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Here are two conjectures. I came up with them, and it seems to me intuitively they must be true.

Conjecture $1$. Let $A$ and $B$ be $n\times m$ matrices with $n>m$. Unless either of $n,m$ is equal to $1$, $AB^T$ is never invertible.

Conjecture $2$. Let $A$ and $B$ be $n\times m$ matrices with $m\geq n$ (Note the difference with $(1)$). Then $AB^T$ is invertible iff both $A$ and $B$ have rank $n$.

If $n=m$ in the last conjecture, then it is obvious. Are these conjectures true generally?

user56834
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  • In the first, if either $n,m$ is $1$, then $m=1$, and then $AB^{T}$ is non-invertible $n\times n$. – Thomas Andrews Jan 26 '18 at 15:35
  • Consider $C=B^T$ and consider the conditions for full rank of $AC$, e.g., here. The first conjecture is proved there. – Dietrich Burde Jan 26 '18 at 15:41
  • But if $A=(1,0,0;0,1,0)$ and $B=A$ i have that $AB^T=I$. Where am i getting wrong? – Skills Jan 26 '18 at 15:44
  • It is sometimes true that $AB^{T}$ is invertible when their ranks are $n$, but it is not always true. So the fact that you can find such examples is not surprising. The question is whether you can find examples where it is not true... – Thomas Andrews Jan 26 '18 at 15:44
  • @Skills if you're trying to disprove the first conjecture, note that your matrices are $2 \times 3$, but we want $n > m$. – Ben Grossmann Jan 26 '18 at 15:45
  • @Omnomnomnom lol, thank you. But we use this for disprove the second one right? – Skills Jan 26 '18 at 15:49
  • @Skills no. In particular, both the matrices you used have full rank, and the product is an invertible $n \times n$ matrix. The conjecture is that for wide matrices, the product will be invertible if and only if the matrices multiplied have full rank. – Ben Grossmann Jan 26 '18 at 15:52

1 Answers1

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Your first conjecture is correct, but the second is false.

Proof of 1: Note that $\operatorname{rank}(AB^T) \leq \min\{\operatorname{rank}(A),\operatorname{rank}(B^T)\}$. Thus, the product $AB^T$ has rank at most $m$, which is less than $n$, and therefore can't be invertible.

Counterexample for 2: consider the product $$ \pmatrix{1 & -1}\pmatrix{1&1}^T = 0 $$ Because of the inequality I mentioned above, it is necessary (but not sufficient) for both $A$ and $B$ to have rank $n$ in this second situation.

Ben Grossmann
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