Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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












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

share|cite|improve this question
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
up vote 1 down vote accepted

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 $ad_y, ad_z$.

share|cite|improve this answer
+1 Well done! Go ahead and accept this, so that the system can put the question to sleep :-) – Jyrki Lahtonen Sep 21 '11 at 6:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.