Worked examples — Eigenvalues and eigenvectors — characteristic polynomial

4.5.29 · D3 · Maths › Linear Algebra (Full) › Eigenvalues and eigenvectors — characteristic polynomial

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The scenario matrix

Har $2\times 2$ (aur chhoti $n$) matrix in case classes mein se bilkul ek mein aati hai, jo do cheezein decide karti hain: characteristic polynomial ke discriminant ka sign (ki do roots kaise nikalte hain) aur matrix mein koi bhi degeneracy.

#	Case class	Kya trigger karta hai	Covered by
A	Do alag real eigenvalues	$\Delta >0$	Ex 1
B	Dono eigenvalues negative / sign check	$\Delta >0$, $\det >0$, $\mathrm{tr}<0$	Ex 2
C	Triangular shortcut	diagonal ke neeche zero	Ex 3
D	Repeated root, defective	$\Delta =0$, geo $<$ alg	Ex 4
E	Repeated root, NON-defective (scalar matrix)	$\Delta =0$, geo $=$ alg	Ex 5
F	Complex conjugate pair (rotation)	$\Delta <0$	Ex 6
G	Zero eigenvalue / singular matrix	$\det (A)=0$	Ex 7
H	$3\times 3$ + real-world word problem	degree-3 polynomial	Ex 8

Us table ki har row ek trap hai jo exam set kar sakta hai. Hum inhe neeche sab spring karte hain.

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Example 1 — Case A: do alag real eigenvalues

Forecast (pehle guess karo): $\mathrm{tr}=7$, $\det =12-2=10$. Do positive reals guess karo jinki sum $7$ aur product $10$ ho. Woh soch apne paas rakho.

1. Characteristic polynomial banao. $p(\lambda )=(4-\lambda )(3-\lambda )-(1)(2)={\lambda }^2-7\lambda +10.$ Yeh step kyun? Yeh ${\lambda }^2-\mathrm{tr}(A)\lambda +\det (A)$ hai — parent ke $2\times 2$ formula se seedha raasta. Dono coefficients ke liye dekho Trace of a Matrix aur Determinants.

2. Discriminant compute karo. $\Delta =7^2-4(10)=49-40=9>0.$ Yeh step kyun? $\Delta >0$ confirm karta hai ki hum Case A mein hain — do alag real roots.

3. Solve karo. $\lambda =\frac{7\pm \sqrt{9}}{2}=\frac{7\pm 3}{2}\Rightarrow {\lambda }_1=5,{\lambda }_2=2.$ Yeh step kyun? Quadratic formula; $9$ ka square root clean hai, isliye roots exact hain.

4. $\lambda =5$ ke liye Eigenvector. $(A-5I)\mathbf{x}=0$ solve karo: $\left(\begin{array}{cc}-1& 1\\ 2& -2\end{array}\right)\mathbf{x}=0\Rightarrow -x_1+x_2=0\Rightarrow x_1=x_2\Rightarrow \mathbf{x}=\left(\begin{array}{c}1\\ 1\end{array}\right).$ Yeh step kyun? Eigenvectors $A-\lambda I$ ke null space mein rehte hain, kabhi determinant mein nahi.

5. $\lambda =2$ ke liye Eigenvector. $(A-2I)=\left(\begin{array}{cc}2& 1\\ 2& 1\end{array}\right)$, row deta hai $2x_1+x_2=0\Rightarrow x_2=-2x_1$, isliye $\mathbf{x}=\left(\begin{array}{c}1\\ -2\end{array}\right)$.

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Example 2 — Case B: dono roots negative ke saath sign-check

Forecast: $\mathrm{tr}=-8$ (roots ki sum negative hai), $\det =16-1=15>0$ (product positive). Do numbers jinki sum negative aur product positive ho, woh dono negative hone chahiye. Guess: do negative reals.

1. Polynomial. $p(\lambda )={\lambda }^2-(-8)\lambda +15={\lambda }^2+8\lambda +15$. Yeh step kyun? $-\mathrm{tr}=+8$; note karo sign flip hota hai kyunki trace negative hai — ek classic galti.

2. Discriminant. $\Delta =8^2-4(15)=64-60=4>0$ → Case A/B real distinct. Yeh step kyun? Signs claim karne se pehle real roots confirm karta hai.

3. Solve karo. $\lambda =\frac{-8\pm 2}{2}\Rightarrow {\lambda }_1=-3,{\lambda }_2=-5$. Dono negative ✓ (forecast se match karta hai). Yeh step kyun? $\sqrt{4}=2$; upar ka $-8$ dono roots ko negative khichta hai, bilkul waise jaise sign argument ne promise kiya tha.

4. $\lambda =-3$ ke liye Eigenvector. $(A+3I)=\left(\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right)\Rightarrow x_1=x_2\Rightarrow \mathbf{x}=\left(\genfrac{}{}{0ex}{}{1}{1}\right)$.

