Linear Algebra
4.79 min read

Rank of a Product

How does multiplication affect rank? Two fundamental inequalities: Rank(AB)Rank(A)\text{Rank}(AB) \leq \text{Rank}(A) and Rank(AB)Rank(B)\text{Rank}(AB) \leq \text{Rank}(B).

Geometrically: Col(AB)Col(A)\text{Col}(AB) \subseteq \text{Col}(A) — the outputs of ABAB are also outputs of AA, so ABAB can't exceed AA's rank. Similarly, Null(AB)Null(B)\text{Null}(AB) \supseteq \text{Null}(B) — anything that BB kills, ABAB also kills.

If either AA or BB has full rank (is injective or surjective respectively), then multiplying by it doesn't reduce rank. In particular, multiplying by an invertible matrix preserves rank.

Formal View

Theorem 4.7 (Rank Inequalities)
For ARm×nA \in \mathbb{R}^{m \times n} and BRn×pB \in \mathbb{R}^{n \times p}: \begin{enumerate} \item Col(AB)Col(A)\text{Col}(AB) \subseteq \text{Col}(A), so Rank(AB)Rank(A)\text{Rank}(AB) \leq \text{Rank}(A) \item Null(B)Null(AB)\text{Null}(B) \subseteq \text{Null}(AB), so Nullity(AB)Nullity(B)\text{Nullity}(AB) \geq \text{Nullity}(B) \item Rank(AB)min(Rank(A),Rank(B))\text{Rank}(AB) \leq \min(\text{Rank}(A), \text{Rank}(B)) \end{enumerate}

Why This Matters

Rank inequalities for products appear throughout matrix analysis and algorithm design.

  • Low-rank matrix approximations: if A=BCA = BC with B,CB, C low-rank, then AA is low-rank
  • In neural networks, weight matrices at each layer bound the effective rank of learned representations
  • PCA: the product of a data matrix with its transpose has rank = number of principal components

Quiz

Question 1

If Rank(AA) = 2 and Rank(BB) = 5, what can we say about Rank(ABAB)?

Common Mistakes

  • Thinking rank is multiplicative — Rank(ABAB) \leq min(RankAA, RankBB), not equal.