# What is the geometric interpretation of the transpose?

I can follow the definition of the transpose algebraically, i.e. as a reflection of a matrix across its diagonal, or in terms of dual spaces, but I lack any sort of geometric understanding of the transpose, or even symmetric matrices.

For example, if I have a linear transformation, say on the plane, my intuition is to visualize it as some linear distortion of the plane via scaling and rotation. I do not know how this distortion compares to the distortion that results from applying the transpose, or what one can say if the linear transformation is symmetric. Geometrically, why might we expect orthogonal matrices to be combinations of rotations and reflections?

To answer your second question first: an orthogonal matrix $O$ satisfies $O^TO=I$, so $\det(O^TO)=(\det O)^2=1$, and hence $\det O = \pm 1$. The determinant of a matrix tells you by what factor the (signed) volume of a parallelipiped is multipled when you apply the matrix to its edges; therefore hitting a volume in $\mathbb{R}^n$ with an orthogonal matrix either leaves the volume unchanged (so it is a rotation) or multiplies it by $-1$ (so it is a reflection).
To answer your first question: the action of a matrix $A$ can be neatly expressed via its singular value decomposition, $A=U\Lambda V^T$, where $U$, $V$ are orthogonal matrices and $\Lambda$ is a matrix with non-negative values along the diagonal (nb. this makes sense even if $A$ is not square!) The values on the diagonal of $\Lambda$ are called the singular values of $A$, and if $A$ is square and symmetric they will be the absolute values of the eigenvalues.
The way to think about this is that the action of $A$ is first to rotate/reflect to a new basis, then scale along the directions of your new (intermediate) basis, before a final rotation/reflection.
With this in mind, notice that $A^T=V\Lambda^T U^T$, so the action of $A^T$ is to perform the inverse of the final rotation, then scale the new shape along the canonical unit directions, and then apply the inverse of the original rotation.
Furthermore, when $A$ is symmetric, $A=A^T\implies V\Lambda^T U^T = U\Lambda V^T \implies U = V$, therefore the action of a symmetric matrix can be regarded as a rotation to a new basis, then scaling in this new basis, and finally rotating back to the first basis.