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Let $0 < \sigma_i \in \mathbb{R}$ ($i=1 \dots d$). Is it true that

$$ c \sum_{i=1}^d (\sigma_i-1)^2 \le \sqrt{\sum_{i=1}^d (\sigma_i^2-1)^2} \tag{1}$$

for some $c>0$ which does not depend on the $\sigma_i$.

(I actually suspect this holds for $c=1$).

Motivation:

This inequality is equivalent to the following

$$c \|\sqrt{A^TA}-I\|^2 \le \|A^TA-I\|, \tag{2}$$

where $A$ is a $d \times d$ real matrix. (The equivalence is obtained by considering SVD).

(Inequality $(2)$ comes from comparing different ways to measure deviation of a linear transformation from being an isometry).

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Isn't this just a direct application of Cauchy-Schwarz inequality? We have \begin{align*} \sum_{i=1}^d (\sigma_i-1)^2 &= \left\langle\left((\sigma_1-1)^2,\ldots,(\sigma_d-1)^2\right),\ (1,\ldots,1)\right\rangle\\ &\le \left\|\left((\sigma_1-1)^2,\ldots,(\sigma_d-1)^2\right)\right\|\cdot \|(1,\ldots,1)\|\\ &= \sqrt{d}\sqrt{\sum_i(\sigma_i-1)^4}\\ &= \sqrt{d}\sqrt{\sum_i\left[(\sigma_i-1)^4-(\sigma_i^2-1)^2+(\sigma_i^2-1)^2\right]}\\ &= \sqrt{d}\sqrt{\sum_i\left[-4\sigma_i^2(\sigma_i-1)^2+(\sigma_i^2-1)^2\right]}\\ &\le \sqrt{d}\sqrt{\sum_i(\sigma_i^2-1)^2}. \end{align*} Equality holds in the first inequality iff all $|\sigma_i-1|$s are equal, and equality holds in the second inequality iff each $\sigma_i$ is $0$ or $1$. Hence ties occur in both inequalities iff all $\sigma_i$ are equal to zero or all $\sigma_i$ are equal to one, i.e. iff $A=0$ or $A$ is real orthogonal.

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  • $\begingroup$ Well spotted!${}$ $\endgroup$ – Omnomnomnom Mar 10 '17 at 4:28
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Sorry for all the edits...

The inequality holds for $c=1/d$. Indeed, $$ \sum(\sigma_i-1)^2 \le d \max_i (\sigma_i-1)^2 \le d \max_i |\sigma_i^2-1|\le d\sqrt{\sum (\sigma_i^2-1)^2}, $$

The inequality does not hold for any $c>1/\sqrt{d}$: Assume $\sigma_i = n+1$ for all $i$. Then the inequality reads $$ c\,d\,n^2 \le \sqrt{d(n^2+2n)^2} = \sqrt{d}(n^2+2n),$$ and by taking $n\to \infty$ you get that $c\le 1/\sqrt{d}$.

In fact, if $\sum \sigma_i \ge d$, then inequality holds with $c=1/\sqrt{d}$. Indeed, using Hölder's inequality, we have $$ \sum(\sigma_i-1)^2 = \sum(\sigma_i^2-1) - 2\sum(\sigma_i-1) \le \sqrt{d}\sqrt{\sum (\sigma_i^2-1)^2} - 2\sum(\sigma_i-1). $$ I suspect $c=1/\sqrt{d}$ is the best constant.

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It's clear that in the case of $\sum_i(\sigma_i-1)^2 \leq 1$, we have $$ \sqrt{\sum_{i=1}^d (\sigma_i^2-1)^2} = \sqrt{\sum_{i=1}^d (\sigma_i-1)^2(\sigma_i+1)^2} \geq \sqrt{\sum_{i=1}^d (\sigma_i-1)^2} \geq \sum_{i=1}^d (\sigma_i-1)^2 $$ I'm not sure about the general case, though.


Consider the case of $\sigma_1 = x+1 >\sigma_2 = y+1 > 1$ $\sigma_3 = \cdots = \sigma_d = 1$. We define $$ f(x,y) = \sqrt{\sum_{i=1}^d (\sigma_i^2-1)^2} = \sqrt{[(x+1)^2 - 1]^2 + [(y+1)^2-1]^2}\\ = \sqrt{(x^2 + 2x)^2 + (y^2 + 2y)^2}\\ g(x,y) = \sum_{i=1}^d(\sigma_i - 1)^2 = x^2 + y^2 $$ Now, if we select $x = \alpha y^2$ for $\alpha > 0$, I think we'll notice something. I have a hunch that $f(x,y)/g(x,y)$ has no lower bound.

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  • $\begingroup$ Thanks, this is nice. I still hope for a solution of the general case. $\endgroup$ – Asaf Shachar Mar 9 '17 at 17:56
  • $\begingroup$ Added a little something to consider. $\endgroup$ – Omnomnomnom Mar 9 '17 at 19:49

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