# Proof of Isometry: Inner Product Preserving Map

For known points $x_i,x_j,\ldots,x_k$, in $\mathbb{R}^n$, consider a mapping $y_i,y_j,\ldots,y_k$ in $\mathbb{R}^n$ produced by minimizing the function $f(y)=\sum_{i,j} \left \langle x_i,x_j \right \rangle d(y_i,y_j)$ over $y$ where $d(\cdot)$ is the squared Euclidean distance. The minimization is obtained over orthonormal constraints. i.e, the matrix $Y_{k \times n}$ whose rows are $y_i,y_j,\ldots,y_k$ are orthonormal. Is the result obtained by this mapping isometric to the points $x$?

Can this be proved for isometry or non-isometry? The usage of inner-products and euclidean distances, (which is a metric induced by the inner product norm) makes me feel that this mapping is isometric. Am not sure and am looking for a proof.

You may instead answer this question over a map that preserves $\|\langle x_i,x_j\rangle- d(y_i,y_j)\|$ instead of the above function $f(\cdot)$ if you find this easier to work around inorder to prove/check for isometry. In a broader sense, am looking for a proof for isometry for a map that preserves the inner-products as euclidean-distances.

-
I changed $<x_i,x_j>$ to $\langle x_i,x_u\rangle$. $\TeX$ is sophisticated. –  Michael Hardy Sep 6 '12 at 4:56
What is the definition of orthogonal matrix when $k \neq n$? –  Max Sep 7 '12 at 4:07
Changed "Orthogonal" to "Orthonormal". Also, made minor progress on this question using the polarization identity. Would like to see, how this would be solved over here. Conditions required over the dimension for orthonormality: Columns of $Y$ can be orthonormal when $n \leq k$ and the rows of Y can be orthonormal when $k \leq n$. In our case, it is the latter-where the rows are orthonormal. –  user23600 Sep 7 '12 at 15:05
add comment