If $M \in SL(n, K)$,then $M$ is a $n\times n$ matrix with entries in $K$ such that $det\ M = +1$. To get the algebra, $\mathfrak sl(n, K)$ I expand as follows (keeping always only first order in $x^a$):
$$det\ M = 1 = det\ (1 + x^a·T_a +\ ...) \simeq det\ (1 + x^a·T_a) \tag1$$
Where $x^a$ are the parameters of the group and $T_a$ the algebra's elements.
Now, what I do is to suppose that $M$ is diagonalizable, so
$$det\ (1 + x^a·T_a) = det\ (1 + x^a·T_a^{diag}) \simeq 1 + x^a·tr[T_a^{diag}] \tag2$$
From here, $\mathfrak sl(n, K)$ is the set of $n\times n$ matrices with entries in $K$ and traceless.
But here is my question: is the assumption of $M$ diagonalizable always true? If not, how can I get the traceless condition of matrices in $\mathfrak sl(n, K)$ that I'm sure it is true (due to literature)?