

Picture dekho: red aur mint arrows eigenvector directions hain. Red wala $3\times$ grow karta hai; mint wala flip hota hai (yahi hai negative eigenvalue) aur same length rehta hai. Plane mein koi aur direction apni direction nahi rakhti.

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Example B — repeated eigenvalue, phir bhi diagonalizable ($g=a$)

Forecast: $\lambda =5$ do baar aata hai. Repeated $\Rightarrow$ danger. Lekin yeh pehle se diagonal hai, toh zaroor bach jaayega. Guess: haan, $g(5)=2$ ke saath.

1. Eigenvalues. Diagonal hone ki wajah se, woh diagonal par hi baithte hain: $\lambda =5$ (do baar), $\lambda =2$. Toh $a(5)=2,a(2)=1$. Yeh step kyun? Ek diagonal (ya triangular) matrix ke liye, $\det (A-\lambda I)=\prod ({\text{diag}}_i-\lambda )$, toh roots = diagonal entries. Koi kaam nahi.

2. Geometric multiplicity of $\lambda =5$. $A-5I=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& -3\end{array}\right)$. Null space mein woh saare vectors hain jinka teesra component $0$ hai — yeh ek 2-dimensional plane hai. Toh $g(5)=2$. Yeh step kyun? Humein $g=a$ check karna hai. Yahan $g(5)=2=a(5)$, toh repeat "healthy" hai.

3. Eigenvectors. $\lambda =5$ ke liye: ${\mathbf{e}}_1=(1,0,0),{\mathbf{e}}_2=(0,1,0)$. $\lambda =2$ ke liye: ${\mathbf{e}}_3=(0,0,1)$. Yeh step kyun? Ek $3\times 3$ matrix ke liye teen independent arrows $\Rightarrow$ woh space span karte hain $\Rightarrow$ diagonalizable.

4. Assemble karo. $P=I$, $D=A$. (Yeh pehle se hi diagonal tha.)

Verify karo: $g(5)+g(2)=2+1=3=n$ ✓. Yahi parent note ka lesson hai — ek repeated eigenvalue theek hai jab woh kaafi arrows deta hai.

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Example C — defective: repeated eigenvalue with $g<a$

Forecast: upper-triangular with repeated $4$ aur ek nonzero corner — shear jaisa lagta hai. Guess: sirf ek eigenvector, toh defective.

1. Eigenvalues. Triangular $\Rightarrow \lambda =4$ do baar, $a(4)=2$. Yeh step kyun? Triangular matrices mein eigenvalues diagonal se seedhe padh lo.

2. Eigenvectors. $A-4I=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Solve karo: doosra component $x_2=0$, $x_1$ free. Sirf ek arrow $\mathbf{v}=(1,0)$, toh $g(4)=1$. Yeh step kyun? Hum $(A-4I)\mathbf{x}=0$ ke independent solutions count karte hain. Sirf ek $\Rightarrow$ mushkil.

3. Compare karo. $g(4)=1<a(4)=2$. Do independent arrows nahi hain, toh koi invertible $P$ exist nahi kar sakti. Yeh step kyun? $P$ ko $n=2$ independent columns chahiye; hamare paas $1$ hai. Diagonalization impossible hai — yeh matrix Jordan Normal Form territory mein jaati hai.

Verify karo: $g<a$ confirm hua. Sanity check: ek shear points ko sideways push karta hai; sirf horizontal axis jagah par rehti hai, toh geometrically genuinely ek fixed direction hai — $g=1$ se match karta hai.

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Example D — complex eigenvalues (ek rotation)

Forecast: rotation har real direction ko turn karta hai, toh koi real eigenvector nahi hai. Guess: eigenvalues complex hain; sirf $\mathbb{C}$ ke upar diagonalizable.

1. Characteristic equation. $\det (A-\lambda I)={\lambda }^2+1=0\Rightarrow \lambda =\pm i$. Yeh step kyun? Jahan ek real matrix ka koi real root nahi hota, complex roots conjugate pair mein aate hain. Yahan $i=\sqrt{-1}$, woh number jiska square $-1$ hai.

2. Eigenvector for $\lambda =i$. $A-iI=\left(\begin{array}{cc}-i& -1\\ 1& -i\end{array}\right)$. Row 2: $x_1-i{x}_2=0\Rightarrow x_1=i{x}_2$. Lo ${\mathbf{p}}_1=\left(\begin{array}{c}i\\ 1\end{array}\right)$. Yeh step kyun? Eigenvector ab ${\mathbb{C}}^2$ mein rehta hai. Yeh allowed hai — hum sirf ise real arrow ke roop mein draw nahi kar sakte.

