Eigenvalues and eigenvectors — characteristic polynomial

4.5.29 · Maths › Linear Algebra (Full)

HUM kya define kar rahe hain?

$\det (A-\lambda I)=0$ eigenvalues kyun dhundhta hai? (Scratch se derivation)

Hum yaad nahi karte — hum ise banate hain.

Step 1 — Defining equation ko rewrite karo. $A\mathbf{x}=\lambda \mathbf{x}⟺A\mathbf{x}-\lambda \mathbf{x}=0.$ Yeh step kyun? Hum sab kuch ek side par laana chahte hain taaki $\mathbf{x}$ ko factor out kar sakein.

Step 2 — Identity use karke $\mathbf{x}$ ko factor out karo. Hum $(A-\lambda )\mathbf{x}$ nahi likh sakte kyunki $A$ ek matrix hai aur $\lambda$ ek scalar hai — yeh dono alag worlds mein rehte hain. Identity insert karo: $\lambda \mathbf{x}=\lambda I\mathbf{x}$. Tab $(A-\lambda I)\mathbf{x}=0.$ Yeh step kyun? Ab $A-\lambda I$ ek genuine matrix hai, isliye yeh ek ordinary homogeneous linear system hai.

Step 3 — Nonzero solution ki maang karo. Ek homogeneous system $M\mathbf{x}=0$ ka nonzero solution hota hai iff $M$ singular hai (not invertible). Agar $M=(A-\lambda I)$ invertible hoti, toh $M^{-1}$ se multiply karne par $\mathbf{x}=0$ force ho jaata, jo hum forbid kar chuke hain. Yeh step kyun? Poori baat hi $\mathbf{x}\ne 0$ ki hai, aur singularity iske liye gateway condition hai.

Step 4 — Singular $\iff$ zero determinant. $\boxed{\det (A-\lambda I)=0.}$ Yeh step kyun? "Not invertible" aur "determinant $=0$" ek hi baat hai. Yeh $\lambda$ mein ek polynomial equation hai — solve karna possible hai!

Do structural facts (derived, memorised nahi)

Ek $2\times 2$ matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ ke liye: $p(\lambda )=\det \left(\begin{array}{cc}a-\lambda & b\\ c& d-\lambda \end{array}\right)=(a-\lambda )(d-\lambda )-bc.$ Expand karo: $p(\lambda )={\lambda }^2-(a+d)\lambda +(ad-bc)={\lambda }^2-\mathrm{tr}(A)\lambda +\det (A).$

Worked Example 1 — ek clean $2\times 2$

Maano $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$.

Characteristic polynomial: $p(\lambda )=(2-\lambda {)}^2-1.$ Yeh step kyun? Direct $(a-\lambda )(d-\lambda )-bc$ with $a=d=2,bc=1$.

$p(\lambda )={\lambda }^2-4\lambda +3=(\lambda -1)(\lambda -3).$ Toh ${\lambda }_1=1,{\lambda }_2=3$.

Check: $\mathrm{tr}=4=1+3$ ✓, $\det =3=1\cdot 3$ ✓.

$\lambda =3$ ke liye Eigenvector: $(A-3I)\mathbf{x}=0$ solve karo: $\left(\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right)\mathbf{x}=0\Rightarrow x_1=x_2\Rightarrow \mathbf{x}=\left(\begin{array}{c}1\\ 1\end{array}\right).$ Yeh step kyun? Dono rows $-x_1+x_2=0$ deti hain; ek free direction bachti hai.

$\lambda =1$ ke liye Eigenvector: $\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\mathbf{x}=0\Rightarrow x_1=-x_2\Rightarrow \mathbf{x}=\left(\begin{array}{c}1\\ -1\end{array}\right)$.

Worked Example 2 — ek triangular matrix (shortcut)

Maano $A=\left(\begin{array}{cc}5& 7\\ 0& -2\end{array}\right)$.

$p(\lambda )=\det \left(\begin{array}{cc}5-\lambda & 7\\ 0& -2-\lambda \end{array}\right)=(5-\lambda )(-2-\lambda )-0.$ Yeh step kyun? Lower-left entry $0$ hai, isliye ek triangular matrix ka determinant sirf diagonal entries ka product hota hai.

Roots: $\lambda =5,\lambda =-2$ — exactly diagonal entries.

Worked Example 3 — repeated root (algebraic vs geometric)

Maano $A=\left(\begin{array}{cc}3& 1\\ 0& 3\end{array}\right)$. $p(\lambda )=(3-\lambda {)}^2\Rightarrow \lambda =3\text{ (twice)}.$ Root $3$ ki algebraic multiplicity 2 hai. Eigenvectors: $(A-3I)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, toh $x_2=0$, sirf $\mathbf{x}=\left(\genfrac{}{}{0ex}{}{1}{0}\right)$ milta hai — ek single independent eigenvector (geometric multiplicity 1).

Common mistakes (Steel-manned)

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Flashcards

Connections

• Determinants — singularity test jo poori method ko power deta hai.
• Null Space and Rank — eigenvectors $\ker (A-\lambda I)$ mein rehte hain.
• Diagonalization — jab independent eigenvectors space fill kar lein, $A=PD{P}^{-1}$.
• Trace of a Matrix — sum-of-eigenvalues sanity check.
• Complex Numbers — roots real $A$ ke liye bhi complex ho sakti hain.
• Cayley–Hamilton Theorem — $A$ apne characteristic polynomial ko satisfy karta hai.

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