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Suppose I have distribution on linear map $D_{\epsilon,\delta}$ over $\mathbb{R}^{d\times d}$ such that for any $0<\epsilon,\delta<1/2$ we have $\forall x$, $$\Pr_{A \sim D_{\epsilon,\delta}}(\lvert \lVert Ax\rVert^2- \lVert x\rVert^2\rvert>\epsilon \lVert x\rVert^2)<\delta$$ Now given this can I say that $A$ drawn from $D_{\epsilon,\delta}$ also preserves inner product in the following sense $$\Pr_{A \sim D_{\epsilon,\delta}}(\left\lvert \langle Ax_1,Ax_2\rangle-\langle x_1,x_2\rangle\right\rvert>\epsilon\lVert x_1\rVert\lVert x_2\rVert)<\delta$$ I tried using polarization identity to get the result, I have following thing so far \begin{align*} \langle Ax_1,Ax_2\rangle&=\frac{1}{4}(\lVert A(x_1+x_2)\rVert^2-A(x_1-x_2)\rVert^2) \\ &\le\frac{1}{4}((1+\epsilon)\lVert x_1+x_2\rVert^2-(1-\epsilon)\lVert x_1+x_2\rVert^2)\\ &=\langle x_1,x_2 \rangle+\frac{\epsilon}{2}(\lVert x_1\rVert^2+\lVert x_2\rVert^2) \end{align*} Now I am not able to go further as $\lVert x_1\rVert^2+\lVert x_2\rVert^2\geq 2\lVert x_1\rVert\lVert x_2\rVert$. Any help, comments, hints are greatly appreciated. Thanks.

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    Your question can be rewritten: is every approximate linear isometry also an approximate orthogonal transformation (https://en.wikipedia.org/wiki/Orthogonal_transformation)? An orthogonal transformation in 2-D and 3-D must preserve angles. So, a good starting direction for your is to try finding a linear approximate isometry on one of those two spaces that is not an orthogonal transformation. – 2017-01-17
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    @ClementC. I thought the question was bit long and therefore decided to be more concise in asking the main problem. – 2017-01-17
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    The first equation really is only for all $x$ of length one? If you set $x_1=x_2=x$ the first and the second seem to be quite different with respect to scaling of $x$. – 2017-01-21
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    @gg yeah yeah you are right, missed that. Fixed. – 2017-01-21

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You have proved that $$|\langle Ax,Ay \rangle - \langle x,y \rangle| \leq \frac{\varepsilon}{2}(\|x\|^2+\|y\|^2) = \varepsilon,$$ for all vectors $x$ and $y$ of length $1$. Hence for all vector of any length : $$|\langle Ax,Ay \rangle - \langle x,y \rangle| \leq \varepsilon.\|x\|.\|y\|.$$

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    I see what you are saying. By linearity of inner products we get what we want. But also note that $||x||^2+||y||^2\ge2||x||||y||$. Which means we are getting something stronger just because of linearity, can u comment something on that?. And typo on statement one, square is inside. Thanks – 2017-01-23
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    I see what you mean, but I have no 'fondamental explanation'. If you have already seen the proof of [Hölder's inequality](https://en.wikipedia.org/wiki/H%C3%B6lder's_inequality), the trick is the same (prove if for vectors of length 1, then use homogeneity), and I have never seen a direct proof without using this trick. – 2017-01-24