3. Eigenvector for $\lambda =-i$. Sab conjugate karo: ${\mathbf{p}}_2=\left(\begin{array}{c}-i\\ 1\end{array}\right)$. Yeh step kyun? Ek real matrix ke liye, $\bar{\lambda }$ ka eigenvector $\lambda$ ke eigenvector ka conjugate hota hai — free mein milta hai.

4. $\mathbb{C}$ ke upar assemble karo. $P=\left(\begin{array}{cc}i& -i\\ 1& 1\end{array}\right),D=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right).$

Verify karo: $A{\mathbf{p}}_1=\left(\begin{array}{c}-1\\ i\end{array}\right)=i\left(\begin{array}{c}i\\ 1\end{array}\right)$ ✓ (kyunki $i\cdot i=-1$). $\mathbb{R}$ ke upar: koi real eigenvector exist nahi karta, toh $\mathbb{R}$ ke upar diagonalizable nahi — parent note ki doosri "steel-man" mistake concrete roop mein.

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Example E — zero eigenvalue (singular matrix)

Forecast: yeh plane ko $x$-axis par flatten karta hai. Ek flattened direction $0$ se multiply hoti hai. Guess: eigenvalues $1$ aur $0$, dono real, diagonalizable.

1. Eigenvalues. Diagonal $\Rightarrow \lambda =1,\lambda =0$. Yeh step kyun? $\lambda =0$ ek bilkul legal eigenvalue hai; iska matlab hai $A$ singular hai ($\det A=0$) aur uska ek nonzero null space hai.

2. Eigenvector for $\lambda =1$. $A-I=\left(\begin{array}{cc}0& 0\\ 0& -1\end{array}\right)\Rightarrow x_2=0$. ${\mathbf{p}}_1=(1,0)$. Yeh step kyun? $x$-axis untouched hai (poori length par rakha gaya), toh uska eigenvalue $1$ hai.

3. Eigenvector for $\lambda =0$. $A-0I=A\Rightarrow x_1=0$. ${\mathbf{p}}_2=(0,1)$. Yeh step kyun? $y$-direction zero par crush ho jaati hai — yahi $\lambda =0$ ka eigenvector hai, aur yeh exactly $\ker A$ hai.

4. Assemble karo. $P=I,D=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$. Do independent arrows $\Rightarrow$ diagonalizable.

Verify karo: $g(1)=1,g(0)=1$, sum $=2=n$ ✓. Ek zero eigenvalue diagonalization ko block nahi karta; sirf yeh kehta hai ki ek axis flat crush ho gayi.

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Example F — symmetric matrix (Spectral Theorem, orthogonal $P$)

Forecast: symmetric $\Rightarrow$ Spectral Theorem real eigenvalues aur perpendicular eigenvectors promise karta hai jinhe hum rotation matrix mein badal sakte hain.

1. Eigenvalues. $\det (A-\lambda I)=(2-\lambda {)}^2-1={\lambda }^2-4\lambda +3=(\lambda -3)(\lambda -1)$. Toh $\lambda =3,1$ — dono real, jaise guarantee tha. Yeh step kyun? Symmetric matrices kabhi complex eigenvalues nahi produce karti; isliye cell F hamesha ✅ hai.

2. Eigenvectors. $\lambda =3$: $x_1=x_2\Rightarrow (1,1)$. $\lambda =1$: $x_1=-x_2\Rightarrow (1,-1)$. Dot product $=0$ ✓ perpendicular. Yeh step kyun? Hum orthonormal basis chahte hain; ek symmetric matrix ke distinct-eigenvalue eigenvectors automatically orthogonal hote hain.

3. Normalize karo (har ek ko uski length $\sqrt{2}$ se divide karo) ek orthogonal $P$ paane ke liye: $P=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right),D=\left(\begin{array}{cc}3& 0\\ 0& 1\end{array}\right).$ Yeh step kyun? Unit, perpendicular columns ke saath, $P^{\top }P=I$, toh $P^{-1}=P^{\top }$ — koi messy inverse nahi chahiye.

Verify karo: $P^{\top }P=I$ (check karo: rows unit aur orthogonal hain) ✓, aur $\det P=-1$ (ek orthogonal matrix, $|\det |=1$) ✓.

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Example G — word problem: population model & $A^n$

Forecast: column sums $=1$ hamesha eigenvalue $\lambda =1$ (steady state) force karta hai. Guess: doosra eigenvalue $0$ aur $1$ ke beech hai, toh repeated powers ke neeche uska contribution khatam ho jaata hai.

