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I know a square matrix is called orthogonal if its rows (and columns) are pairwise orthonormal

But is there a deeper reason for this, or is it only an historical reason? I find it is very confusing and the term would let me assume, that a matrix is called orthogonal if its rows (and columns) are orthogonal and that it is called orthonormal if its rows (and columns) are orthonormal but apparently that's not conventional.

I know that square matrices with orthogonal columns have no special interest, but thats not the point. If I read the term orthogonal matrix my first assumption is, that its rows (and columns) are orthogonal what is correct of course, but the more important property is that they are also orthonormal


So, Question: Why do you call an orthogonal matrix orthogonal and not orthonormal? Wouldn't this be more precisely and clearly?

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    $\begingroup$ "It might be tempting to suppose a matrix with orthogonal (not orthonormal) columns would be called an orthogonal matrix, but such matrices have no special interest and no special name." en.wikipedia.org/wiki/Orthogonal_matrix#Matrix_properties $\endgroup$ – oldrinb Jan 24 '14 at 22:37
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    $\begingroup$ Orthonormal would have been a better name. $\endgroup$ – littleO Jan 24 '14 at 23:10
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    $\begingroup$ Not claiming this is the historical origin of the term, but note that multiplication by an $n \times n$ matrix $A$ is an isometry, i.e., preserves (length and) orthogonality of arbitrary vectors, if and only if the columns of $A$ are orthonormal. That is, I suspect it's the transformation $T(x) = Ax$ that's referred to by the term "orthogonal", not the rows/columns of $A$. $\endgroup$ – Andrew D. Hwang Jan 25 '14 at 1:03
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    $\begingroup$ the comment tells you why: they are of no particular interest... if a matrix has orthogonal columns then it necessarily is of the form $UD$ where $U$ is orthogonal and $D$ is diagonal $\endgroup$ – oldrinb Jan 25 '14 at 1:41
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    $\begingroup$ @Tunococ what about $\lambda\cdot I,\ |\lambda|\notin\{0,1\}$ then? That's a matrix which obviously has orthogonal, yet not orthonormal column AND row vectors. $\endgroup$ – Sora. Jul 27 '16 at 20:11
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A affine transformation which preserves the dot-product on $\mathbb{R}^n$ is called an isometry of Euclidean $n$-space. In fact, one can begin without the assumption of an affine map and derive it as a necessary consequence of dot-product preservation. See the Mazur Ulam Theorem which shows this result holds for maps between finite dimensional normed spaces where the notion of isometry is that the map preserves the norm. In particular, an isometry of $\mathbb{R}^n$ can be expressed as $T(v)=Rv+b$ where $R^TR=I$. The significance of such a transformation is that it provides rigid motions of Euclidean $n$-space. Two objects are congruent in the sense of highschool geometry if and only if some rigid motion carries one object to the other.

My Point? this is the context from which orthogonal matrices gain their name. They correspond to orthogonal transformations. Of course, these break into reflections and rotations according to $\text{det}(R)= -1,1$ respective.

Likely thought: we should just call such transformations orthonormal transformations. I suppose that would be a choice of vernacular. However, we don't, so... they're not orthonormal matrices. But, I totally agree, this is just a choice of terminologies. Here's another: since $R^TR=I$ implies $R$ is either a rotation or reflection let's call the set of such matrices rotoreflective matrices. In any event, I would advocate the change of terminology you advocate, but, the current terminology is pretty well set at this point, so, good luck if you wish to change this culture.

