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

$$1 - \frac{\left\| \frac{\rm A - A^\top}{2} \right\|_{\text{F}}^2}{\| \rm A \|_{\text{F}}^2} = 1 -  \left( \frac 12 \cdot\frac{\| \rm A - A^\top \|_{\text{F}}}{\| \rm A \|_{\text{F}}} \right)^2$$