Functionally, what does a symmetric matrix say about the linear transformation it represents? I understand the definition of a symmetric matrix in terms of how it’s components are related. But functionally, what does it entail about the linear transformation it represents?
For example, block tri-diagonal matrices have special relations between entries but they also, functionally, tell us that some non-trivial vector subspace is invariant under the linear transformation with respect to a particular basis.
Incidentally, what do skew-symmetric matrices represent, functionally ?
 A: In the comments (and in the linked discussion) on the question, I make the following claim:

$M$ is symmetric relative to at least one choice of (possibly oblique) basis if and only if $M$ is diagonalizable with real eigenvalues. $M$ is skew-symmetric relative to at least one choice of basis if and only if $M$ is a direct sum of scaled $90^\circ $ rotations and zero transformations.

First, the symmetric case. If $M$ is symmetric, then the spectral theorem states that $M$ is diagonalizable with real eigenvalues. Conversely, if $M$ is diagonalizable with real eigenvalues, then there is a basis relative to which the matrix of $M$ is diagonal with real diagonal entries. Since this diagonal matrix is symmetric, $M$ is symmetric relative to this choice of basis.
For case where $M$ is skew-symmetric, there are two common approaches. For the easy direction: if $M$ is a direct sum of $90^\circ$ rotations and zero transformations, then there is a basis relative to which the matrix of $M$ is the block-diagonal skew-symmetric matrix
$$
\pmatrix{0 & \kappa_1 \\-\kappa_1 & 0 \\ && \ddots \\ &&& 0 & \kappa_p\\ &&&-\kappa_p & 0
\\ &&&&&0 \\ &&&&&&\ddots\\ &&&&&&& 0}. 
$$
There are two approaches for the converse. One is essentially to apply the spectral theorem for Hermitian matrices, noting that if $M$ is skew-symmetric then the complex matrix $iM$ is Hermitian. Alternatively, we can systematically construct a basis relative to which the matrix of $M$ has the above block-diagonal form as is outlined in this post and the proof linked therein.
