# Rank product of matrix compared to individual matrices. [duplicate]

Possible Duplicate:
How to prove $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))$?

If $A$ is an $m\times n$ matrix and $B$ is a $n \times r$ matrix, prove that the rank of matrix $AB$ is at most $\mathrm{rank}(A)$.

I asked a similar question earlier phrased incorrectly. The above is closer to the actual question generalised.

## marked as duplicate by Jyrki Lahtonen, William, user31373, Chris Eagle, Rudy the ReindeerSep 9 '12 at 8:51

• A duplicate (actually at least a triplicate!). Handled here and here – Jyrki Lahtonen Sep 7 '12 at 5:50

$rank(A)$ is the dimension of the column space of $A$. The product $Ab$, where $b$ is any column vector, is a column vector that lies in the column space of $A$. Therefore, all columns of $AB$ must be in the column space of $A$.

• @JyrkiLahtonen Thank you for pointing out my mistake. – Tunococ Sep 7 '12 at 20:36
• is there any mistake currently in this answer? – user166305 Jan 18 '16 at 16:08
• +1 This answer is more understandable than the answers to the duplicate ones – Mayur Kulkarni Apr 23 '18 at 1:48

$$\mathcal{C}(A) = \{Ax | x \in \mathbb{R}^n\}$$

$$\mathcal{C}(AB) = \{ABy | y \in \mathbb{R}^r\}$$

Notice how $By$ is an n by 1 vector.

Assume to the contrary that the rank of the latter is strictly greater than the first. Then there exists some $y \in \mathbb{R}^m$ that is in $\mathcal{C}(AB)$, but not $\mathcal{C}(A)$. This implies $\exists z \in \mathbb{R}^r$ such that

$$ABz = y.$$

But since $Bz$ is an n by 1 vector, this $y$ will also necessarily lie in $\mathcal{C}(A)$. This is a contradiction.

The rank of $AB$ is equal to the dimension of the image of $AB,$ and similarly for the rank of $A.$ The image of $A$ contains the image of $AB.$