Chapter 27 Quiz — The Spectral Theorem
Twelve conceptual checks. Try each before opening the answer. One-line explanations follow every answer. Remember: the Spectral Theorem applies to symmetric ($A = A^{\mathsf{T}}$) or Hermitian ($A = A^{*}$) matrices — never to a general matrix.
Q1. State the three guarantees the Spectral Theorem gives for a real symmetric matrix.
Answer
(1) All eigenvalues are **real**; (2) there is an **orthonormal basis of eigenvectors**; (3) $A$ is **orthogonally diagonalizable**, $A = Q\Lambda Q^{\mathsf{T}}$ with $Q^{\mathsf{T}}Q = I$. *All three hold for every symmetric matrix, with no exceptions.*Q2. A classmate writes "every matrix is diagonalizable, and its eigenvectors are orthogonal." What is wrong?
Answer
Both claims are **false in general**. Only matrices with enough independent eigenvectors are diagonalizable (defective matrices are not), and only **symmetric/Hermitian** matrices are guaranteed orthogonal eigenvectors. *The correct statement names the condition: symmetric ⇒ orthogonally diagonalizable.*Q3. Geometrically, what does a symmetric transformation do to space?
Answer
A **pure stretch along mutually perpendicular axes** — its eigenvectors — with **no rotation and no shear**. *The eigenvalues are the stretch factors along those orthogonal eigen-axes.*Q4. Why can a real symmetric matrix never have a complex eigenvalue?
Answer
Because $A^{*} = A$ forces the scalar $\mathbf{v}^{*}A\mathbf{v} = \lambda\lVert\mathbf{v}\rVert^2$ to equal its own conjugate, so it is real; dividing by the positive real $\lVert\mathbf{v}\rVert^2$ makes $\lambda$ real (§27.4). *A complex eigenvalue would mean a hidden rotation, which symmetry forbids.*Q5. Eigenvectors of a symmetric matrix for eigenvalues $3$ and $8$ — what is the angle between them?
Answer
**$90°$** (they are orthogonal), for free, because the eigenvalues are distinct (§27.5). *Symmetry slides $A$ across the dot product, forcing $(\lambda - \mu)(\mathbf{u}\cdot\mathbf{w}) = 0$; since $\lambda \ne \mu$, the dot product is zero.*Q6. In $A = Q\Lambda Q^{\mathsf{T}}$, why do we get a transpose where general diagonalization $A = PDP^{-1}$ has an inverse?
Answer
Because the eigenvectors are **orthonormal**, so the change-of-basis matrix $Q$ is **orthogonal** ($Q^{-1} = Q^{\mathsf{T}}$). *The orthogonality of §27.5 is exactly what upgrades $P^{-1}$ (expensive elimination) to a free transpose.*Q7. What is each term $\mathbf{q}_i\mathbf{q}_i^{\mathsf{T}}$ in the spectral decomposition $A = \sum_i\lambda_i\mathbf{q}_i\mathbf{q}_i^{\mathsf{T}}$?
Answer
A **rank-one orthogonal projector** onto the line spanned by the unit eigenvector $\mathbf{q}_i$ (symmetric and idempotent, $P_i^2 = P_i$). *A symmetric matrix is a weighted sum of perpendicular projections, weighted by the eigenvalues.*Q8. Which numpy function should you use for a symmetric or Hermitian matrix, and why not the general eig?
Answer
**`np.linalg.eigh`** (the "h" is for Hermitian). *It is faster and more accurate, and it is guaranteed to return real eigenvalues and orthonormal eigenvectors — exactly the Spectral Theorem's promise — whereas `eig` may return spurious tiny imaginary parts and non-orthonormal vectors.*Q9. The matrix $M = \begin{psmallmatrix}2 & 1\\ 0 & 3\end{psmallmatrix}$ has two distinct real eigenvalues. Is it orthogonally diagonalizable?
Answer
**No.** It is diagonalizable (distinct eigenvalues), but it is **not symmetric**, and its eigenvectors $(1,0)$ and $(1,1)$ are **not orthogonal** (dot product $1$). *Diagonalizable does not imply orthogonally diagonalizable — only symmetry guarantees perpendicular eigenvectors.*Q10. What is the complex analogue of a symmetric matrix, and what does the Spectral Theorem guarantee about it?
Answer
A **Hermitian** matrix, $A = A^{*}$ (conjugate transpose). It has **real eigenvalues** and an orthonormal eigenbasis, and is **unitarily diagonalizable**, $A = U\Lambda U^{*}$ with $U^{*}U = I$. *The most general unitarily diagonalizable class is the **normal** matrices, $AA^{*} = A^{*}A$.*Q11. Why must a quantum observable be a Hermitian matrix?
Answer
Because the possible measurement outcomes are its **eigenvalues**, and measured quantities (energy, spin, position) are **real numbers** — so the eigenvalues must be real, which §27.4 guarantees precisely when $A = A^{*}$. *The orthogonal eigenstates (§27.5) are the states with definite values of the observable.*Q12. The eigenvalues of the symmetric matrix of a quadratic form $q(\mathbf{x}) = \mathbf{x}^{\mathsf{T}}A\mathbf{x}$ are $6$ and $-2$. Is the level set $q = 1$ an ellipse or a hyperbola, and is the origin a minimum or a saddle?