5. $\lambda =-5$ ke liye Eigenvector. $(A+5I)=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\Rightarrow x_1=-x_2\Rightarrow \mathbf{x}=\left(\genfrac{}{}{0ex}{}{1}{-1}\right)$.

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Example 3 — Case C: triangular shortcut

Forecast: diagonal ke neeche sab kuch $0$ hai. Parent note ne kaha tha ki triangular matrices diagonal free mein de deti hain. Guess $\{6,-1,2\}$.

1. $\det (A-\lambda I)$ likho. $\det \left(\begin{array}{ccc}6-\lambda & -3& 4\\ 0& -1-\lambda & 5\\ 0& 0& 2-\lambda \end{array}\right)=(6-\lambda )(-1-\lambda )(2-\lambda ).$ Yeh step kyun? Ek triangular matrix ka determinant uske diagonal ka product hota hai — upar ke off-diagonal entries kabhi enter nahi karte, kyunki expansion ke har term ko ek zero column se factor chahiye.

2. Roots padho. Har factor $=0$ deta hai $\lambda =6,-1,2$. Yeh step kyun? Product tab zero hota hai jab ek factor zero ho — koi algebra nahi chahiye.

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Example 4 — Case D: repeated root, DEFECTIVE

Forecast: triangular, isliye $\lambda =2$ do baar. Lekin kitni directions? Padhne se pehle guess karo.

1. Polynomial. $p(\lambda )=(2-\lambda {)}^2$, isliye $\lambda =2$ ke saath algebraic multiplicity 2. Yeh step kyun? $(\lambda -2)$ par exponent count karta hai ki root kitni baar repeat hota hai.

2. Eigenvectors nikalo. $(A-2I)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. System force karta hai $x_2=0$, $x_1$ free, isliye $\mathbf{x}=\left(\genfrac{}{}{0ex}{}{1}{0}\right)$ ek hi direction hai. Yeh step kyun? Free variables ki sankhya = null space ki dimension = geometric multiplicity $=1$.

3. Diagnose karo. Geometric ($1$) $<$ algebraic ($2$) ⇒ defective, isliye $A$ diagonalizable nahi hai (dekho Diagonalization).

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Example 5 — Case E: repeated root, NON-defective

Forecast: yeh ek scalar matrix hai; yeh sab kuch $2$ se scale karta hai. Toh shayad har vector ek eigenvector hai? Guess karo.

1. Polynomial. $p(\lambda )=(2-\lambda {)}^2$, $\lambda =2$, algebraic multiplicity $2$ — Ex 4 jaise abhi tak same. Yeh step kyun? Dikhata hai ki polynomial akela nahi bata sakta defective se non-defective — tumhe null space inspect karna hoga.

2. Eigenvectors. $(A-2I)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$, zero matrix. Har vector $0\mathbf{x}=0$ satisfy karta hai, isliye $\left(\genfrac{}{}{0ex}{}{1}{0}\right)$ aur $\left(\genfrac{}{}{0ex}{}{0}{1}\right)$ dono independent eigenvectors hain. Yeh step kyun? Zero matrix ka null space pura plane hota hai — geometric multiplicity $=2$.

3. Diagnose karo. Geometric $=$ algebraic $=2$ ⇒ non-defective, fully diagonalizable (yeh already diagonal hai).

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Example 6 — Case F: complex conjugate pair (ek rotation)

Forecast: rotation har real vector ko naye direction mein le jaata hai, isliye koi real eigenvector exist nahi kar sakta. Jo bhi roots aayenge woh non-real hone chahiye. Figure dekho.

Red vector aur uska image $A\mathbf{x}$ (chalk-blue) perpendicular hain — koi real vector apni line par nahi rehta, isliye koi real $\lambda$ nahi.

1. Polynomial. $p(\lambda )=(0-\lambda )(0-\lambda )-(-1)(1)={\lambda }^2+1$. Yeh step kyun? Seedha ${\lambda }^2-\mathrm{tr}\lambda +\det$ se $\mathrm{tr}=0,\det =1$ ke saath.

2. Discriminant. $\Delta =0^2-4(1)=-4<0$ → Case F, complex roots. Yeh step kyun? Root ke neeche negative hone se hum Complex Numbers mein jaate hain.

3. Solve karo. ${\lambda }^2=-1\Rightarrow \lambda =\pm i$, jahan $i$ woh number hai jiske liye $i^2=-1$. Yeh step kyun? $\sqrt{-4}=2i$, isliye $\lambda =\frac{0\pm 2i}{2}=\pm i$. Dono roots conjugates hain, jaise woh hamesha real matrix ke liye hote hain.

4. Geometry padho. Magnitude $|\lambda |=|\pm i|=1$ matlab rotation length preserve karta hai (koi stretch nahi), aur roots non-real hone ka matlab yeh hai ki woh vectors ko sach mein turn karta hai — picture se match karta hai.

