Orthonormal Matrices-Intuition Why is it, geometrically, that the row space is ALSO orthonormal? What exactly does the transpose LOOK like? Normally the row and column space are two separate things, but in the case of an orthonormal matrix, you have that the row space is actually the inverse of the column space, so also, what does an inverse space look like? I feel there should be a very good geometric interpretation, but I don't see it.
I think this is actually a good question. We can talk about two things: the geometric intuition of an orthogonal matrix and the relationship between row and column space. The latter is slightly more general.

In the first place, let $O(n) = \{A \in M_n(\mathbb{R}) : A^T A = 1\}$, and let $SO(n)$ be the restriction of $O(n)$ to matrices with determinant 1. Note that this basically gets rid of the $A$ with negative determinant, since the condition $A^TA = I$ implies
\begin{gather*}
1 = \det(I_n) = \det(A^T A) = \det(A^T)\det(A) = \det(A)^2,
\end{gather*}
since the transpose does not affect the determinant (prove it or trust me, it's probably not the key idea on a first reading, though it is a consequence of and implies non-trivial things). Therefore $\det(A) = \pm 1$.

Always give yourself simple examples (and counter-examples!). For instance, take 
$$A = \begin{pmatrix} 1 &amp; 0 \\ 0 &amp; -1\end{pmatrix}.$$
This matrix is not in $SO(n)$, since $\det(A) \neq 1$. Geometrically, this matrix is a reflection across the $x$-axis. To see this, see what happens to the basis vectors $(1,0)$ and $(0,1)$. The $x$-coordinate (1,0) stays fixed, while $y \rightarrow -y$. Note that this does not preserve orientation. That is,



Flips aren't all bad though. They still preserve volume. Matrices in $SO(n)$, however, preserve both orientation and volume and can be viewed as rotations. Consider for instance the matrix
$$R = \begin{pmatrix} \cos\theta &amp; -\sin\theta \\ \sin\theta &amp; \cos\theta \end{pmatrix}$$
This is a rotation matrix in $\mathbb{R}^2$. Its characteristic polynomial is $\lambda^2 -2\lambda \cos\theta + 1$, and the discriminant of this polynomial is $b^2 - 4ac = -4\sin^2(\theta).$ Therefore, whenever $\theta \neq 0 (mod 2\pi)$, the eigenvalues are imaginary. The eigenvalues of an orthogonal matrix always have norm $1$, so they are of the form $e^{i\theta_k}$ for various $k$. 

If you want to talk about row and column space, you're going to have to think about the adjoint map a bit, i.e. $A \rightarrow A^*$, where $A^*$ is the matrix which satisfies $(Au,v) = (u,A^*v)$ for every $u, v$ in the vector space. Here $( \cdot,\cdot)$ denotes the inner product on the space (say the Euclidean inner product $x \cdot y$). 

It turns out (one can prove by straight linear algebra by expanding sums and re-arranging terms) that the matrix which satisfies precisely that condition is always the transpose of the matrix. One can then ask what the relationship between these matrices is, or how one can think of them.

One interpretation is that if $A$ is a map from the vector space $V$ to v.s. $W$, then $A^* : W^* \rightarrow V^*$. Here, $V^*$ denotes the dual space of $V$ (ditto $W$). It can be seen as the map on row vectors ('dual vectors') induced by the map on column vectors.

Note: the reason I call row vectors 'dual vectors' is because one can associate to any row a linear functional on the vector space: write the row as $v^T$ and act on vectors $u$ by $u \rightarrow v^Tu \in \mathbb{R}$. 

If you want I can get into this, depends on how much of this is below/above your level.
A real orthogonal matrix is geometrically a rotation, a reflection or a permutation of the axes.  For example, the matrix M =
$\begin{pmatrix}
cos(t) &amp; -sin(t)\\
sin(t) &amp; cos(t)\\
\end{pmatrix}$ rotates any vector through an angle of t.  $M^{-1}$ undoes what M did, which is to say, it rotates every vector back through the angle (-t) and leaves it where it started.  The matrix 
$\begin{pmatrix}
1 &amp; 0\\
0&amp; -1\\
\end{pmatrix}$
reflects a vector across the x axis and its inverse reflects it back. This matrix can also be expressed in terms of sines and cosines if you wish.

The matrix
$\begin{pmatrix}
0 &amp; 1\\
1&amp; 0\\
\end{pmatrix}$ maps the x-axis into the y-axis and the y-axis into the x-axis.  It inverse does the opposite.  It also can be expressed in terms of sines and cosines.

To be sure this is right, just compute what each matrix does to a sample of vectors; figure out what its inverse is, and see that it undoes what M did to your sample.