Statistical Odds & Ends

Let’s say you have 2 matrices $A$ and $B$

which you can multiply together to get $AB$ . Knowing the the rank of

$A$ and $B$ does not in generally give you the rank of $AB$ . However, the ranks of these 3 matrices are governed by the following inequality:

$\text{rank} (AB) \leq \min \{ \text{rank}(A), \text{rank}(B) \}.$

The proof is pretty easy. The rank of a matrix is the dimension of its column space. For any vector $v$ in the column space of $AB$ , there is some vector $x$ such that $v = (AB)x = A(Bx)$ . The last equality implies that $v$ is also in the column space of $A$ . Thus, the column space of $AB$ lies in the column space of $A$ , and so $\text{rank}(AB) \leq \text{rank} (A)$ . Repeating the same argument with $(AB)^T$ and noting that the rank of a matrix and its transpose is the same, we have $\text{rank}(AB) \leq \text{rank} (B)$ .

In the special case where $B$ is an invertible matrix, we can say a lot more:

$\text{rank}(AB) = \text{rank}(A)$ .

The first theorem above proves that the LHS is less than or equal to the RHS; we just need to show that $\text{rank}(AB) \geq \text{rank}(A)$ . To do so, note that

$\text{rank}(A) = \text{rank}(AB B^{-1}) \leq \min \{\text{rank}(AB),\text{rank}(B^{-1})\} \leq\text{rank}(AB).$

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