Orthogonal Diagonalization

The spectral theorem states that symmetric matrices can be orthogonally diagonalized. This canonical form provides optimal coordinates for understanding the matrix's action.

TheoremSpectral Theorem for Real Symmetric Matrices

Let $A$ be a real $n \times n$ symmetric matrix. Then:

All eigenvalues of $A$ are real
$A$ has $n$ orthonormal eigenvectors
$A$ is orthogonally diagonalizable: there exists an orthogonal matrix $Q$ and diagonal matrix $\Lambda$ such that: $A = Q\Lambda Q^T$

where the columns of $Q$ are orthonormal eigenvectors and the diagonal entries of $\Lambda$ are the corresponding eigenvalues.

The complex version: A Hermitian matrix is unitarily diagonalizable: $A = U\Lambda U^*$ .

This result is remarkable: every symmetric matrix, regardless of repeated eigenvalues, admits an orthonormal eigenvector basis. The orthogonality condition $Q^T = Q^{-1}$ makes computations particularly elegant.

ExampleOrthogonal Diagonalization

Diagonalize $A = \begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix}$ .

Eigenvalues: $\det(A - \lambda I) = (3-\lambda)^2 - 1 = 0$ gives $\lambda_1 = 4, \lambda_2 = 2$ .

Eigenvectors: For $\lambda_1 = 4$ : $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ , normalized: $\mathbf{q}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$

For $\lambda_2 = 2$ : $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ , normalized: $\mathbf{q}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$

Then: $A = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = Q\Lambda Q^T$

DefinitionSpectral Decomposition

The orthogonal diagonalization $A = Q\Lambda Q^T$ can be written as a sum: $A = \sum_{i=1}^n \lambda_i \mathbf{q}_i\mathbf{q}_i^T$

where $\mathbf{q}_i$ are the orthonormal eigenvectors and $\lambda_i$ are eigenvalues. Each term $\lambda_i\mathbf{q}_i\mathbf{q}_i^T$ is a rank-one projection matrix.

This spectral decomposition expresses $A$ as a weighted sum of orthogonal projections onto eigenspaces.

TheoremProperties of Orthogonal Diagonalization

If $A = Q\Lambda Q^T$ with $Q$ orthogonal:

$A^k = Q\Lambda^k Q^T$ (easy to compute powers)
$\det(A) = \prod_{i=1}^n \lambda_i$ (product of eigenvalues)
$\text{tr}(A) = \sum_{i=1}^n \lambda_i$ (sum of eigenvalues)
$A$ is invertible iff all $\lambda_i \neq 0$ ; then $A^{-1} = Q\Lambda^{-1}Q^T$
The eigenvectors give principal axes; eigenvalues measure variance along these axes

Remark

Orthogonal diagonalization is optimal for computation: the orthogonality condition $Q^T = Q^{-1}$ means we don't need to invert $Q$ explicitly. Moreover, orthogonal transformations preserve lengths and angles, making them numerically stable. The spectral decomposition perspective—viewing the matrix as a sum of scaled projections—is particularly powerful in applications like principal component analysis.