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Let $A,B \in \text{SL}(n,\mathbb{R})$. Define $s_1=AA^T,s_2=BB^T$.

Write $$ \text{det}(ts_1-s_2) = (t-\lambda_1)(t-\lambda_2)\cdots(t-\lambda_n) $$ where each $\lambda_i$ is positive and they satisfy $\lambda_1\lambda_2\cdots\lambda_n=1$.

Now, suppose that $\lambda_i=1$ for every $1 \le i \le n$.

Why does this imply $s_1=s_2$?

Comment:

This question comes from an attempt to see directly why the distance on $\text{SL}(n,\mathbb{R})/\text{SO}(n)$ induced by the unique $\text{SL}(n,\mathbb{R})$-invariant metric is positive.

(It turns out that when representing elements $m=A\cdot \text{SO}(n)$ via $s=\sigma(m)=AA^T$, then up to a constant multiple, one has $$ dist(s_1,s_2) = \left(\sum_{i=1}^n (\log\lambda_i)^2\right)^{1/2}. $$

(For further details see this answer by Robert Bryant)

Edit:

Denote $C=A^{-1}BB^TA^{-T}$.

As observed in the answer below,

$$ \det(tI-C)= (t-\lambda_1)(t-\lambda_2)\cdots(t-\lambda_n), $$

so the eigenvalues of the symmetric matrix $C$ are exactly $\lambda_1,...,\lambda_n$.

Since $C$ is orthogonally diagonalizable, there exist $P \in O_n$ s.t

$$ PCP^{-1}=\operatorname{diag}(\lambda_1,...,\lambda_n).$$

Thus, the closer $\Lambda=\operatorname{diag}(\lambda_1,...,\lambda_n)$ to $I$, the closer $C$ is to $I$. Explicitly:

$$ \| C-I\|=\| P^{-1}\Lambda P-I\|=\| P^{-1}(\Lambda-I)P\|=\|\Lambda-I \|$$

or, by noting that $C \in Psym_n$, and hence has a unique symmetric logarithm,

$$ d(C,I)=\| \log C\|=\| \log (P^{-1}\Lambda P)\|=\| P^{-1}(\log \Lambda)P\|=\|\log \Lambda \|=d(\Lambda,I),$$ where $\| \|$ is the standard Euclidean (Frobenius) norm, and we endow $Psym_n$ with its left translation metric).

So, indeed $\|\log \Lambda \|$ can serve as a reasonable measure for the distance between $s_1,s_2$.

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  • $\begingroup$ Since $ts_1-s_2$ is symmetric and thus diagonalizable, by the end result you know $P(ts_1-s_2)P^{-1}=t-I$. The desired result thus follow from a comparision of the coefficient of $t$. $\endgroup$ – Bombyx mori Mar 11 '17 at 14:34
  • $\begingroup$ Thanks for the hint. I am not sure how to deduce the exact claim you are implying though. (I was able to deduce something weaker, please see my edited part in the question). What am I missing? $\endgroup$ – Asaf Shachar Mar 11 '17 at 16:00
  • $\begingroup$ Do you know the determinant of a diagonal matrix? $\endgroup$ – Bombyx mori Mar 11 '17 at 19:31
  • $\begingroup$ maybe one way to see it, consider s1 $\endgroup$ – Bombyx mori Mar 11 '17 at 19:47
  • $\begingroup$ I am still not sure what exactly do you mean in your suggested approach. However, a nice solution (which seems somewhat in the spirit of your argument) was added below. $\endgroup$ – Asaf Shachar Mar 11 '17 at 20:11
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Write $A^{-T}=(A^T)^{-1}$. Then $$ \det(ts_1-s_2)=\det(A)\det(tI-A^{-1}BB^TA^{-T})\det(A^T)=\det(tI-A^{-1}BB^TA^{-T}). $$ So, the given conditions mean that all eigenvalues of the symmetric matrix $A^{-1}BB^TA^{-T}$ are equal to $1$. Hence $A^{-1}BB^TA^{-T}=I$, i.e. $AA^T=BB^T$.

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