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?


2 Answers 2


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).


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.

  • $\begingroup$ aren't the only integer orthogonal matrices signed permutations? $\endgroup$
    – Turbo
    Sep 14, 2017 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. $\endgroup$ Sep 14, 2017 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.