Let A be $ m \times n$ and $B$ be $n \times p$, and suppose $AB = 0$. Explain why $\mathrm{rank}(A) + \mathrm{rank}(B) \leq n.$

Question

Let $A$ be $m \times n$ and $B$ be $n \times p$, and suppose $AB = 0$. Explain why $\text{rank}(A) + \text{rank}(B) ≤ n$?

Answer 1

1) $AB=0$ means that the range of $B$ is a subspace of the kernel of $A$. This means that the rank of $B$ must be less than dimensionality of kernel of $A$.

2) The sum of the dimensionality of kernel of $A$ and the dimensionality of orthogonal complement of kernel of $A$ must be less than $n$. (Fundamental theorem of linear algebra)

3) The dimensionality of orthogonal complement of the kernel must equal the row-rank of a matrix, (and since row-rank = column-rank,) therefore the rank of a matrix.

Putting 1, 2, and 3 together, $\textrm{rank}(A) + \textrm{rank}(B) < n$

Answer 2

Sylvester's rank inequality (see Sylvester rank inequality) states that if $A$ is an $m \times n$ matrix and $B$ is an $n \times p$ matrix, then: $$\text{rank}(A) + \text{rank}(B) - n \leq \text{rank}(AB)$$ Since in your case $AB = 0$, you have that $\text{rank}(AB) = 0$ and hence $$\text{rank}(A) + \text{rank}(B) \leq n$$