Suppose $A$ is a $k \times n$ matrix and $B$ is an $n \times k$ matrix. Suppose $n \geq k$. Both $A$ and $B$ have rank $k$. Can we say $AB$ is invertible? Also, what happens if $n<k$?
I'm a beginner in linear algebra. I know this question has been answered many times but most answers are too technical for me (involving kernel etc.) and I could not understand them. I'm familiar with the terminology of rank, linear dependence and invertibility, but not much beyond that.
Thank you for your help.
No, if one of the columns of $B$ happens to be orthogonal to all the rows of $A$ then that column of $AB$ will be all zeros.
I expect the rank has to be at least $2k-n$. Extend $B$ to an $n\times n$ matrix $C$ so its columns are a basis of $\mathbb{R}^n$. $C$ is invertible, so $AC$ has rank $k$. Now remove the $n-k$ extra columns of $AC$, and what's left has rank between $k$ and $k-(n-k)$.