Which linear transformations between f.d. Hilbert spaces contract the inner product?

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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?

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All partial isometries - and also all multiples of them by scalars of modulus at most 1. If two partial isometries have orthogonal support and range then adding multiples of them will also contract the inner product. – Ollie Margetts Nov 25 2010 at 6:34
Sorry, a little more thought (and Faisal's answer) shows that the above comment is wrong! – Ollie Margetts Nov 25 2010 at 6:38

1 Answer

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

Chmieliński, Linear mappings approximately preserving orthogonality. J. Math. Anal. Appl. 304 (2005), no. 1, 158–169.

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 $|\alpha| \leq 1$.

Edit: 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 `$\{e_1, \ldots, e_n\}$` for $U$. Observe that $e_i + e_j \perp e_i - e_j$. Thus `$$ 0 = \langle T(e_i + e_j), T(e_i - e_j) \rangle = \langle Te_i, Te_i \rangle - \langle Te_j, Te_j \rangle. $$` 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 `$\{Te_1, \ldots, Te_n\}$` is an orthogonal basis for the range of $T$. In the latter situation, an easy computation yields `$$ \|Tx\|^2 = \sum_i \frac{|\langle Tx, Te_i \rangle|^2}{\langle Te_i, Te_i \rangle} = \sum_i \frac{|\langle \sum_j \langle x, e_j \rangle Te_j, Te_i \rangle|^2}{\langle Te_i, Te_i \rangle} = \sum_i |\langle x,e_i \rangle|^2 \langle Te_i, Te_i \rangle = \alpha \|x\|^2 $$` for all $x \in U$. It follows that $\frac{1}{\sqrt{\alpha}}T$ is an isometry.

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