For all $ m$ and $ n,$ there exists an $n \times m$ matrix $ A$ such that $ A^TA=\frac{n}{m}\,I$ and every element of the diagonal of $ AA^T$ is $ 1$?

Question

The following speculation comes from this thread, which was inspired from a Putnam competition problem.

For all $ m$ and $ n,$ there exists an $n \times m$ matrix $ A$ such that $ A^TA=\frac{n}{m}\,I_m$ and every element of the diagonal of $ AA^T$ is $ 1.$

I personally care more about the cases when $m<n$. I have tried a couple of cases for small $n$'s and tried using induction to prove it in general but was not successful. Solving matrix equations is very foreign to me.

Update: I added the tag algebraic geometry. If anyone can show the existence of solution (essentially polynomial equations) with any advanced theory of algebraic geometry, I am also happy to know, although I prefer to see elementary solutions.

@MartinR Yes I did as I mentioned in the question body. – No One May 26 '22 at 06:43 — No One, May 26 '22 at 06:43
@P.Quinton Yes, that is the trivial case. – No One May 26 '22 at 06:43 — No One, May 26 '22 at 06:43
Do you have a solution for $m=2$, $n=3$ for instance ? – P. Quinton May 26 '22 at 06:45 — P. Quinton, May 26 '22 at 06:45

score 5 · Accepted Answer · answered May 26 '22 at 07:08

Presumably the underlying field is real. The required $A$ exists if and only if $n\ge m$. The necessity is obvious, as the rank of the LHS of $A^TA=\frac{n}{m}I_m$ is at most $\min\{n,m\}$. For sufficiency, we first pick any matrix $B\in\mathbb R^{n\times m}$ such that $B^TB=\frac{n}{m}I_m$, such as $B^T=\sqrt{\frac{n}{m}}\pmatrix{I_m&0}$. Then $BB^T-I_n$ is traceless. Hence it is orthogonally similar to a matrix $M=Q(BB^T-I)Q^T$ with a zero diagonal (see the second paragraph of my other answer for a proof). In turn, $M+I=QBB^TQ^T$ has a diagonal of ones. Now take $A=QB$ and we are done.

For all $ m$ and $ n,$ there exists an $n \times m$ matrix $ A$ such that $ A^TA=\frac{n}{m}\,I$ and every element of the diagonal of $ AA^T$ is $ 1$?

1 Answers1