# Determine matrix associated with lie algebra representation

Consider $$A_1$$ with basis $$\{e,f,h\}$$ and let $$\phi$$ be a three-dimension representation of $$A_1$$. Let $$\phi(h)=\begin{pmatrix} 0 & 2 & 0 \\ 0 & -2 & 0 \\ 2 & -2 & 2 \end{pmatrix}$$ and $$\phi(f)=\begin{pmatrix} 0 & -1 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & -1 \end{pmatrix}$$. I want to determine $$\phi(e)$$.

My first guess is to use that $$\phi$$ is a representation of a Lie algebra so

$$\phi(h)=\phi([e,f])=[\phi(e),\phi(f)]=\phi(e)\phi(f)-\phi(f)\phi(e)$$, so one get 9 equations determining $$\phi(e)$$. However there is not a unique solutions to these equation, so I get in doubt, if this is the correct way to do it. What is the approach here?

You can first find the weight spaces, i.e. the eigenspaces of $$\phi(h)$$. We have $$v_2 = (0, 0, 1), \quad v_0 = (1, 0, -1), \quad v_{-2} = (-1, 1, 1)$$ forming an eigenbasis, where $$\phi(h) v_2 = 2 v_2$$, and $$\phi(h) v_0 = 0 v_0$$, and $$\phi(h) v_{-2} = -2 v_{-2}$$. We can summarise this by saying that $$\phi(h) v_\lambda = \lambda v_\lambda$$.
The operator $$\phi(f)$$ must decrease weights $$2$$ each time, meaning that $$\phi(f) v_{\lambda}$$ must be a multiple of $$v_{\lambda + 2}$$. Indeed, we have $$\phi(f) v_2 = v_0$$ and $$\phi(f) v_0 = v_{-2}$$ and $$\phi(f) v_{-2} = 0$$.
The operator $$\phi(e)$$ must increase weights by $$2$$ each time, so we already know that $$\phi(e) v_2 = 0$$, since $$2$$ is the largest weight in this representation. We will also have $$\phi(e) v_0 = a v_2, \quad \phi(e) v_{-2} = b v_0,$$ for some complex numbers $$a$$ and $$b$$ which we need to determine. You can use the bracket to determine these, since we will have reduced that large matrix equation down to just a simple equation of numbers. For example, we have $$-2 v_{-2} = \phi(h) v_{-2} = \phi(e) \phi(f) v_{-2} - \phi(f) \phi(e) v_{-2} = 0 - b v_{-2}$$ and hence $$b = 2$$.