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I suspect the following is well-known, but don't know of a reference (and it is not close to the area I normally work in).

I have two sequences of matrices $Q_{1},\ldots,Q_{k}$ and $R_{1},\ldots,R_{k}$ in $\mathbb{R}^{n^{2}}$ and would like to bound $|| \prod_{i=1}^{k} Q_{i} - \prod_{i=1}^{k} R_{i} ||_{\mathrm{HS}}$ in terms of the $k$ numbers $|| Q_{i} - R_{i} ||_{\mathrm{HS}}$.

If we have $Q_{i}, R_{i} \in SO(n)$ for all $i$, it is easy to check that

$|| \prod_{i=1}^{k} Q_{i} - \prod_{i=1}^{k} R_{i} ||_{\mathrm{HS}} \leq \sum_{i=1}^{k} || Q_{i} - R_{i} ||_{\mathrm{HS}}$.

That isn't true more generally (e.g. if the matrices don't all have norm 1, this must appear in the upper bound), and I am looking for a standard reference for any similar fact (especially if it includes this bound as a special case).

I am particularly interested in:

  1. $k$ large, $|| Q_{i} - R_{i} ||_{\mathrm{HS}}$ small.

  2. The case where, for each $i$, either $Q_{i}, R_{i} \in SO(n)$ or $Q_{i}, R_{i}$ are in the associated Lie algebra and have the same (moderate) norm.

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I am not sure that you can get anything better than the following triangle inequality bound:

$$\left\|\prod_{i=1}^k Q_i - \prod_{i=1}^k R_i\right\|_{HS} \leq \sum_{h=1}^k \left\|\prod_{i=1}^{h-1}Q_i\right\|_2 \, \left\|Q_h-R_h\right\|_{HS}\, \left\|\prod_{i=h+1}^k R_k\right\|_2.$$

EDIT: I have converted the outer HS norms to (smaller) operator norms, using the fact that $\|{AB}\|_{HS} \leq \|A\|_{HS}\|B\|_2$ and $\|{AB}\|_{HS} \leq \|A\|_{2}\|B\|_{HS}$. Note that this bound reduces to the one you found in the case when the $Q_i$ and $R_i$ are orthogonal.

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  • $\begingroup$ Thank you! This looks good, and seems to have essentially the same short proof. Do you know of any place that this is written down, or does one just re-prove as necessary? $\endgroup$
    – QAMS
    Mar 5, 2015 at 19:04
  • $\begingroup$ @QAMS No, I don't know of a place where that identity and triangle inequality argument appear. For the inequalities combining HS-norm and 2-norm, you can see for instance Hogben, Handbook of Linear Algebra, 37-5 (it is a mammoth-sized book full of facts to cite in linear algebra) $\endgroup$ Mar 5, 2015 at 19:25
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    $\begingroup$ In any case, the proof is a one-liner: the same formula without the norms is an identity (telescoping sum). $\endgroup$ Mar 5, 2015 at 19:38

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