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  • $\begingroup$ +1 Yes it definitely comes from this context. (Although I would hesitate to lump "rotoreflective" transformations and "rigid motions" in with each other. My impression is that preserving orientation is a requirement for rigid motions.) $\endgroup$ – rschwieb Jun 18 '14 at 14:53
  • $\begingroup$ @rschwieb that may be, let me check... well, not in O'neill's Elementary Differential Geometry (which I happen to have sitting here) on page 100 he indicates rigid motion as just another name for isometry. I suppose, the question is, is a reflection a rigid motion? $\endgroup$ – James S. Cook Jun 18 '14 at 15:00
  • $\begingroup$ Yeah, it's just a terminology decision. If "rigid" just means "doesn't change the distance relationships between points," then it's just an isometry, but for some authors rigid implies orientation preservation too. (Is the motion that carries you onto your mirror reflection rigid? :) ) Either way is OK, I just wanted to mention the topic. I wouldn't go so far as to say anything is "wrong" with either one. $\endgroup$ – rschwieb Jun 18 '14 at 15:04
  • $\begingroup$ @rschwieb it could be worse, at least we only have two components to worry about here :) $\endgroup$ – James S. Cook Jun 18 '14 at 15:10
  • $\begingroup$ Well, reflections are limited to having determinant $\pm 1$ since $\det(T)^2=1$ only has two solutions over any field, and rotations are products of reflections, so it can't be any worse, right? $\endgroup$ – rschwieb Jun 18 '14 at 15:17
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A matrix $V$ with mutually orthogonal columns is called orthogonal because it maps each of the standard orthogonal coordinate directions to a new set of mutually orthogonal coordinate directions. For example, spherical coordinates are orthogonal because $$ x=r\sin\phi\cos\theta,\;\; y=r\sin\phi\sin\theta,\;\; z = r\cos\phi $$ maps the coordinate lines where $r$ alone varies, $\phi$ alone varies, and $\theta$ alone varies into mutually orthogonal curves in $\mathbb{R}^{3}$. This is evidenced in the Jacobian matrix whose columns are mutually orthogonal: $$ \frac{\partial(x,y,z)}{\partial(r,\phi,\theta)} = \left[\begin{array}{ccc} \sin\phi\cos\theta & r\cos\phi\cos\theta & -r\sin\phi\sin\theta \\ \sin\phi\sin\theta & r\cos\phi\sin\theta & r\sin\phi\cos\theta \\ \cos\phi & -r\sin\phi & 0 \end{array}\right] $$ It is well known that the spherical coordinate system is orthogonal because the three different coordinate lines always intersection orthogonally. The determinant of the Jacobian matrix for this orthogonal transformation is easy to determine: it is the product of the lengths of the three columns of the Jacobian matrix, which is the product of the standard spherical distance scale factors $\frac{dl}{dr}=1$, $\frac{dl}{d\phi}=r$, $\frac{dl}{d\theta}=r\sin\phi$. The standard spherical coordinate volume element is the the product of these $dV = r^{2}\sin\phi\,dr\,d\phi\,d\theta$.

A non-zero unitary matrix goes one step further: any set of orthogonal vectors is mapped to another, which is much stronger than the standard coordinate directions being mapped to mutually orthogonal directions. This condition implies that there is a constant $C$ such that the unitary matrix maps all unit vectors to vectors of length $C$. After renormalizing by dividing by $C$, the unitary matrix is distance preserving and preserve angles: $(Vx,Vy)/\|Vx\|\|Vy\|=(x,y)/\|x\|\|y\|$.

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  • $\begingroup$ Isn't unitary just the analogue of orthogonal in the case complex case? I don't think unitary is stronger at all since both preserve scalar product completely so all properties enhanced in a vector... $\endgroup$ – C-Star-W-Star Jun 18 '14 at 8:22
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    $\begingroup$ -1: A matrix with mutually orthogonal columns is not called orthogonal. That is the entire point of the question. $\endgroup$ – Rahul Jun 18 '14 at 9:38
  • $\begingroup$ @Rahul: You can argue over how the terminology is used for a matrix, but I answered the reason behind calling a coordinate transformation orthogonal. I described what the transformation being orthogonal means--regardless of your preferred definition--and how it relates to mutually orthogonal columns. $\endgroup$ – DisintegratingByParts Jun 18 '14 at 20:49
  • $\begingroup$ @Freeze_S : If you're only talking about matrices, then orthogonal maps coordinate lines in one orthogonal frame to orthogonal lines in another. Unitary does the same thing, but in complex spaces. For smooth coordinate systems, orthogonal is a little different. Calculus orthogonal coordinate changes are very old, and important since the time of Fourier. Most significant linear theory--in one way or another--grew out of Fourier analysis. Abstract linear (esp inner-product) spaces definitely have their roots in Fourier's Principle of Superposition, orthogonal expansion, Separation of Variables. $\endgroup$ – DisintegratingByParts Jun 18 '14 at 21:01
  • $\begingroup$ I'm afraid you're mistaken. The definition of an orthogonal transformation is well established (Wikipedia, MathWorld, Encyclopedia of Mathematics), and it's not what you say it is. Your post is correct insofar as you're talking about orthogonal coordinate systems, but they're not what the question is about, and the Jacobian of an orthogonal coordinate system is not necessarily an orthogonal matrix. $\endgroup$ – Rahul Jun 18 '14 at 22:42
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Preservation of Structure

First of all note that any matrix represents a linear transformation: $$T(x):=Ax\quad$$ (That is what is of most interest.)

Now a quick calculation shows: $$\langle T(x),T(y)\rangle=\langle Ax,Ay\rangle=(Ax)^\intercal(Ay)=x^\intercal A^\intercal Ay$$

So a linear operator preserves scalar product iff it's adjoint is a left inverse: $$\langle T(x),T(y)\rangle\equiv\langle x,y\rangle\iff A^\intercal A=1$$ That is it is linear and preserves angles and lengths, especially orthogonality and normalization. These transformation are the morphisms between scalar product spaces and we call them orthogonal (see orthogonal transformations).

