Questions: The Spectral Theorem

Question 1

A matrix A is diagonalizable (A = PDP⁻¹) with real eigenvalues, but A is not symmetric. What does the Spectral Theorem say about orthogonal diagonalizability?

• A. A can still be orthogonally diagonalized since its eigenvalues are real
• B. A cannot be orthogonally diagonalized in general — that requires symmetry
• C. The Spectral Theorem guarantees orthogonal diagonalizability for all diagonalizable matrices
• D. A's eigenvectors are automatically orthogonal if the eigenvalues are distinct

Answer: (B)

The Spectral Theorem applies specifically to symmetric matrices — it does NOT hold for general diagonalizable matrices. A non-symmetric matrix can have real eigenvalues and be diagonalizable without being orthogonally diagonalizable. Orthogonal diagonalizability (A = QDQᵀ with Qᵀ = Q⁻¹) is a strictly stronger guarantee. Option D is the key misconception: real distinct eigenvalues do not force eigenvector orthogonality unless A is symmetric.

Question 2

A real symmetric matrix has eigenvalues λ₁ = 3 and λ₂ = 5 with corresponding eigenvectors u₁ and u₂. Without computing u₁ and u₂ explicitly, what can you conclude about their dot product?

• A. Some nonzero value determined by the specific entries of A
• B. Zero — they must be orthogonal
• C. One — eigenvectors are normalized by convention
• D. Undefined without knowing the specific entries of A

Answer: (B)

The Spectral Theorem guarantees that eigenvectors of a real symmetric matrix corresponding to *distinct* eigenvalues are always orthogonal. The proof uses symmetry: if Au = λu and Av = μv with λ ≠ μ, then ⟨u,v⟩ = 0 follows from uᵀAv = (Aᵀu)ᵀv = (Au)ᵀv. This holds regardless of the specific entries — it is a consequence of symmetry, not of the particular values of A.

Question 3

Nearly every real matrix with distinct eigenvalues is orthogonally diagonalizable.

• T. True
• F. False

Answer: False

Distinct real eigenvalues guarantee diagonalizability (enough linearly independent eigenvectors to write A = PDP⁻¹), but not orthogonal diagonalizability. A non-symmetric matrix with distinct real eigenvalues has eigenvectors that may not be orthogonal — you can write A = PDP⁻¹ but cannot write A = QDQᵀ with Q orthogonal. The Spectral Theorem's guarantee of orthogonal diagonalizability is exclusive to symmetric matrices.

Question 4

The columns of Q in the decomposition A = QDQᵀ are orthonormal eigenvectors of A.

• T. True
• F. False

Answer: True

This is a direct statement of orthogonal diagonalizability. Q is orthogonal (Qᵀ = Q⁻¹), which means its columns are mutually orthogonal unit vectors. Each column is an eigenvector of A (by the structure of the diagonalization), and together they form an orthonormal basis for ℝⁿ. The diagonal entries of D are the corresponding real eigenvalues. This is what the Spectral Theorem guarantees: not just any diagonalizing matrix, but one whose columns form a complete orthonormal eigenbasis.

Question 5

Why does the symmetry condition Aᵀ = A force eigenvectors corresponding to different eigenvalues to be orthogonal? Sketch the key step of the argument.

Model answer: Suppose Au = λu and Av = μv with λ ≠ μ. Compute uᵀAv two ways: uᵀ(Av) = uᵀ(μv) = μ(uᵀv), and (Aᵀu)ᵀv = (Au)ᵀv = (λu)ᵀv = λ(uᵀv), using Aᵀ = A. So μ(uᵀv) = λ(uᵀv), giving (μ − λ)(uᵀv) = 0. Since λ ≠ μ, we conclude uᵀv = 0.

Symmetry does the crucial work: it lets you move A from one side of the inner product to the other (uᵀAv = (Aᵀu)ᵀv = (Au)ᵀv when Aᵀ = A). This algebraic identity, combined with the two eigenvalue equations, forces the inner product of distinct eigenvectors to be zero. Without symmetry, this step fails — which is why non-symmetric matrices can have non-orthogonal eigenvectors even with distinct eigenvalues.