Questions: Symmetric Matrices and Their Properties
5 questions to test your understanding
Score: 0 / 5
Question 1 Multiple Choice
A 3×3 matrix has all positive entries on its main diagonal and appears 'roughly symmetric.' A student concludes it must be positive definite. What is wrong with this reasoning?
APositive definiteness requires the matrix to be exactly symmetric first, and roughly symmetric is not sufficient
BPositive definiteness is determined by whether all eigenvalues are positive — not by diagonal entries. A matrix can have positive diagonals but still have a negative eigenvalue
CPositive definiteness only applies to 2×2 matrices, not 3×3
DThe matrix must be orthogonal, not just symmetric, for positive definiteness to be defined
Positive definiteness means xᵀAx > 0 for all nonzero x, which is equivalent to all eigenvalues being positive. Positive diagonal entries do not guarantee this: the off-diagonal entries can still create a direction in which the quadratic form is negative. A simple counterexample: [[1, 10], [10, 1]] has positive diagonal entries but determinant 1−100 = −99 < 0, implying a negative eigenvalue. Always check eigenvalues (or leading principal minors) to verify positive definiteness.
Question 2 Multiple Choice
What guarantees that eigenvectors of a real symmetric matrix corresponding to distinct eigenvalues are orthogonal?
AThe fact that symmetric matrices always have integer eigenvalues
BThe symmetry condition Aᵀ = A, which forces vᵀAw to equal λ(v·w) in one computation and μ(v·w) in another — when λ ≠ μ, this requires v·w = 0
CThe fact that all real matrices with distinct eigenvalues produce orthogonal eigenvectors
DPositive definiteness, which forces all eigenvectors to be unit vectors
The proof uses symmetry directly. Let Av = λv and Aw = μw with λ ≠ μ. Compute vᵀAw two ways: using Av = λv gives λ(vᵀw), and using Aᵀ = A gives (vᵀA)w = (Aᵀv)ᵀw = (Av)ᵀw = λ(vᵀw) — wait, let's be precise: vᵀAw = (Av)ᵀw (by symmetry) = (λv)ᵀw = λ(vᵀw). But also vᵀAw = vᵀ(μw) = μ(vᵀw). So λ(vᵀw) = μ(vᵀw), which with λ ≠ μ forces vᵀw = 0. This argument works only because Aᵀ = A.
Question 3 True / False
Any square matrix with most real eigenvalues is expected to be symmetric.
TTrue
FFalse
Answer: False
Symmetry is a sufficient condition for real eigenvalues, but not necessary. There exist non-symmetric matrices with all real eigenvalues — for example, any upper triangular matrix with real diagonal entries has real eigenvalues (they are its diagonal entries) but is generally not symmetric. The implication runs one way: symmetric ⟹ real eigenvalues. The converse is false.
Question 4 True / False
For any matrix A, the product AᵀA is always symmetric.
TTrue
FFalse
Answer: True
To verify: (AᵀA)ᵀ = Aᵀ(Aᵀ)ᵀ = AᵀA. The transpose of the product equals itself, confirming symmetry. This is why covariance matrices in statistics — which are computed as (1/n)XᵀX — are always symmetric, regardless of what X looks like. This fact connects symmetric matrix theory to its most important practical domain: statistics, PCA, and quadratic optimization.
Question 5 Short Answer
Covariance matrices in statistics are always symmetric. Explain why this is the case and what it implies about their eigenvalues.
Think about your answer, then reveal below.
Model answer: A covariance matrix Σ is computed as XᵀX (up to scaling), and (XᵀX)ᵀ = XᵀX, so it is always symmetric. Because it is a real symmetric matrix, all of its eigenvalues are guaranteed to be real. Furthermore, covariance matrices are positive semidefinite (xᵀΣx ≥ 0 for all x), meaning all eigenvalues are non-negative. These eigenvalues represent the variance captured in each principal component direction.
This connection between the algebra of symmetric matrices and statistics is why PCA (principal component analysis) works. The eigenvectors of the covariance matrix are the principal components (orthogonal directions of maximum variance), and the eigenvalues tell you how much variance each direction captures. The guarantee of real, non-negative eigenvalues and orthogonal eigenvectors comes entirely from the symmetry and positive semidefiniteness of the covariance matrix.