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Rank from the Spectral Decomposition

The rank of a symmetric matrix $A$ equals the number of nonzero eigenvalues. This follows from the outer product form $A = \sum_{i=1}^n \lambda_i \mathbf{u}_i \mathbf{u}_i^\top$ : each nonzero term contributes 1 to the rank, and the terms are "orthogonal" (in a precise sense), so the total rank is exactly the count of nonzero $\lambda_i$ .

Equivalently, the null space of $A$ is spanned by the eigenvectors corresponding to zero eigenvalues. If $\lambda_i = 0$ , then $A\mathbf{u}_i = 0$ , so $\mathbf{u}_i \in \ker(A)$ . And since the eigenvectors span $\mathbb{R}^n$ , the nullity equals the number of zero eigenvalues. By the rank-nullity theorem, rank = $n$ - nullity = number of nonzero eigenvalues.

Formal View

Theorem 8.4 — Rank from Spectral Decomposition

Let

A

be a real symmetric

n \times n

matrix with eigenvalues

\lambda_1, \ldots, \lambda_n

. Then

\operatorname{rank}(A) = \#\{i : \lambda_i \neq 0\}

, the number of nonzero eigenvalues.

The null space $\ker(A) = \operatorname{span}\{\mathbf{u}_i : \lambda_i = 0\}$ , and the column space $\operatorname{col}(A) = \operatorname{span}\{\mathbf{u}_i : \lambda_i \neq 0\}$ .

Interactive Visualization

Rank-Nullity Theorem

Why This Matters

Rank tells you the "effective dimension" of a matrix — how many independent directions it can act along.

A covariance matrix with a zero eigenvalue indicates a redundant feature — a linear combination of features is constant.
Rank deficiency in $A^\top A$ signals that the regression problem has infinitely many solutions.
Low-rank approximation: keeping only the top- $k$ eigenvalues gives the best rank- $k$ approximation of the matrix.

Learning Resources

Rank and nullity

Khan Academy

Rank-nullity theorem and how to compute rank.

12 min

Column space and null space

3Blue1Brown

Geometric meaning of column space and null space.

10 min

Quiz

Question 1

A symmetric $4 \times 4$ matrix has eigenvalues $\{3, -1, 0, 0\}$ . What is its rank?

Question 2

If a symmetric matrix has rank $k$ , exactly $k$ of its eigenvalues are nonzero.

Common Mistakes

Counting the number of distinct eigenvalues instead of the number of nonzero eigenvalues — a repeated nonzero eigenvalue still contributes its full multiplicity to the rank.
Forgetting that nullity = number of zero eigenvalues (counting multiplicity), not the count of distinct zero eigenvalues.