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 $F=\mathbb{C}$?

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

1 Answer 1


"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 $B = \left[ \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right]$.


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