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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    $\begingroup$ None that I can think of. How about Scaled orthogonal matrix? Just made that up. $\endgroup$ – Inquest Jul 2 '12 at 8:53
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    $\begingroup$ It's a great pity that the name "orthogonal matrix" is already taken. These kinds of matrices are quite common in my business (geometric modeling). They correspond to non-uniform scaling operations. Maybe I'll have to call them "non-uniform scaling matrices" if the world of mathematics can't offer anything better. $\endgroup$ – bubba Jul 2 '12 at 9:06
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    $\begingroup$ I would probably reserve that name for the diagonal matrices used in coordinate transformation for scaling with unequal scaling coeffecients in the coordinate directions but that's just me. $\endgroup$ – Inquest Jul 2 '12 at 9:13
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    $\begingroup$ Pseudorthogonal? $\endgroup$ – rschwieb Jul 2 '12 at 11:07
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    $\begingroup$ So, since the 0 vector is orthogonal to everything, some of the columns could be 0? ... what is so bad about saying "matrix with orthogonal columns"? $\endgroup$ – GEdgar Dec 21 '12 at 20:40

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.


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?

  • $\begingroup$ No, orthogonality of rows bears no relation to shear. Scaling the first vector by $2$ has orthogonal rows, then composing with a rotation by $45^\circ$ produces non-orthogonal rows $\sqrt2({1\atop1}~{0.5\atop-0.5})$, but this is not in any way a shear transformation. $\endgroup$ – Marc van Leeuwen Apr 2 '13 at 5:01
  • $\begingroup$ @Marc: Yes, you are quite right. $\endgroup$ – TonyK Apr 2 '13 at 8:43

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