Can I decompose an arbitary matrix as the sum of antisymmetric, identity, and outer product parts? I'm trying to see if I could represent any arbitrary $3\times 3$ matrix by the following matrix:
$$\begin{bmatrix}0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{bmatrix}+\begin{bmatrix}e \\ f \\ g\end{bmatrix}\begin{bmatrix}e & f & g\end{bmatrix} + \begin{bmatrix}d & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & d\end{bmatrix} = \begin{bmatrix}d+e^2 & ef-c & eg+b \\ ef+c & d+f^2 & fg-a \\ eg-b & fg+a & d+g^2\end{bmatrix}$$
where $a,b,c,d,e,f,g \in \Bbb R$.
So I guess I need to see if the three columns span $\Bbb R^3$.  All those variables, though, are looking pretty intimidating.  Is there an easy way to check?
 A: The answer is no. You can not make such as decomposition.
Here the explanation:


*

*You can decompose every matrix $M$ into a symmetric and antisymmetric part by $$ M = \frac{M + M^T}{2} + \frac{M - M^T}{2}.$$ Note, that this 
decomposition is unique, i.e. if you have $M = S + A$ with $S$ symmetric and $A$ antisymmetric, then $$ S =  \frac{M + M^T}{2} \text{ and } A = \frac{M - M^T}{2}.$$

*Hence your question boils down to the question if any symmetric $3 \times 3$ matrix can be written as $$\begin{pmatrix} e^2 + d & ef & eg \\ ef & f^2 + d & fg \\ eg & fg & g^2 + d \end{pmatrix},$$ which is the symmetric part of your proposed decomposition.

*An easy counter example for this decomposition is $$ M = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}. $$ From $M_{1,3} = 0$ it follows that $e = 0$ or $g = 0$, but $e = 0$ is a contradiction to $M_{1,2} = 1$ and $g = 0$ is a contradiction to $M_{2,3} = 1$.

A: You can decompose any matrix into a symmetric and antisymmetric part (just by addition and subtraction), and you can further decompose the symmetric part into a traceless part and a traceful part that is proportional to the identity.
You can think of the antisymmetric part as a generalized curl of a linear function represented by using the matrix in multiplication.  The trace is a generalized divergence.
The symmetric part in particular can be written as a symmetric sum of outer products, but since the trace is subtracted off into its own piece, I'm not sure if the traceless symmetric part has such an interpretation.