1. Eigenvalues. $\det (A-\lambda I)=(0.9-\lambda )(0.8-\lambda )-0.02={\lambda }^2-1.7\lambda +0.7=(\lambda -1)(\lambda -0.7)$. Toh $\lambda =1,0.7$. Yeh step kyun? $\lambda =1$ steady state confirm karta hai; $\lambda =0.7<1$ "transient" hai jo shrink hota hai.

2. Steady-state eigenvector ($\lambda =1$). $A-I=\left(\begin{array}{cc}-0.1& 0.2\\ 0.1& -0.2\end{array}\right)\Rightarrow x_1=2x_2$. Toh ${\mathbf{p}}_1=(2,1)$; population ke fraction ke roop mein, $(\frac{2}{3},\frac{1}{3})$. Yeh step kyun? Yahi woh distribution hai jis par sheher converge karta hai, start se regardless.

3. Transient eigenvector ($\lambda =0.7$). $A-0.7I=\left(\begin{array}{cc}0.2& 0.2\\ 0.1& 0.1\end{array}\right)\Rightarrow x_1=-x_2$, toh ${\mathbf{p}}_2=(1,-1)$. Yeh step kyun? $A^n$ ke under, yeh direction $0.7^n\to 0$ se multiply hoti hai, toh vanish ho jaati hai — sheher apna initial imbalance bhool jaata hai.

4. Long run. $A^n=P{D}^n{P}^{-1}$ aur $D^n=\left(\begin{array}{cc}1& 0\\ 0& 0.7^n\end{array}\right)\to \left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$. Toh har start $(2:1)$ split mein funnel hota hai, yaani $\approx 66.7\%$ city, $33.3\%$ suburb.

Verify karo: $A$ ke column sums $1$ hain ✓, toh $\lambda =1$ zaroor ek eigenvalue hona chahiye. Eigenvector fractions $\frac{2}{3}+\frac{1}{3}=1$ ✓. Units: people/people (ek fraction) — dimensionless ✓.

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Example H — exam twist: $D$ aur $P$ se $A$ rebuild karo

Forecast: diagonalization ko ulta chalana — tumhare paas pehle se eigen-data hai, bas multiply out karo. Yahi jagah hai jahan Similar Matrices aur Change of Basis milte hain.

1. $P^{-1}$ compute karo. $2\times 2$ ke liye: $P^{-1}=\frac{1}{\det P}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)$ jahan $\det P=-2$ hai, giving $P^{-1}=\left(\begin{array}{cc}1/2& 1/2\\ 1/2& -1/2\end{array}\right)$. Yeh step kyun? $A=PD{P}^{-1}$ ko inverse chahiye; $2\times 2$ inverse formula (diagonal swap karo, off-diagonal negate karo, det se divide karo) ise fast karta hai.

2. $PD$ multiply karo. $\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{cc}2& 0\\ 0& 4\end{array}\right)=\left(\begin{array}{cc}2& 4\\ 2& -4\end{array}\right)$ (column 1 ko $2$ se scale karo, column 2 ko $4$ se). Yeh step kyun? Right side par diagonal matrix se multiply karna sirf columns scale karta hai — sasta hai.

3. $P^{-1}$ se multiply karo. $\left(\begin{array}{cc}2& 4\\ 2& -4\end{array}\right)\left(\begin{array}{cc}1/2& 1/2\\ 1/2& -1/2\end{array}\right)=\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)$. Yeh step kyun? Yeh $PD{P}^{-1}$ finish karta hai, original transformation ko standard coordinates mein wapas deta hai.

Verify karo: $A=\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)$. Trace check karo: $3+3=6=2+4$ (eigenvalues ka sum) ✓. Determinant check karo: $9-1=8=2\cdot 4$ (eigenvalues ka product) ✓. Symmetric hai, jaise $P$ ke orthogonal-ish columns se force hota hai.

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Recall Kaun sa cell, kaun sa verdict?

Distinct real eigenvalues — diagonalizable? ::: Haan, hamesha (Cell A). Repeated eigenvalue with $g=a$ — diagonalizable? ::: Haan (Cell B). Repeated eigenvalue with $g<a$ (defective) — diagonalizable? ::: Nahi (Cell C). Real rotation with eigenvalues $\pm i$ — $\mathbb{R}$ ke upar diagonalizable? ::: Nahi; sirf $\mathbb{C}$ ke upar (Cell D). Ek zero eigenvalue ($\det A=0$) — kya yeh diagonalization block karta hai? ::: Nahi; sirf ek squashed direction mark karta hai (Cell E). Symmetric matrix — diagonalizable? ::: Hamesha, aur orthogonally, Spectral Theorem se (Cell F). Ek final $A$ par sabse quick sanity checks? ::: Trace $=\sum {\lambda }_i$ aur $\det =\prod {\lambda }_i$.

Parent topic par wapas jao.