# Name for matrices with orthogonal (not necessarily orthonormal) rows

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.

• None that I can think of. How about Scaled orthogonal matrix? Just made that up. Jul 2, 2012 at 8:53
• 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. Jul 2, 2012 at 9:06
• 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. Jul 2, 2012 at 9:13
• Pseudorthogonal? Jul 2, 2012 at 11:07
• 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"? Dec 21, 2012 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?

• 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. Apr 2, 2013 at 5:01
• @Marc: Yes, you are quite right. Apr 2, 2013 at 8:43