Spectral Theorem

Symmetric matrices: real eigenvalues, orthogonal eigenvectors, A = Q Q^T.

The idea

Theorem (Spectral theorem).

Let $A$ be a real symmetric matrix, meaning $A = A^{T}$. Then every eigenvalue of $A$ is real, eigenvectors belonging to distinct eigenvalues are orthogonal, and $A$ can be written as $A = Q \Lambda Q^{T},$ where $\Lambda$ is diagonal, holding the eigenvalues, and $Q$ is orthogonal, holding unit eigenvectors as its columns.

Each conclusion removes a failure a general matrix can suffer. The eigenvalues of a real matrix are roots of a real polynomial and may be complex — a rotation of the plane has no real eigenvalue at all. The eigenvectors of a general matrix can sit at any angle to each other, or be too few to span the space, in which case no diagonalization exists. Symmetry rules out all of these failures.

The factorization $A = Q\Lambda Q^{T}$ is the form we compute with. Ordinary diagonalization reads $A = P \Lambda P^{-1}$ and requires inverting $P$; here the eigenvectors are orthogonal, so scaling each to unit length makes the columns of $Q$ orthonormal, and $Q^{T}Q = I$ gives $Q^{-1} = Q^{T}$. Geometrically, a symmetric matrix stretches space along mutually perpendicular axes — the columns $q_{1}, q_{2}, \ldots$ of $Q$ — by the eigenvalues $\lambda_{1}, \lambda_{2}, \ldots$ as factors: applied to the unit circle, $A$ produces an ellipse whose axes still lie along $q_{1}$ and $q_{2}$.

Ways to work on it

• Walkthrough. Why symmetric matrices have real eigenvalues and orthogonal eigenvectors, on a 2 × 2.
• Practice. Confirm two eigenvectors of a symmetric matrix are orthogonal.
• Hardest. Build an orthonormal eigenbasis and verify it assembles into an orthogonal matrix.

Not sure where to start? Take the ten-question placement test.