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Let $X,Y \in St(n,p) :=\{ A: A \in \mathbb{R}^{p\times n}, A^\top A = I_n \}$. I'm curious to see whether $$d(X,Y)=\arccos\left|\frac{\text{tr}(X^\top Y)}{n\text{det}(X^\top Y)^{\frac{1}{n}}}\right|$$ is a metric. I verified that $d\geq 0,$ $d(X, X)=0,$ and the symmetry, but I don't know how to prove the triangle inequality; I tried to apply Jensen's Inequality but it doesn't seem to work. Any ideas?

ASA
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  • Did you also manage to show that $d(X,Y)=0$ implies $X=Y$? Would you mind sharing how one gets symmetry? – Severin Schraven Oct 05 '22 at 02:18
  • @SeverinSchraven Yes $d(X,Y)=0$ implies $\text{tr}(X^\top Y)/n = \text{det}(X^\top Y)^\frac{1}{n}$, meaning that $X^\top Y= I$. Thus, $X=Y$. – ASA Oct 05 '22 at 02:22
  • Why does $tr(X^T Y)/n =det(X^T Y)^{1/n}$ imply that $X^T Y=I$? Is there some trick to see that? – Severin Schraven Oct 05 '22 at 02:26
  • In fact, even $X^T Y=I$ implying $X=Y$ is not clear to me. – Severin Schraven Oct 05 '22 at 02:28
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    @SeverinSchraven Consider the eigenvalues of $X^T Y$, the equation can be rewritten as an equality between the arithmetic mean and the geometric mean of the eigenvalues. Since the equality holds, the eigenvalues must be equal and they should be 1 because otherwise $X^\top Y$ would be $\alpha I$ for some $\alpha \neq 1,$ you see a contradiction if you start from $X^\top X=Y^\top Y=I$ – ASA Oct 05 '22 at 02:37
  • @SeverinSchraven Now given that $X^\top Y=I$ means that $I=X^\top X= Y^{-1} X$. Thus $X=Y$ – ASA Oct 05 '22 at 02:41
  • This is a nice application of the geometric mean! Thanks for sharing. For your second comment. What is $Y^{-1}$? The matrix $Y$ is not square. – Severin Schraven Oct 05 '22 at 02:44
  • Thank you! Also, sorry for inverting $Y$. It should be $X^\top Y=I$ implies $XX^\top Y= X$, thus $X=Y$. Thanks for noticing it! – ASA Oct 05 '22 at 02:48
  • $X^TY$ is orthogonal, so your denominator is always $n$. Then apply Jensen's, no? – layabout Oct 05 '22 at 02:55
  • @ASA Sorry, I do not understand your argument. I though that $X^T X= I_n$ need not imply that $X X^T=I_p$ (see for example here https://math.stackexchange.com/questions/3696747/does-ata-i-imply-aat-i). One would need to see that $X X^T$ is the idenity on the range of $Y$. By the way, don't we need an absolute value for the determinant? The determinant might be negative, right? – Severin Schraven Oct 05 '22 at 05:05
  • @layabout How does one see that $X^T Y$ is orthogonal. I get $$ (X^T Y)^T (X^T Y) = Y^T (X X^T) Y.$$ But then I would need to have a better understanding of $X X^T$ (maybe this is easy, but I do not quite see it right away). – Severin Schraven Oct 05 '22 at 05:09

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