Could someone please tell me what I am missing in the following argument. Either my understanding of the exact statement of Ado's theorem is wrong, or there is a flaw in my argument below. 


For a finite dimensional Lie algebra ${\bf g}$ consider the adjoint representation $\phi:{\bf g}\rightarrow {\bf gl}\left({\bf g}\right); \phi\left(X\right)={\bf ad}\_X$. We have $\ker\left(\phi\right)={\cal z}\left({\bf g}\right)$ the centre of ${\bf g}$, so we can think of ${\bf g}$ as the direct sum $\left({\bf g} / {\cal z}\left({\bf g}\right)\right) \bigoplus {\cal z}\left({\bf g}\right)$ of the matrix Lie algebra of little ad matrices and the centre ${\cal z}\left({\bf g}\right)$. But then, suppose the centre is m-dimensional; since Abelian it can be represented as the algebra of $m\times m$ diagonal matrices. So the whole Lie algebra can be realised as the algebra of block-diagonal matrices of the form:


$\left(\begin{array}{cc}{\bf ad}\_X & {\bf 0} \\\\ {\bf 0} & {\bf \Lambda}\end{array}\right)$


where ${\bf \Lambda}$ is an $m\times m$ diagonal matrix.



Ado's Theorem Proof
Your argument fails because the bracket can (sometimes) take pairs of elements into the centre. Therefore the direct sum as vector spaces isn't necessarily a direct sum of Lie algebras. For nilpotent Lie algebras, such as the Heisenberg algebra, you can't make the claim about block-diagonal form.


