One encounters often in numerics non-square matrices with orthonormal columns, i.e., $U\in\mathbb{R}^{m\times n}$, with $m > n$, such that $U^TU=I$ (but, clearly, $UU^T \neq I$).

Is there a name for these matrices?


Orthonormal $\boldsymbol n$-frames :  https://en.wikipedia.org/wiki/Stiefel_manifold.

Added: This terminology of Hirzebruch (1966), Steenrod (1951) translates the $\boldsymbol n$-Systeme of Stiefel (1936), $\boldsymbol n$-podes, $\boldsymbol n$-pèdes or multipèdes of Einstein (1931), Waelsch (1907), $\boldsymbol n$-Beine or Vielbeine of Hirzebruch (1956), Einstein (1928), Blaschke (1920), or Waelsch (1906).

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Francois Ziegler has already provided a very relevant answer.

Let me point out three further relevant things:

1. While the OP is not expressly interested in matrices with integer entries, a very relevant article nevertheless is W. Plesken: Solving $XX^{\mathrm{tr}}=A$ Over the Integers. Linear Algebra and its Applications. Volumes 226–228, September–October 1995, Pages 331-344. Therein, in is in particular proved (let $A=I$ and $X=U^{\mathrm{t}}$) that in the context of the OP

$U^{\mathrm{t}}U = I\quad$ ${}\quad$ if and only if${}\qquad\qquad$ $(U U^{\mathrm{t}})^2 = U U^{\mathrm{t}}$ $\quad$and$\quad$ $\mathrm{tr}( UU^{\mathrm{t}})=n$

$\color{white}{( U^{\mathrm{t}}U = I )}{}\quad$ if and only if${}\qquad\qquad$ $UU^{\mathrm{t}}$ is idempotent and has its trace equal to its rank

$\color{white}{( U^{\mathrm{t}}U = I )}{}\quad $ if and only if${}\qquad\qquad$ $UU^{\mathrm{t}}$ is idempotent,

where the first equivalence holds by Proposition 2.2 in loc. cit., the penultimate holds by a mere reformulation of the first-mentioned equivalence (note that the OP's hypotheses imply that $U U^{\mathrm{t}}$ has rank $n$), and the last step being a widely-known, not-quite-obvious fact from linear algebra (every idempotent matrix has its trace equal to its rank).

So we have shown:

The matrices of the OP are precisely those $U\in\mathbb{R}^{m\times n}$ for which $UU^T$ is an idempotent in the matrix ring $\mathrm{Mat}(m\times n;\mathbb{R})$.

2. Moreover, in the context defined by the OP we have:

If the OP's hypotheses are satisfied, then the Moore-Penrose pseudoinverse of $U$ equals the transpose of $U$. Conversely, if $U$ is a real matrix whose Moore-Penrose pseudoinverse equals its transpose, then the sum of the squares of the rank-sized minors equals $1$. 1

Proof of 2. By the usual formula,

$U^+ = \biggl(\overline{U}^{\mathrm{t}}\cdot U\biggr)^{-1}\cdot \overline{U}^{\mathrm{t}}\qquad\qquad$ (0).

Sufficiency: If the OP's hypotheses are satisfied, then $\overline{U}^{\mathrm{t}}\cdot U = \mathrm{Id}$ and $\overline{U}^{\mathrm{t}} = U^{\mathrm{t}}$, so (0) implies $U^+=U^{\mathrm{t}}$.

Necessity: Suppose conversely that $U^+=U^{\mathrm{t}}$. Then (0) implies that $U^{\mathrm{t}} = (U^{\mathrm{t}}U) U^{\mathrm{t}}$. This implies $U^{\mathrm{t}}U = ((U^{\mathrm{t}}U) U^{\mathrm{t}})\cdot(U(U^{\mathrm{t}}U))$ $=$ (by associativity) $=$ $( U^{\mathrm{t}}U)^3$, hence, applying the homomorphism $\det\colon \mathbb{R}^{m\times m}\to\mathbb{R}$ it follows that, abbreviating $d:=\mathrm{det}(U^{\mathrm{t}}U)$, we have $d = d^3$. Since the OP's hypotheses imply that $U^{\mathrm{t}}U\in\mathbb{R}^n$ has full rank $n$, we know that $d\neq 0$, and hence it follows that $1=d^2$, and hence $d\in\{-1,+1\}$; for further reference,

$\det(U^{\mathrm{t}}U)\in\{-1,1\}\qquad\qquad (1)$.

By the Cauchy-Binet-theorem, it follows from (1) that (in particular since sums of squares of real numbers are non-negative)

$$1 = \sum_{S:\quad\textsf{$n$-element subsets of $m$}} \det( U|_{S\times n})^2 $$

The completes the proof of 2.

1 This is not a "name", yet may lead the OP to useful relevant literature.

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  • $\begingroup$ aren't the only integer orthogonal matrices signed permutations? $\endgroup$ – 1.. Sep 14 '17 at 14:27
  • $\begingroup$ @Turbo: yes, but Plesken's Proposition 2.2 is valid (and explicitly stated) for any matrix $X\in\mathbb{R}^{n\times k}$, the specific title of the paper notwithstanding. $\endgroup$ – Peter Heinig Sep 14 '17 at 15:04

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