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)?
Best Answer
No, a matrix $M\in\mathrm{SL}(n,\mathbb{K})$ doesn't have to be diagonalizable; take $M=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$, for instance.
If $G$ is a Lie subgroup of $\mathrm{GL}(n,\mathbb{K})$, its Lie algebra is the set of the $n\times n$ matrices $M$ such that$$(\forall t\in\mathbb{K}):e^{tM}\in G.$$If $\operatorname{tr}M=0$, then$$\det e^{tM}=e^{\operatorname{tr}(tN)}=e^0=1.$$Therefore, $M\in\mathrm{SL}(n,\mathbb{K})$. On the other hand, if $(\forall t\in\mathbb{K}):e^{tM}\in\mathrm{SL}(n,\mathbb{K})$, that is, if $(\forall t\in\mathbb{K}):\det e^{tM}=1$, then, differentiating both sides at $t=0$, you get that $\operatorname{tr}M=0$.