Which linear transformations between f.d. Hilbert spaces contract the inner product? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T18:17:32Z http://mathoverflow.net/feeds/question/47301 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47301/which-linear-transformations-between-f-d-hilbert-spaces-contract-the-inner-produ Which linear transformations between f.d. Hilbert spaces contract the inner product? Mike Stay 2010-11-25T05:49:38Z 2010-11-25T13:00:41Z <p>Given two finite-dimensional Hilbert spaces $U, V,$ a linear transformation $T:U\to V$ contracts the inner product if for all $x,y \in U,$ $$\langle x,y \rangle_U \ge \langle Tx, Ty\rangle_V.$$ All unitary transformations satisfy this criterion; is there a larger class of linear transformations that do?</p> http://mathoverflow.net/questions/47301/which-linear-transformations-between-f-d-hilbert-spaces-contract-the-inner-produ/47303#47303 Answer by Faisal for Which linear transformations between f.d. Hilbert spaces contract the inner product? Faisal 2010-11-25T06:28:15Z 2010-11-25T13:00:41Z <p>Such a map will preserve orthogonality, and any such map must be a scalar multiple of an isometry. This is true in great generality, e.g. the map $T$ doesn't have to be linear, and $U$ and $V$ don't have to be finite-dimensional; see Theorem 1 in</p> <blockquote> <p>Chmieliński, <em>Linear mappings approximately preserving orthogonality.</em> J. Math. Anal. Appl. <strong>304</strong> (2005), no. 1, 158–169.</p> </blockquote> <p>Consequently, a map $T \colon U \to V$ contracts the inner product if and only if $T = \alpha S$, where S is an isometry and $|\alpha| \leq 1$.</p> <hr> <p><strong>Edit:</strong> Here's a simple proof of the assertion that an orthogonality-preserving linear map between finite-dimensional inner product spaces is a scalar multiple of an isometry. Let $T \colon U \to V$ be such a map and fix an orthonormal basis <code>$\{e_1, \ldots, e_n\}$</code> for $U$. Observe that $e_i + e_j \perp e_i - e_j$. Thus <code>$$0 = \langle T(e_i + e_j), T(e_i - e_j) \rangle = \langle Te_i, Te_i \rangle - \langle Te_j, Te_j \rangle.$$</code> So we may set $\alpha = \langle Te_i, Te_i \rangle$; this is a nonnegative constant independent of $i$. In particular, if $T$ kills one $e_i$, it kills all the others. It follows that either $T=0$ or else <code>$\{Te_1, \ldots, Te_n\}$</code> is an orthogonal basis for the range of $T$. In the latter situation, an easy computation yields <code>$$\|Tx\|^2 = \sum_i \frac{|\langle Tx, Te_i \rangle|^2}{\langle Te_i, Te_i \rangle} = \sum_i \frac{|\langle \sum_j \langle x, e_j \rangle Te_j, Te_i \rangle|^2}{\langle Te_i, Te_i \rangle} = \sum_i |\langle x,e_i \rangle|^2 \langle Te_i, Te_i \rangle = \alpha \|x\|^2$$</code> for all $x \in U$. It follows that $\frac{1}{\sqrt{\alpha}}T$ is an isometry.</p>