Size of Derivation of Lie algebra If $L$ is a Lie algebra in ${\rm gl}\ (n,{\bf C})$ then $$ {\rm Der}\ (L)\subset {\rm gl}\ (n,{\bf C})$$
(If $L$ is semisimple then $ L = {\rm ad}\ L ={\rm Der}\ L$)
Is this true ? Thank you.
 A: No.  Consider the 3-dimensional Lie algebra $\mathfrak b$ spanned by $\{c,h,e\}$ with $z$ central and $[h,e]=2e$.  It embeds into $\mathfrak{gl}(2)$ as the upper-triangular matrices by $c = \bigl( \begin{smallmatrix} 1 & 0 \\ & 1 \end{smallmatrix}\bigr)$, $h = \bigl( \begin{smallmatrix} 1 &0 \\ & -1 \end{smallmatrix}\bigr)$, and $e = \bigl( \begin{smallmatrix} 0 & 1 \\ & 0\end{smallmatrix}\bigr)$.  Its Lie algebra $\operatorname{Der}\mathfrak b$ of derivations is spanned, in the ordered basis $\{c,h,e\}$, by the matrices
$$ X = \begin{pmatrix} 1 && \\ &0&\\&&0 \end{pmatrix}, \quad Y = \begin{pmatrix} 0 && \\ &0&\\&&1\end{pmatrix}, \quad Z = \begin{pmatrix} 0 & 1 & \\ &0&\\&&0 \end{pmatrix}, \quad W = \begin{pmatrix} 0 &  & \\ &0&\\&1&0 \end{pmatrix}$$
In particular, $[X,Y] = [X,W] = [Y,Z] = [Z,W] = 0$.  (And $[X,Z] = Z$ and $[Y,W] = W$.)  Thus $\operatorname{Der}\mathfrak b$ is not isomorphic to $\mathfrak{gl}(2)$ (they have different derived subalgebras), and since they have the same dimension, there cannot be an embedding.
