Intuition behind row vectors of orthonormal matrix being an orthonormal basis 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?
 A: 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.
A: 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.
