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Is there a name for a matrix whose rows (or columns) are non-zero orthogonal vectors ?

It seems to me that "orthogonal matrix" would be a good name, but this is already taken -- it refers to a matrix whose rows (or columns) form an orthonormal set of vectors.

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    This question has been asked a few times. The term "orthogonal matrix" for $A\in M_n(\mathbb{R})$ such that $A^TA=(AA^T)=I_n$ is a bit unfortunate, athough widespread. Unitary is less ambiguous, and works in the real case, like in the complex case. Then you could call orthogonal a matrix whose columns are orthogonal. But that's too late for a change of habits.2013-07-13

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An orthogonal matrix refers to a matrix whose rows and columns are orthonormal. This is a key property of orthogonal matrices, one which ultimately requires these matrices to be square.

Suppose $A$ is rectangular matrix ($n > m$) with row and column vectors which are [a] non-zero and [b] orthogonal to one another. We know,

  1. Orthogonal vectors are also linearly independent.
  2. The row rank of $A$ equals the column rank of $A$, $\textrm{rank}(A') = \textrm{rank}(A)$.

Then, (1) and (2) together suggest $\textrm{rank}(A') = \textrm{rank}(A)$, or $m = n$. But this a contradiction.

Restricting the rows or columns to be orthogonal and non-zero is a departure of sorts. A semi-orthogonal matrix $B$ is a non-square matrix with real entries having the property that either (1) $BB' = I_m$ or (2) $B'B = I_n$, with the respective true case representing an orthonormal basis.

The case you speak of, a matrix whose rows or columns are orthogonal (not orthonormal), could be described as a semi-orthogonal matrix under a scaling transformation.

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Geometrically, two non-orthogonal rows mean a shear transformation, don't they? If this is right (and I'm sure somebody will let me know if I'm wrong), then you could call them shear-free matrices. Updated to add: Somebody let me know I was wrong, as predicted...

But a pure shear is volume-preserving, so this might be misleading. How about row-orthogonal matrices?

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    @Marc: Yes, you are quite right.2013-04-02