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By definition, in an orthonormal matrix, all the column vectors are unit vectors and mutually orthogonal. However, the row vectors also turn out to be an orthonormal basis. I know how to prove it mathematically, but is there any intuition or geometric interpretation behind this observation?

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    $\begingroup$ Nice question! The statement the the columns of a matrix $A$ are orthonormal is equivalent to the equation $A^T A = I_n$. In general, if $A,B$ are square matrices and $BA = I_n$ then $AB = I_n$ so this implies that $A A^T = I_n$ so the columns of $A^T$ (i.e the rows of $A$) are also orthonormal. Unfortunately, the proof that $BA = I_n$ implies that $AB = I_n$ is not very geometric so I don't know how to extract a geometric interpretation from that. $\endgroup$
    – levap
    Aug 13 '17 at 16:21
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It sufficies to observe that, given an orthogonal operator in $f \colon V \to V$, where $V$ is a finite-dimensional real euclidean space, its transpose operator ${}^tf \colon V^* \to V^*$ is also orthogonal.

In fact we have, for all $\varphi, \psi \in V^*$,
$$\langle \varphi, \, \psi \rangle: = \langle a, \, b \rangle,$$ where $a$ and $b$ are the elements in $V$ representing the functionals $\varphi, \, \psi$ with respect to the given scalar product. So we obtain, using the orthogonality of $f$, $$\langle {}^t f(\varphi), \, {}^tf(\psi) \rangle = \langle f(a), \, f(b) \rangle = \langle a, \, b \rangle = \langle \varphi, \, \psi \rangle$$ and we are done.

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  • $\begingroup$ I think what you mentioned is just another way to restate my original statement. What am I really looking for is a geometric interpretation behind why the transpoe of an orthogonal operator in f: V --> V is also orthogonal. $\endgroup$ Aug 13 '17 at 16:58
  • $\begingroup$ What do you mean exactly by "geometric interpretation"? The fact that the transpose of an operator respecting the scalar product respects the scalar product as well is a purely geometric statement. $\endgroup$ Aug 13 '17 at 17:03
  • $\begingroup$ If you prefer, the natural pairing $$\Phi \colon V \to V^*, \quad v \mapsto \langle v, \, \cdot \rangle$$ associate to a vector $v \in V$ the hyperplane $\ker \Phi(v)$, and orthogonality of two vectors in $V$ corresponds to the orthogonality of the respective hyperplanes. $\endgroup$ Aug 13 '17 at 17:08
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The transposition operation of a nonsingular real matrix corresponds to the geometric operation of passing from a basis of $\mathbb R^n$ to the dual one. That is, if $v_1\ldots v_n$ are the columns of $A$, then the columns $w_1\ldots w_n$ of $A^T$ form a basis of $\mathbb R^n$ and they satisfy the condition $$ v_i\cdot w_j= \delta_{i{}j}.$$ (That's why the two bases are called dual to each other). The theorem that the transpose of an orthogonal matrix is orthogonal can be restated by saying that the dual to an orthonormal basis is an orthonormal basis.

The problem is therefore to geometrically characterize the transformation of a basis into its dual one. In $\mathbb R^3$ this can be done via the cross product as follows: setting $g=(v_1\times v_2)\cdot v_3$, you have that $$ \begin{array}{ccc} w_1=g^{-1} v_2\times v_3, & w_2=g^{-1}v_3\times v_1, & w_3=g^{-1}v_1\times v_2\end{array}.$$ Source: Itskov's book, pag.10.

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  • $\begingroup$ This is very much the same answer as Francesco Polizzi's one, with the simpler notion of duality given by the dot product. $\endgroup$ Aug 13 '17 at 17:24

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