Calculate percentage of symmetry of a given matrix Is it possible to calculate a percentage that quantifies how symmetric a given matrix is?
For example, even if a given matrix is not symmetric, instead of just classifying it as "not symmetric", one can say that it's "$80\%$ symmetric". 
My overall goal is to get a kind of a "symmetry score" adding up values of different kinds of symmetry (e.g., translational, rotational). I appreciate any help. Thanks!
 A: This reference could be what you are looking for:
Digraph Laplacian and the Degree of Asymmetry:

We introduce a metric – the largest singular value $\delta$ of $(\Gamma − \Gamma^T )/2$, where $\Gamma$ is the Laplacian of a directed graph – to quantify
  and measure the degree of asymmetry in the graph. The degree of asymmetry
  captures the overall "directedness" of the graph.

A: Consider the following $2 \times 2$ matrix and its decomposition in a "natural" orthonormal basis.
$$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix} + \begin{bmatrix} 0 & 0\\ 0 & 1\end{bmatrix} + \frac{1}{\sqrt 2} \left( \frac{1}{\sqrt 2} \begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix} \right) + \frac{1}{\sqrt 2} \left( \frac{1}{\sqrt 2} \begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix} \right)$$
Let us use the squared Frobenius norm to measure energy. The total energy of the matrix is the sum of the squared coefficients
$$1^2 + 1^2 + \left(\frac{1}{\sqrt 2}\right)^2 + \left(\frac{1}{\sqrt 2}\right)^2 = 3$$
while the energy of the skew-symmetric part is $\left(1/\sqrt 2\right)^2 = 0.5$. Hence, the fraction of total energy that is not skew-symmetric is
$$1 - \frac{0.5}{3} = \frac 56 \approx 83.3 \%$$
However, neglecting the entries on the main diagonal, the fraction of the total off-diagonal energy that is symmetric is (only) $50\%$.
More generally, given a matrix $\rm A$, the fraction of total energy that is not skew-symmetric is
$$1 - \frac{\left\| \frac{\rm A - A^\top}{2} \right\|_{\text{F}}^2}{\| \rm A \|_{\text{F}}^2} = 1 -  \left( \frac 12 \cdot\frac{\| \rm A - A^\top \|_{\text{F}}}{\| \rm A \|_{\text{F}}} \right)^2$$
