Symmetric Matrices and the Spectral Theorem

The spectral theorem (real, symmetric version)

If $A$ is a real symmetric matrix ($A^T = A$), then:

1. $A$ has $n$ real eigenvalues (counted with algebraic multiplicity).
2. Eigenvectors for distinct eigenvalues are orthogonal.
3. $A$ can be diagonalized by an orthogonal matrix $Q$: $A = Q D Q^T$

with $Q^T Q = I$ and $D$ diagonal with the eigenvalues.

This is sometimes called the orthogonal diagonalization theorem.

Step-by-step

$A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ — symmetric.

Step 1 — Eigenvalues.

Characteristic polynomial: $(2 - \lambda)^2 - 1 = \lambda^2 - 4 \lambda + 3 = (\lambda - 3)(\lambda - 1)$. $\lambda_1 = 3, \quad \lambda_2 = 1$

Both real (as the theorem promises). ✓

Step 2 — Eigenvectors.

For $\lambda = 3$: $(A - 3I) \mathbf{v} = 0 \implies \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \mathbf{v} = 0$. Solution: $\mathbf{v}_1 = (1, 1)^T$.

For $\lambda = 1$: $(A - I) \mathbf{v} = 0 \implies \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \mathbf{v} = 0$. Solution: $\mathbf{v}_2 = (1, -1)^T$.

Check orthogonality: $\mathbf{v}_1 \cdot \mathbf{v}_2 = 1 \cdot 1 + 1 \cdot (-1) = 0$ ✓ (Exactly as the theorem guarantees.)

Step 3 — Normalize.

$\|\mathbf{v}_1\| = \sqrt{2}$, $\|\mathbf{v}_2\| = \sqrt{2}$: $\mathbf{q}_1 = \dfrac{1}{\sqrt{2}} (1, 1)^T, \quad \mathbf{q}_2 = \dfrac{1}{\sqrt{2}} (1, -1)^T$

Step 4 — Assemble. $Q = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \quad D = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}$

$Q$ is orthogonal ($Q^T Q = I$), so $Q^{-1} = Q^T$.

Step 5 — Verify $A = Q D Q^T$.

$D Q^T = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 3 & 3 \\ 1 & -1 \end{pmatrix}$.

$Q D Q^T = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \cdot \dfrac{1}{\sqrt{2}} \begin{pmatrix} 3 & 3 \\ 1 & -1 \end{pmatrix} = \dfrac{1}{2} \begin{pmatrix} 4 & 2 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix} ✓$

Wait — I get $(2, 1; 2, 2)$, not $(2, 1; 1, 2)$? Let me recompute column 1 of $QDQ^T$: row 2 of $QD$ is $\dfrac{1}{\sqrt 2}(3, -1)$; dotted with column 1 of $Q^T$ which is $\dfrac{1}{\sqrt 2}(1, 1)$: $\dfrac{1}{2}(3 + (-1)) = 1$. So it's $(2, 1; 1, 2) = A$ ✓. Transcription verification lesson — always double-check the dot product.

Why symmetric matrices are special

1. Real eigenvalues guaranteed. Non-symmetric matrices can have complex eigenvalues; symmetric ones can't.
2. Orthogonal eigenvectors. Even across repeated eigenvalues, you can always choose an orthogonal eigenvector basis.
3. Always diagonalizable. Never defective.
4. Geometric structure. The level sets of $\mathbf{x}^T A \mathbf{x}$ are aligned with the eigenvectors — the principal axes.

This is the "cleanest" diagonalization scenario in all of linear algebra.

Applications

• Principal Component Analysis (PCA): diagonalize the sample covariance matrix; its (orthogonal) eigenvectors are the principal components.
• Quadratic forms: $\mathbf{x}^T A \mathbf{x}$ simplifies to $\lambda_1 y_1^2 + \lambda_2 y_2^2 + \cdots$ in the eigenvector coordinate system.
• Rigid-body mechanics: the inertia tensor is symmetric; its eigenvectors give the principal axes of rotation.
• Fourier analysis: the discrete Fourier transform diagonalizes circulant matrices; the continuous version diagonalizes translation-invariant operators.
• Spectral graph theory: the adjacency matrix and Laplacian are symmetric; their eigenvectors encode graph structure.

Extension: Hermitian matrices

Over $\mathbb{C}$, the analogue is Hermitian matrices ($A^* = A$ where $*$ denotes conjugate transpose). Same conclusions: real eigenvalues, unitary diagonalization $A = U D U^*$.

Generalization: normal matrices

The spectral theorem applies most broadly to normal matrices ($A A^* = A^* A$) over $\mathbb{C}$: they're unitarily diagonalizable. This class includes Hermitian, skew-Hermitian, and unitary matrices.

Common mistakes

• Forgetting to normalize. Orthogonality isn't enough; $Q$ must have unit columns to be orthogonal as a matrix.
• Using $P$ (general) instead of $Q$ (orthogonal). You can diagonalize as $A = PDP^{-1}$ with non-orthogonal eigenvectors, but the whole point of the spectral theorem is that symmetric matrices admit orthogonal diagonalization — which is cheaper and numerically stable.
• Assuming repeated eigenvalues cause defects. Symmetric matrices are still diagonalizable even with repeated eigenvalues; you just pick an orthonormal basis for the eigenspace.

Try it in the visualization

The matrix acts on the unit circle, producing an ellipse; the eigenvectors emerge as the semi-major and semi-minor axes. Sliders move the off-diagonal entry; eigen-directions rotate in response while always staying perpendicular — confirming the spectral theorem.