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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.

Theo Bendit
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Canine360
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  • When $n<k$ the matrices cannot have rank $k$. In any matrix the rank cannot exceed either the number of rows or the number of columns. – Oscar Lanzi Nov 18 '18 at 02:09
  • May be relevant: https://math.stackexchange.com/questions/674310/proving-the-product-of-two-non-singular-matrices-is-also-non-singular and https://math.stackexchange.com/questions/1136495/multiplication-of-inverse-and-non-inverse-matrices – NoChance Nov 18 '18 at 03:06

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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 $k-(n-k)$.

Empy2
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