Symmetric matrices — spectral theorem (real eigenvalues, orthogonal eigenvectors)

4.5.38 · Maths › Linear Algebra (Full)

THEOREM kya hai?

Decomposition $A=Q\Lambda Q^{\top }$ ko spectral decomposition kehte hain. Isko sum ke roop mein likhne par: $A={\sum }_{i=1}^n{\lambda }_i{\mathbf{q}}_i{\mathbf{q}}_i^{\top }$ har ${\mathbf{q}}_i{\mathbf{q}}_i^{\top }$ ek rank-1 projection hai $i$-th eigen-axis par, jo ${\lambda }_i$ se scale hoti hai.

Eigenvalues real KYU hote hain? (Scratch se derivation)

Hum yeh assume nahi karte ki eigenvalues real hain — hum prove karte hain. Maano $A\mathbf{x}=\lambda \mathbf{x}$ jahan $\lambda$ aur $\mathbf{x}\ne 0$ possibly complex hain. $\bar{\mathbf{x}}$ complex conjugate hai aur conjugate transpose ${\mathbf{x}}^{\ast }={\bar{\mathbf{x}}}^{\top }$ use karo.

Step 1 — Scalar ${\mathbf{x}}^{\ast }A\mathbf{x}$ banao. ${\mathbf{x}}^{\ast }A\mathbf{x}={\mathbf{x}}^{\ast }(\lambda \mathbf{x})=\lambda ({\mathbf{x}}^{\ast }\mathbf{x})$ Yeh step kyun? $A\mathbf{x}=\lambda \mathbf{x}$ plug karne se $\lambda$ aage aa jaata hai. Yahaan ${\mathbf{x}}^{\ast }\mathbf{x}=\sum |x_i{|}^2>0$ ek positive real number hai.

Step 2 — Usi scalar ka conjugate transpose lo. Ek $1\times 1$ quantity apne conjugate transpose ke barabar hoti hai: $({\mathbf{x}}^{\ast }A\mathbf{x}{)}^{\ast }={\mathbf{x}}^{\ast }{A}^{\ast }\mathbf{x}.$ Kyunki $A$ real aur symmetric hai, $A^{\ast }={\bar{A}}^{\top }=A^{\top }=A$. Toh $({\mathbf{x}}^{\ast }A\mathbf{x}{)}^{\ast }={\mathbf{x}}^{\ast }A\mathbf{x}.$ Yeh step kyun? Symmetry + realness hi woh cheez hai jo ${\mathbf{x}}^{\ast }A\mathbf{x}$ ko apne conjugate ke barabar banati hai — matlab yeh ek real number hai.

Step 3 — Compare karo. Step 1 se, ${\mathbf{x}}^{\ast }A\mathbf{x}=\lambda ({\mathbf{x}}^{\ast }\mathbf{x})$, aur iska conjugate $\bar{\lambda }({\mathbf{x}}^{\ast }\mathbf{x})$ hai. Kyunki value real hai: $\lambda ({\mathbf{x}}^{\ast }\mathbf{x})=\bar{\lambda }({\mathbf{x}}^{\ast }\mathbf{x})⟹(\lambda -\bar{\lambda })({\mathbf{x}}^{\ast }\mathbf{x})=0.$ Kyunki ${\mathbf{x}}^{\ast }\mathbf{x}>0$, hume milta hai $\lambda =\bar{\lambda }$, toh $\boxed{\lambda \in \mathbb{R}}$. ∎

Eigenvectors orthogonal KYU hote hain? (Scratch se derivation)

Do eigenpairs lo jinka distinct eigenvalue ho: $A\mathbf{u}=\lambda \mathbf{u}$, $A\mathbf{v}=\mu \mathbf{v}$, $\lambda \ne \mu$.

Step 1 — ${\mathbf{u}}^{\top }A\mathbf{v}$ ko do tarike se compute karo. ${\mathbf{u}}^{\top }A\mathbf{v}={\mathbf{u}}^{\top }(\mu \mathbf{v})=\mu ({\mathbf{u}}^{\top }\mathbf{v}).$ Kyun? Daayein $A$ ko $\mathbf{v}$ par apply karo.

Step 2 — Symmetry use karke $A$ ko left side pe laao. ${\mathbf{u}}^{\top }A\mathbf{v}=(A^{\top }\mathbf{u}{)}^{\top }\mathbf{v}=(A\mathbf{u}{)}^{\top }\mathbf{v}=(\lambda \mathbf{u}{)}^{\top }\mathbf{v}=\lambda ({\mathbf{u}}^{\top }\mathbf{v}).$ Kyun? ${\mathbf{u}}^{\top }A=(A^{\top }\mathbf{u}{)}^{\top }$, aur $A^{\top }=A$ se hum $A$ ko $\mathbf{u}$ par apply kar sakte hain.

Step 3 — Barabar karo. $\mu ({\mathbf{u}}^{\top }\mathbf{v})=\lambda ({\mathbf{u}}^{\top }\mathbf{v})⟹(\lambda -\mu )({\mathbf{u}}^{\top }\mathbf{v})=0.$ Kyunki $\lambda \ne \mu$, toh zaroori hai ki ${\mathbf{u}}^{\top }\mathbf{v}=0$, yaani $\mathbf{u}\perp \mathbf{v}$. ∎

$A=Q\Lambda Q^{\top }$ KYU aata hai?

Orthonormal eigenvectors ko $Q$ ke columns ke roop mein stack karo, toh $Q^{\top }Q=I$ (yaani $Q^{-1}=Q^{\top }$). Eigen-equations $A{\mathbf{q}}_i={\lambda }_i{\mathbf{q}}_i$ ko ek saath likhen toh: $AQ=Q\Lambda \text{(column }i\text{ is }A{\mathbf{q}}_i={\lambda }_i{\mathbf{q}}_i).$ Daayein $Q^{-1}=Q^{\top }$ se multiply karo: $\boxed{A=Q\Lambda Q^{\top }.}$ Yeh special kyun hai: ek general diagonalizable matrix $A=P\Lambda P^{-1}$ deti hai jahan $P^{-1}\ne P^{\top }$. Symmetry $P^{-1}$ ko $Q^{\top }$ mein upgrade kar deti hai — yahi orthogonality ka bonus hai.

Worked Examples

Forecast-then-Verify

Common Mistakes (Steel-manned)

Flashcards

Connections

• Eigenvalues and Eigenvectors — woh base concept jis par yeh build karta hai.
• Orthogonal Matrices — $Q$ yahaan bilkul ek yahi hai ($Q^{\top }Q=I$).
• Diagonalization — symmetric case sabse saaf special case hai.
• Quadratic Forms — ${\mathbf{x}}^{\top }A\mathbf{x}$ eigenvalues ke signs se classify hota hai.
• Positive Definite Matrices — saare ${\lambda }_i>0$ wali symmetric matrix.
• Singular Value Decomposition — SVD symmetric $A$ ke liye spectral theorem mein reduce ho jaata hai.
• Gram-Schmidt Process — repeated eigenspaces ko orthonormalize karne ke liye use hota hai.

Concept Map