8.97 min read

Freedom in the Spectral Decomposition

The spectral decomposition $A = U \Lambda U^\top$ is not unique — there are several sources of freedom in how we write it. First, you can reorder the eigenvalues (and correspondingly reorder the columns of $U$ ). Second, for each eigenvalue, you can flip the sign of the corresponding eigenvector (since $(-\mathbf{u}_i)$ is also an eigenvector). Third, if an eigenvalue has multiplicity greater than 1 (repeated eigenvalue), you can choose any orthonormal basis for the corresponding eigenspace.

Despite this freedom, the set of eigenvalues is unique — it is a fundamental property of the matrix. The eigenvalues are also called the spectrum of the matrix (hence "spectral theorem"). The eigenvectors are unique up to the freedoms listed above.

Formal View

Remark 8.2 — Non-Uniqueness of Spectral Decomposition

The spectral decomposition

A = U\Lambda U^\top

is non-unique due to:\n1. Reordering: permuting

\lambda_1, \ldots, \lambda_n

with a corresponding permutation of columns of

U

.\n2. Sign flip: replacing

\mathbf{u}_i

with

-\mathbf{u}_i

for any

i

.\n3. Eigenspace basis choice: if

\lambda_i = \lambda_j

, any orthonormal basis for

\operatorname{span}(\mathbf{u}_i, \mathbf{u}_j)

works.\nHowever, the multiset of eigenvalues

\{\lambda_1, \ldots, \lambda_n\}

is uniquely determined by

A

Why This Matters

Understanding the freedoms prevents confusion when different software packages return different-looking but equally valid decompositions.

MATLAB and NumPy may return eigenvectors with different signs or orderings — all are correct.
In PCA, the sign of a principal component is arbitrary; only its direction matters.
Repeated eigenvalues occur in isotropic materials; any orthonormal basis for the eigenspace is physically valid.

Learning Resources

Eigenvalue multiplicity and eigenspaces

Khan Academy

Explains geometric and algebraic multiplicity and basis choices for eigenspaces.

12 min

Symmetric matrices and orthogonal eigenvectors

MIT OpenCourseWare

Strang discusses repeated eigenvalues and the choice of eigenvectors.

45 min

Quiz

Question 1

The eigenvalues of a symmetric matrix are uniquely determined (as a multiset), even though the eigenvectors are not.

Question 2

If $\mathbf{u}$ is an eigenvector of $A$ with eigenvalue $\lambda$ , which of the following is also an eigenvector with eigenvalue $\lambda$ ?

Common Mistakes

Thinking there is a unique spectral decomposition — the eigenvalues are unique, but the eigenvectors have multiple valid choices.
Assuming different software will return the same eigenvectors — they may differ by signs, ordering, or basis choice for repeated eigenvalues.