# Express $ad_x$ in terms of the basis elements

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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You are exactly down to the linear algebra problem you mentioned. You have calculated how the linear transformation $ad_x$ acts on the basis $\{x,y,h\}$, so you can write the matrix for $ad_x$ in terms of that basis. Similarly for $ad_y$ and $ad_h$. (At least this is how I'm interpreting the problem when you say "represent $ad_x$, $ad_y$, $ad_h$ in terms of the basis elements $x,y,z$", since I can't make sense of it any other way.) –  Ted Sep 21 '11 at 5:00
Given that you A) have successfully computed how the three mappings act on the elements of a basis, and B) know how to represent a linear mapping as a matrix with respect to a given basis, it is a bit hard to imagine what the remaining problem is? For example, just call $x$ the first basis element, $y$ the second and $h$ the third, and write down the damn 3x3 matrices already :-) –  Jyrki Lahtonen Sep 21 '11 at 5:01
Sorry for going off-topic here, but I don't want to give the answer away in the other thread (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. –  t.b. Dec 22 '11 at 6:48
@t.b. thank you :) –  mathmath8128 Dec 22 '11 at 6:57

$ad_x(v)$ 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 $ad_y, ad_z$.