Unfortunately I guess that is not where the name comes from historically. But one should keep in mind that a statement about column and row vectors is fancy but also special and hides what is really happening...

Invertibility

First of all note that if it bijective, so invertible, its inverse will be linear too and also preserves scalar product automatically. Thus it is an isomorphism then.

Now, a linear transformation that preserves scalar product is necessarily injective: $$T(x)=0\implies\|T(x)\|=\langle T(x),T(x)\rangle=\langle 0,0\rangle\implies x=0$$ However it might fail to be surjective in general - take as an example the rightshift operator.
If it happens to be surjective too, so bijective, then it has an inverse matrix: $$A^{-1}A=1\text{ and }AA^{-1}=1$$ But since the inverse is unique we have in this case: $$A^\intercal=A^{-1}$$ Concluding that the isomorphisms are given by matrices that satisfy: $$A^\intercal A=1\text{ and }AA^\intercal=1$$

In the finite dimensional case surjectivity directly follows by the rank nullity theorem. Thus it is enough then to check that the transpose matrix is either left inverse or right inverse instead of checking both. That check goes for any matrix to conclude injectivity by surjectivity or vice versa.

Annotation

The rank nullity theorem states that: $$\dim\mathrm{dom}T=\dim\ker T+\dim\mathrm{ran}T$$

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    $\begingroup$ None of this seems to address the actual question. $\endgroup$ – Santiago Canez Jun 18 '14 at 14:08
  • $\begingroup$ Yes I was afraid already that somebody might doubt that this actually answers the question... However what is hidden behind all this is that the essence should be that the name was more or less a historical accident so that the crucial point one should keep in mind is that these matrices preserve structure especially orthogonality. $\endgroup$ – C-Star-W-Star Jun 18 '14 at 14:52
  • $\begingroup$ @Freeze_S To give some more detailed feedback: the first point is highly relevant, but not very clearly explained. Things starting at "invertibility" and downward look irrelevant. $\endgroup$ – rschwieb Jun 18 '14 at 15:06
  • $\begingroup$ Things starting after "invertibility look irrelevant however happen to be very much relevant as orthogonal transformation (and their associated orthogonal matrices) are the ones that preserve the structure and surely that's where their name comes from (orthogonality and normalization is preserved iff the transformation is linear and orthogonal) but the definition is not taken as $A^\intercal A=1$ only but together with $AA^\intercal=1$. This being no restriction in the finite dimensional case compared to only requiring $A^\intercal A=1$ is explained in the part on invertibility. $\endgroup$ – C-Star-W-Star Jun 19 '14 at 6:09
  • $\begingroup$ After all transformations that preserve orthonormality in any case are given precisely by $A^\intercal A=1$ and not more... $\endgroup$ – C-Star-W-Star Jun 19 '14 at 6:10
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I would like to explain orthogonality in terms of vectors which explains why orthogonal matrix is called orthogonal matrix.

Dot product of 2 orthogonal vectors is 0.

Assume vectors,$\ \vec a= \left[ \begin{matrix} a_1 \\ a_2 \\ a_3 \\ ... \\ a_n \end{matrix} \right] \ and \ \vec b = \left[ \begin{matrix} b_1 \\ b_2 \\ b_3 \\ ... \\ b_n \end{matrix} \right] $.

Now dot product between the vectors can be computed as: $ (\vec a)^T.(\vec b) = [a_1*b_1+a_2*b_2+a3*b3....+a_n*b_n] $

Assume each of the columns in matrix Q as a vector. A matrix Q is an orthogonal matrix if each column vector is orthogonal to the other column vectors in the matrix Q.

So, for every column i and column j in matrix Q, if they have to be orthogonal to each other, the dot product across every column i with every column j should be 0, when i is not equal to j.

Also, the magnitude of every column needs to be same. So, let's say that when i=j, the value is constant 1.

Therefore dot product between column i of matrix Q and column j of matrix Q can be written as $ <Q_i,Q_j> = \begin{cases} 0, i \ne j \\ 1, i = j \end{cases} $

So, computing dot products across all columns can be done simultaneously using $ Q^T.Q$

The result will be $ I_n $

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  • $\begingroup$ Welcome to MSE! $\endgroup$ – José Carlos Santos Jan 26 at 15:11
  • $\begingroup$ "A matrix Q is an orthogonal matrix if each column vector is orthogonal to the other column vectors in the matrix Q." No, it's not. A matrix is orthogonal if the columns are orthonormal. That is the entire point of the question. $\endgroup$ – Rahul Jan 26 at 17:37
  • $\begingroup$ Thanks for welcoming, @JoséCarlosSantos $\endgroup$ – Madhusudan N May 2 at 6:38

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