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

True — proof mein real scalar ${\mathbf{x}}^{\ast }A\mathbf{x}$ banta hai; symmetry usse apne conjugate ke barabar force karti hai, isliye $\lambda =\bar{\lambda }$.

False — $\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)$ ke real eigenvalues $1,2$ hain lekin ye symmetric nahi hai; real spectrum symmetry ke liye necessary hai, sufficient nahi.

False — orthogonality ka proof tabhi kaam karta hai jab $\lambda \ne \mu$ ho; ek hi eigenspace ke andar tumhe Gram-Schmidt Process se orthogonal basis choose karni padti hai.

False — symmetric matrices kabhi defective nahi hoti; har eigenvalue ki geometric multiplicity uski algebraic multiplicity ke barabar hoti hai, isliye full eigenbasis hamesha exist karti hai.

True — orthonormal columns se $Q^{\top }Q=I$ milta hai, jo ek orthogonal matrix ki exact definition hai jiska transpose uska inverse hota hai.

False — dono $A$ ko diagonalize karte hain, lekin $Q^{\top }=Q^{-1}$ sirf orthonormal columns ke liye hota hai; ek generic $P$ unnormalized/non-orthogonal eigenvectors se $P^{\top }\ne P^{-1}$ hoga.

True — ye normal matrices hain; $\mathbb{C}$ par ye unitarily diagonalizable hain, lekin ek real normal matrix ko complex eigenvectors ki zaroorat pad sakti hai (jaise ek rotation).

True — $I$ ka ek eigenvalue $1$ full multiplicity ka hai, isliye har orthonormal basis ek eigenbasis hai; $Q$ bilkul free hai.

False — repeats ko koi roka nahi hai (jaise $2I$ mein $2$ repeat hota hai); ye poora orthonormal eigenset guarantee karta hai, distinctness nahi.

Error yeh hai: symmetric matrices ke non-real eigenvalues nahi ho sakte, isliye $2i$ impossible hai — sawal mein jo matrix hai wo truly symmetric nahi hai.

Error yeh hai: tum eigenvectors ko normalize karna bhool gaye; unit-length orthogonal columns se $Q^{\top }=Q^{-1}$ milta hai, aur lengths theek karne ke baad theorem sahi rahega.

Error yeh hai: wo matrix symmetric nahi hai ($a_{12}=1\ne 0=a_{21}$), isliye orthogonality ki guarantee pehle se apply hi nahi hoti thi.

Error yeh hai: ek repeated eigenspace ke andar non-perpendicular basis vectors normal hain — unhe orthogonalize karne ke liye Gram-Schmidt Process apply karo; matrix phir bhi fully diagonalizable hai.

Error yeh hai: ${\mathbf{q}}_i{\mathbf{q}}_i^{\top }$ ek outer product hai (column times row) jo ek $n\times n$ rank-1 matrix deta hai; ${\mathbf{q}}_i^{\top }{\mathbf{q}}_i$ scalar $1$ hota.

Error yeh hai: kisi bhi square matrix ke liye trace $=\sum {\lambda }_i$; ek symmetric $A$ ke liye ye compute kiye gaye eigenvalues ko turant sanity-check karne deta hai.

Error yeh hai: agar $A=A^{\top }$ aur $A$ invertible hai, to $A^{-1}=(A^{-1}{)}^{\top }$ bhi hoga — inverse lena symmetry ko preserve karta hai (eigenvalues bas $1/{\lambda }_i$ ban jaate hain).

Kyunki $\lambda$ a priori complex ho sakta hai, isliye $\mathbf{x}$ complex hai; ${\mathbf{x}}^{\ast }\mathbf{x}=\sum |x_i{|}^2$ real aur positive sirf conjugation se rahta hai, jo $\lambda =\bar{\lambda }$ force karta hai.

Proof $A$ ko inner product ke across $A^{\top }=A$ se move karne par pivot karta hai; wo swap hi $(\lambda -\mu )({\mathbf{u}}^{\top }\mathbf{v})=0$ deta hai, aur symmetry ke bina fail ho jaata hai.

Iske columns orthonormal hain, isliye $Q^{\top }Q=I$; geometrically ye stretch kiye bina rotate/reflect karta hai, matlab eigen-coordinates mein jaane par lengths aur angles preserve hote hain.

Right se left padhte hain: $Q^{\top }$ eigen-axes mein rotate karta hai, $\Lambda$ sirf un axes ke along stretch karta hai, $Q$ wapas rotate karta hai — perpendicular directions ke along pure stretch, koi shear nahi.

${\mathbf{q}}_i{\mathbf{q}}_i^{\top }$ us axis par orthogonal projector tab hi hai jab ${\mathbf{q}}_i^{\top }{\mathbf{q}}_i=1$; warna projector galat scale hoga aur pieces $A$ mein sum nahi karenge.

Symmetric (aur isliye normal) matrices kabhi defective nahi hoti, isliye geometric multiplicity algebraic ke barabar hoti hai; ek general matrix jaise Jordan block ki algebraic multiplicity $2$ lekin sirf $1$-dimensional eigenspace ho sakta hai.

Symmetric $A$ ke saath quadratic form ${\mathbf{x}}^{\top }A\mathbf{x}$ eigen-coordinates $\mathbf{y}=Q^{\top }\mathbf{x}$ mein $\sum {\lambda }_i{y}_i^2$ ban jaata hai, isliye eigenvalues form ke pure stretch coefficients hain.

Har eigenvalue $0$ hai ($\Lambda =0$), aur har orthonormal basis $Q$ ki tarah kaam karti hai; "stretch" saari directions ko ek point par collapse kar deta hai.

Haan trivially — ye symmetric hai, iska akela eigenvalue $a\in \mathbb{R}$ hai, aur $Q=[1]$; ye general statement ka degenerate base case hai.

Kuch nahi badalta — $\lambda =0$ ek bilkul valid distinct eigenvalue hai, isliye iske eigenvectors doosre eigenvalues ke eigenvectors se orthogonal rehte hain; matrix bas singular ho jaati hai.

Haan — negative ${\lambda }_i$ ka matlab hai ki $A$ us eigen-axis ke along sign flip karta hai ($|{\lambda }_i|$ se stretch plus ek reflection); ye Positive Definite Matrices se juda hai jo kisi ${\lambda }_i<0$ hone par fail hoti hai.

Ye zaroori $cI$ hogi — full-multiplicity eigenspace poora ${\mathbb{R}}^n$ hai, isliye $A$ uniform scaling ki tarah act karta hai aur $Q$ koi bhi orthogonal ho sakta hai.

Saare eigenvalues strictly positive hain, isliye $A=Q\Lambda Q^{\top }$ ka ek invertible square root $A^{1/2}=Q{\Lambda }^{1/2}{Q}^{\top }$ hota hai, jo Positive Definite Matrices aur Singular Value Decomposition se connect karta hai.

Singular Value Decomposition $A=U\Sigma V^{\top }$ do alag orthonormal bases aur non-negative singular values use karta hai, $A=A^{\top }$ ki zaroorat ke bina "orthogonal stretch" picture recover karta hai.

Unhe simultaneously orthogonally diagonalize kiya ja sakta hai — ek shared orthonormal eigenbasis exist karta hai, isliye ek hi $Q$ dono ko diagonalize karta hai.