An arbitrary matrix can be decomposed into its symmetric and anti-symmetric components.
But the diagonal components are hidden in the symmetric components.
I want to eject the diagonal components from the symmetric components.
How can I decompose that symmetric matrix into the sum of its diagonal components and another matrix.
On
Let $M\in \mathbb R^{n\times n}$ be the generic matrix; first you have the decomposition in symmetric and skew-symmetric parts: $$ M=A+S,\quad A:=\frac{1}{2}(M-M^\top),\quad S:=\frac{1}{2}(M+M^\top) $$ Now you can decompose $S$ in a spherical part (i.e. a multiple of the identity) and a deviatoric part, i.e. a symmetric matrix with vanishing trace: $$ S=\Sigma + \overset{\circ}{S}, \quad \Sigma := \frac{1}{n}\mathrm{Tr}(S)\cdot \mathbf I, \quad \overset{\circ}{S}:=S-\frac{1}{n}\mathrm{Tr}(S)\cdot \mathbf I $$ This is not what you asked, because in general $\overset{\circ}{S}$ is only such that $$ \mathrm{Tr}(\overset{\circ}{S})=0, $$ but diagonal elements can be non vanishing. The point is that this decomposition is invariant under ortogonal transformations (orthogonal changes of bases): in a different orthogonal base, the transformed $R^\top \overset{\circ}{S} R $ is again symmetric with vanishing trace. Of course you can go further and write $ \overset{\circ}{S}$ as the sum of a diagonal matrix $D$ and a symmetric matrix with vanishing diagonal $S_0$. However, this decomposition is no longer invariant under orthogonal change of bases: the transformed $ R^\top S_0 R$ in general has not vanishing diagonal.
PS. I see that this was already pointed out in the answer of Rugh...
Diagonal elements depend upon the chosen basis. The symmetric/anti-symmetric decomposition only depends upon the scalar product (thus independent of the chosen orthonormal basis). Anyway, if you really want to do the extraction, let $e_i$ denote the canonical basis. Then $$ \Delta = \sum_i e_i e_i^T A e_i e_i^T $$ does the job.