Currently I am studying orthogonal vectors and matrices. Although I got the theory, i.e. the connections between dot product, angle, orthonormal basis etc., I miss the intuition that lies behind certain formulas.

My question: How could you interpret it (geometrically) that the transpose of the orthogonal matrix A equals its inverse, viz. $A^{t} = A^{-1}$ or that for the determinant it is $det(A) = 1$.

Of course, one can prove all that analytically just fine. But how could I imagine the linear map and develop an intuition with respect to the 'right angle feature' of the vectors? I hope this is not too vague a question. If so, I will rephrase.

  • $\begingroup$ I use to think of orthogonal (and unitary) matrices as norm preserving (essentially, they are rotations and reflections) first, and just think that transpose and inverse happen to be coinciding notions for them. That the determinant is one tells you that volumes (and orientations) are preserved through multiplication by these matrices, which should really feel comfortable together with the "rotation/reflection" interpretations. However, the determinant could be -1 as well, mind you, in which case orientations are reversed. See Wikipedia: en.wikipedia.org/wiki/Orthogonal_matrix $\endgroup$ – A.Sh Apr 20 '16 at 18:24
  • $\begingroup$ Okay, I hold the notion of reflections and rotations useful and intuitive too. Thanks for the explanation of the determinant. This makes sense now. However, I cannot grasp just now what the visual connection is to the equality of transposed and inversed matrix. These are two different kinds of 'mapping reversion' or 'mapping change' and yet they are equal in the context of rectangular vectors (matrices). Why is that? $\endgroup$ – Taufi Apr 20 '16 at 18:42
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    $\begingroup$ @Taufi In this case, you can think of the transpose of a transformation as its “mirror image.” For a rotation, this reverses the angle and thus undoes the rotation. For a reflection, it makes no difference (reflection matrices are symmetric), but reflections are their own inverses, anyway. $\endgroup$ – amd Apr 20 '16 at 19:12

Think of 2x2 orthogonal matrix. If it's determinant 1, then it will be a matrix of the form $$ \begin{pmatrix} cos(\theta)&-sin(\theta)\\ sin(\theta)&cos(\theta)\\ \end{pmatrix} $$

So it transpose is $$ \begin{pmatrix} cos(\theta)&sin(\theta)\\ -sin(\theta)&cos(\theta)\\ \end{pmatrix} $$

First matrix is rotation by $\theta$ counter-clockwise and another is rotation by $\theta$ clockwise, so it makes sense they are the inverse of each other.

If it's determinant -1, think of it as a reflection matrix.

Higher dimensional matrices can be thought in a similar manner I believe.


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