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I'm working with the set of trace zero matrices, $\mathfrak{sl}(V)\subseteq\mathfrak{gl}(V)$ of endomorphisms of a vector space $V$.

The problem asks us to represent $ad_x, ad_y, ad_h$ in terms of the basis elements

$x = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$

$y = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$

$h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

I have computed that

$ad_x(y)=h$,

$ad_x(h)=-2x$,

$ad_x(x)=0$,

Similarly,

$ad_y(y)=0$,

$ad_y(x)=-h$,

$ad_y(h)=2y$

finally,

$ad_h(h)=0$,

$ad_h(x)=2x$,

$ad_h(y)=-2y$.

This is my first course in Lie algebras, so I'm kind of stuck. In linear algebra, if I could show how a transformation acted on a basis, it'd be easy to write the result in terms of the basis vectors and write down the linear transformation in a matrix. I computed the adjoints and represented them in terms of the basis matrices, but I'm stuck now. If someone could give me a hint, or show how to do it for one of the adjoints, I'd be okay from there. Thanks

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    Sorry for going off-topic here, but I don't want to give the answer away in the [other thread](http://math.stackexchange.com/q/93382/) (that's why I asked the question about what was known): the set of subsequential limit points of a sequence is closed while $\mathbb{Q} \cap [0,1]$ isn't closed, so the answer to that question is: there is indeed no such sequence.2011-12-22

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I'm a bit embarrassed about this question, but I think the comments helped me realize.

$ad_x(v)$ for $v\in\mathfrak{sl}(V)$ is given by $\begin{pmatrix} 0 & 0 & -2 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}v$ and similarly for $ad_y, ad_z$.

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    +1 Well done! Go ahead and accept this, so that the system can put the question to sleep :-)2011-09-21