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The Spectral
Theorem

Discover how every symmetric matrix can be decomposed into its eigenvalues and orthogonal eigenvectors: a cornerstone of linear algebra.

Linear Algebra

What Is a Symmetric Matrix?

A matrix \( A \) is symmetric if it equals its own transpose: \( A = A^T \). This means the entry in row \( i \), column \( j \) always matches the entry in row \( j \), column \( i \).

A 2 7 1 7 3 5 1 5 4 =
\(A^T\)
 2 7 1 7 3 5 1 5 4 Mirror across the diagonal

Quick Review: Eigenvalues & Eigenvectors

For a square matrix \( A \), a nonzero vector \( \mathbf{v} \) is an eigenvector with eigenvalue \( \lambda \) if:

\( \lambda \) : Eigenvalue
The scalar that tells you how much the eigenvector is stretched or compressed.
\( \mathbf{v} \) : Eigenvector
A special direction that \( A \) acts on by pure scaling: no rotation.

Why Are Symmetric Matrices Special?

Not every matrix behaves nicely when we look for eigenvalues. Symmetric matrices have three remarkable properties that make them the best-behaved matrices in all of linear algebra.

1
All eigenvalues are real. No complex numbers: every \( \lambda \in \mathbb{R} \).
2
Eigenvectors from different eigenvalues are orthogonal. They form a natural coordinate system at right angles.
3
There are always enough eigenvectors to form a basis. An \( n \times n \) symmetric matrix always has \( n \) linearly independent eigenvectors.

Orthogonal Eigenvectors Visualised

\( \mathbf{v}_1 \)
\( \lambda_1 = 5 \)
\( \mathbf{v}_2 \)
\( \lambda_2 = 2 \)
\( 90^\circ \)
 Eigenvectors of a symmetric matrix are always perpendicular

The Spectral Theorem: Formal Statement

Because \( Q \) is orthogonal, \( Q^{-1} = Q^T \). This decomposition is called the spectral decomposition or eigendecomposition of \( A \).

Inside the Decomposition

A Symmetric = Q Eigenvectors (orthogonal) · Λ Eigenvalues (diagonal) ·
\(Q^T\)
 Transpose
\((= Q^{-1})\)

Symmetric vs Non-Symmetric Matrices

Property Symmetric Non-Symmetric
Eigenvalues Always real May be complex
Eigenvectors Orthogonal basis Not necessarily orthogonal
Diagonalisable? Always Not always
Decomposition \( Q\Lambda Q^T \) \( P\Lambda P^{-1} \) (if exists)
\( A = A^T \)? ✓ Yes ✕ No

How to Find the Spectral Decomposition

1
Verify symmetry. Check that \( A = A^T \).
2
Find eigenvalues. Solve \( \det(A - \lambda I) = 0 \).
3
Find eigenvectors. For each \( \lambda_i \), solve \( (A - \lambda_i I)\mathbf{v} = \mathbf{0} \).
4
Normalise. Scale each eigenvector to unit length: \( \hat{\mathbf{v}}_i = \frac{\mathbf{v}_i}{\|\mathbf{v}_i\|} \).
5
Assemble. Form \( Q = [\hat{\mathbf{v}}_1 \mid \hat{\mathbf{v}}_2 \mid \cdots] \) and \( \Lambda = \text{diag}(\lambda_1, \lambda_2, \dots) \). Then \( A = Q\Lambda Q^T \).

Worked Example: \( 2 \times 2 \) Matrix

Let us spectrally decompose:

Step 2: Eigenvalues. Solve \( \det(A - \lambda I) = 0 \):

Example (continued): Eigenvectors

Harder Example: \( 3 \times 3 \) Symmetric Matrix

The characteristic polynomial gives eigenvalues \( \lambda_1 = 4 \), \( \lambda_2 = 2 \), \( \lambda_3 = 1 \). Each eigenvalue yields a unit eigenvector, and they are automatically orthogonal.

Geometric Interpretation

Unit circle (before A) A· Ellipse (after A)
\( \lambda_1 \)
\( \lambda_2 \)

Where the Spectral Theorem Appears

PCA
Principal Component Analysis uses the spectral decomposition of the covariance matrix to find the most important directions in data.
Quantum Mechanics
Observables are modelled as Hermitian operators (the complex analogue of symmetric): eigenvalues are the measurable values.
Graph Theory
The adjacency matrix of an undirected graph is symmetric: its eigenvalues reveal connectivity and clustering structure.

Common Mistakes

✕ Forgetting to normalise
The columns of \( Q \) must be unit vectors. Using un-normalised eigenvectors means \( Q \) is not orthogonal and \( Q^{-1} \neq Q^T \).
✕ Applying to non-symmetric matrices
The spectral theorem guarantees \( A = Q\Lambda Q^T \) only for symmetric (or Hermitian) matrices. A general matrix needs \( P\Lambda P^{-1} \).
✕ Wrong order in \( Q \) and \( \Lambda \)
Column \( i \) of \( Q \) must correspond to eigenvalue \( i \) in \( \Lambda \). Mismatching columns and diagonal entries gives a wrong decomposition.

Summary at a Glance

Concept Details
Symmetric matrix \( A = A^T \)
Eigenvalues All real: \( \lambda_i \in \mathbb{R} \)
Eigenvectors Mutually orthogonal
Spectral decomposition \( A = Q\Lambda Q^T \)
\( Q \) is orthogonal \( Q^T Q = I \), so \( Q^{-1} = Q^T \)
Geometric meaning Stretch along eigenvector axes

Question 1 of 5

Which condition must a real matrix \( A \) satisfy for the Spectral Theorem to guarantee \( A = Q\Lambda Q^T \)?

Question 2 of 5

If a \( 3 \times 3 \) symmetric matrix has eigenvalues \( \lambda_1 = 4 \), \( \lambda_2 = 2 \), \( \lambda_3 = -1 \), what is the trace of the matrix?

Question 3 of 5

In the spectral decomposition \( A = Q\Lambda Q^T \), what special property does \( Q \) have?

Question 4 of 5

Eigenvectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) of a symmetric matrix correspond to different eigenvalues. What can you conclude?

Question 5 of 5

A symmetric matrix \( A \) transforms the unit circle into an ellipse. The eigenvalues determine which property of the ellipse?