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Example 7 — Case G: ek zero eigenvalue (singular matrix)

Forecast: $\det =2\cdot 2-4\cdot 1=0$. Eigenvalues ka product $=\det =0$, isliye kam se kam ek eigenvalue $0$ hai. Roots guess karo.

1. Polynomial. $\mathrm{tr}=4,\det =0$, isliye $p(\lambda )={\lambda }^2-4\lambda +0=\lambda (\lambda -4)$. Yeh step kyun? $\det =0$ hone se constant term vanish ho jaata hai, isliye $\lambda$ immediately factor out ho jaata hai.

2. Roots. $\lambda =0$ aur $\lambda =4$. Yeh step kyun? Ek zero eigenvalue ek singular matrix ka algebraic fingerprint hai — $A$ ek puri direction ko origin par crush kar deta hai.

3. $\lambda =0$ ke liye Eigenvector. $(A-0I)\mathbf{x}=A\mathbf{x}=0$ solve karo: $2x_1+4x_2=0\Rightarrow x_1=-2x_2\Rightarrow \mathbf{x}=\left(\genfrac{}{}{0ex}{}{-2}{1}\right)$. Yeh bilkul $A$ ka null space hai. Yeh step kyun? $\lambda =0$ eigenvectors null space HI hain — kyunki $A\mathbf{x}=0\mathbf{x}=0$.

4. $\lambda =4$ ke liye Eigenvector. $(A-4I)=\left(\begin{array}{cc}-2& 4\\ 1& -2\end{array}\right)\Rightarrow -2x_1+4x_2=0\Rightarrow x_1=2x_2\Rightarrow \mathbf{x}=\left(\genfrac{}{}{0ex}{}{2}{1}\right)$.

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Example 8 — Case H: ek real-world story se $3\times 3$

Forecast: har column ka sum $1$ hai (ek stochastic matrix), isliye ek "do-nothing" eigenvalue $\lambda =1$ hona chahiye (steady state). Bottom row decoupled hai, ek aur $\lambda =1$ de rahi hai. Teesra guess karo.

1. Block structure exploit karo. Teesra row/column alag hai, isliye $\lambda =1$ seedha usse aata hai, aur top-left $2\times 2$ block $B=\left(\begin{array}{cc}0.7& 0.2\\ 0.3& 0.8\end{array}\right)$ baki handle karta hai. Yeh step kyun? Ek block-diagonal (yahan block-triangular) matrix ka characteristic polynomial blocks ke polynomials mein factor hota hai — pura $3\times 3$ expand karne ki zarurat nahi.

2. Block polynomial. $\mathrm{tr}(B)=1.5,\det (B)=0.56-0.06=0.5$, isliye $p_B(\lambda )={\lambda }^2-1.5\lambda +0.5=(\lambda -1)(\lambda -0.5).$ Yeh step kyun? $\det (B)=0.7\cdot 0.8-0.2\cdot 0.3=0.56-0.06=0.5$; Vieta clean factoring deta hai.

3. Sabhi eigenvalues. Block se: $\lambda =1,0.5$; decoupled entry se: $\lambda =1$. Isliye pura spectrum hai $\{1,1,0.5\}$. Yeh step kyun? Har block se roots collect karo.

4. Interpret karo. $\lambda =1$ wala mode steady state hai (amounts settle ho jaate hain, kabhi fade nahi hote). $\lambda =0.5$ wala mode roz aadha hota hai ($0.5^k\to 0$), isliye koi bhi imbalance khatam ho jaata hai — talaab equalise ho jaate hain. $\lambda =1$ ke liye uska eigenvector: $(B-I)=\left(\begin{array}{cc}-0.3& 0.2\\ 0.3& -0.2\end{array}\right)\Rightarrow 3x_1=2x_2\Rightarrow \left(\genfrac{}{}{0ex}{}{2}{3}\right)$, equilibrium split.

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Recall

Recall

$\Delta$ ka kaunsa sign kaunse roots deta hai? $\Delta >0$ real distinct ::: $\Delta =0$ repeated real ::: $\Delta <0$ complex conjugate pair.

Recall Same repeated

$\lambda$, Ex 4 defective kyun ho sakta hai lekin Ex 5 nahi? Defectiveness $A-\lambda I$ ki rank se decide hoti hai (geometric multiplicity), polynomial se nahi. ::: Ex 4 ki rank 1 hai (ek eigenvector); Ex 5 ki rank 0 hai (do eigenvectors).

Recall

$\lambda =0$ eigenvalue tumhe instantly kya batata hai? Matrix singular hai ($\det =0$) aur uske $\lambda =0$ eigenvectors uska null space hain.

Dekho bhi Cayley–Hamilton Theorem: in mein se har ek matrix apna khud ka characteristic polynomial satisfy karta hai.