I'm looking for a basis of the vector space formed by the set $$S = \left\{M \in \mathcal{M}_3(\mathbb{K})\ |\ \text{Tr}(M) = 0 \wedge \forall i \in [\![1, 3]\!]: \sum_{j=1}^3 M_{i,j} = 0\right\}$$ together with matrix addition and scalar multiplication, where $\mathbb{K}$ is a commutative field. In practice, what I try to find is a basis that, for any $a,b,c,d \in \mathbb{K}$, produces a matrix of the following types:
$$ \begin{pmatrix}a & b & -a-b \\c & d & -c-d \\a & d & -a-d \end{pmatrix}, \begin{pmatrix}a & b & -a-b \\c & d & -c-d \\d & a & -a-d \end{pmatrix}, \begin{pmatrix}0 & b & -b \\c & 0 & -c \\a & -a & 0 \end{pmatrix} $$
but
I'm not sure these three matrices are a actually prototype for of any possible matrix $M \in S$, and
even if it was the case, the only method I have to look for a base is trial and error, that until now has brought me nothing but generating sets of not linearly independent matrices or sets of linearly independent matrices that do not generate every $M\in S$.
Is there any methodical way to look for a basis of a vector space whose elements are matrices?
Note that\begin{align}S&=\left\{\begin{bmatrix}a&b&-a-b\\c&d&-c-d\\e&f&-e-f\end{bmatrix}\,\middle|\,a,b,c,d,e,f\in\mathbb K\wedge a+d-e-f=0\right\}\\&=\left\{\begin{bmatrix}a&b&-a-b\\c&d&-c-d\\e&a+d-e&-a-d\end{bmatrix}\,\middle|\,a,b,c,d,e\in\mathbb K\right\}.\end{align}So, if$$M(a,b,c,d,e)=\begin{bmatrix}a&b&-a-b\\c&d&-c-d\\e&a+d-e&-a-d\end{bmatrix},$$a basis of this space is\begin{multline}\bigl\{M(1,0,0,0,0),M(0,1,0,0,0),M(0,0,1,0,0),M(0,0,0,1,0),M(0,0,0,0,1)\bigr\}=\\=\left\{\begin{bmatrix}1&0&-1\\0&0&0\\0&1&-1\end{bmatrix},\begin{bmatrix}0&1&-1\\0&0&0\\0&0&0\end{bmatrix},\begin{bmatrix}0&0&0\\1&0&-1\\0&0&0\end{bmatrix},\begin{bmatrix}0&0&0\\0&1&-1\\0&1&-1\end{bmatrix},\begin{bmatrix}0&0&0\\0&0&0\\1&-1&0\end{bmatrix}\right\}.\end{multline}