# If A is a matrix of full rank, then will it be true for rank(AB)=rank(B) always? or does it depend on the field?

Let $F$ be a field of characteristic zero. In Wikipidia, it is given that if $A\in M_{m\times n}(F)$ is a matrix of full rank, then rank(AB)=rank(B) for any matrix B in $M_{n\times p}(F)$ conformable for multiplication with A. But taking $F=\mathbb{C}$, I have counter example namely $A=\begin{pmatrix} 2 & \dfrac{1+\sqrt(3/5)i}{2} & \dfrac{1-\sqrt(3/5)i}{2}\\ 1 & 1 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1/2 & 1\\ \dfrac{2}{1+\sqrt(3/5)i} & 1\\ \dfrac{2}{1-\sqrt(3/5)i} & 1 \end{pmatrix}$. See that rank(A)=rank(B)=2, But rank(AB)=1. Is the result not true for $F=\mathbb{C}$?

• Can you add a link to the wikipedia article? Commented Nov 27, 2017 at 11:21
• WolframAlpha says that your product has rank $2$. Are you certain of your calculations? Commented Nov 27, 2017 at 11:26
• Wikipedia says that $\text{rank}(CA) = \text{rank}(A)$ if $\text{rank}(C)$ is the number of columns of $C$. Commented Nov 27, 2017 at 12:20

"Full rank" is a potentially confusing phrase, and it has confused you. Let $$A$$ be an $$m \times n$$ matrix, meaning $$m$$ rows and $$n$$ columns. Then
• $$A$$ is injective if $$\mathrm{rank}(A) = n$$ and
• $$A$$ is surjective if $$\mathrm{rank}(A) = m$$
It is true that, if $$A$$ is injective, then $$\mathrm{rank}(AB) = \mathrm{rank(B)}$$ and, if $$B$$ is surjective, then $$\mathrm{rank}(AB) = \mathrm{rank}(A)$$.
I have seen "$$A$$ has full rank" used by various people to mean that $$A$$ has rank $$m$$, has rank $$n$$ or has rank $$\min(m,n)$$. Your matrix $$A$$ has rank $$2 = m = \min(m,n)$$, so you might or might not want to say it has full rank. But it is not injective, so $$\mathrm{rank}(AB)$$ need not be $$\mathrm{rank}(B)$$.
The field $$\mathbb{C}$$ is not important; you can see the same phenomenon with $$A = \left[ \begin{smallmatrix} 0 & 1 \end{smallmatrix} \right]$$ and $$B = \left[ \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right]$$.