Consider the homomorphism $\phi : \mathfrak{sl}_2(\mathbb C) \rightarrow \mathfrak{sl}_3(\mathbb C)$ sending $\left(\begin{matrix} a&b \\ c&d \end{matrix}\right)$ to $\left(\begin{matrix} a&b&0 \\ c&d&0 \\ 0&0&0 \end{matrix}\right)$
Consider additionally the adjoint representation $ad : \mathfrak{sl}_3(\mathbb C) \rightarrow \mathfrak{gl}(\mathfrak{sl}_3(\mathbb C))$ of $\mathfrak{sl}_3(\mathbb C)$.
Then the map $ad \circ \phi : \mathfrak{sl}_2(\mathbb C) \rightarrow \mathfrak{gl}(\mathfrak{sl}_3(\mathbb C))$ is a representation of $\mathfrak{sl}_2(\mathbb C)$. I am trying to show that it is completely reducible, and here is what I thought of:
Notice that the action of $h = \left(\begin{matrix} 1&0 \\ 0&-1 \end{matrix}\right)$ determines the representation, and that:
$h \cdot \left(\begin{matrix} a&b&c \\ d&e&f \\ g&h&i \end{matrix}\right) = \left(\begin{matrix} 0&2b&c \\ -2d&0&-f \\ -g&h&0 \end{matrix}\right)$
$\Rightarrow X = \left(\begin{matrix} 0&1&0 \\ 0&0&0 \\ 0&0&0 \end{matrix}\right) \in (\mathfrak{sl}_3(\mathbb C))_2 = \{A \in \mathfrak{sl}_3(\mathbb C) \mid h \cdot A = 2A\}$
Similarly $Y = \left(\begin{matrix} 0&0&1 \\ 0&0&0 \\ 0&0&0 \end{matrix}\right) \in (\mathfrak{sl}_3(\mathbb C))_1$
Notice however that $X, Y$ are also in the kernel of the action of $e = \left(\begin{matrix} 0&1\\ 0&0 \end{matrix}\right)$ and so they are highest-weight vectors of weights $2$ and $1$ respectively.
I think this then tells us that in the decomposition of $\mathfrak{sl}_3(\mathbb C)$ we will see $V(1)$ and $V(2)$ where $V(n)$ is the irreducible representation of $\mathfrak{sl}_2(\mathbb C)$ of degree $n+1$ over $\mathbb C$.
However, this only accounts for $5$ of the $8$ dimensions that $\mathfrak{sl}_3(\mathbb C)$ has as a vector space of $\mathbb C$.
I wanted to ask how I could find the remaining subrepresentations, as well as if everything I've done so far is correct. Thank you!
As usual, write $\mathfrak{sl}_2(\mathbb{C})=\text{span}_\mathbb{C}\{h,e,f\}$, where $$h:=\begin{bmatrix}+1&0\\0&-1\end{bmatrix}\,,\,\,e:=\begin{bmatrix}0&1\\0&0\end{bmatrix}\,,\,\,f:=\begin{bmatrix}0&0\\1&0\end{bmatrix}\,.$$Let $M$ denote the $\mathfrak{sl}_2(\mathbb{C})$-module $\mathfrak{sl}_3(\mathbb{C})$ given by the representation $\text{ad}_{\mathfrak{sl}_3(\mathbb{C})}\circ\phi$. Extend this representation to $\mathfrak{gl}_3(\mathbb{C})$, and call this $\mathfrak{sl}_2(\mathbb{C})$-module $N$. Note that $$h\cdot\begin{bmatrix}p&q&r\\s&t&u\\x&y&z\end{bmatrix}=\begin{bmatrix}0&+2q&+r\\-2s&0&-u\\-x&+y&0\end{bmatrix}\,,$$ $$e\cdot\begin{bmatrix}p&q&r\\s&t&u\\x&y&z\end{bmatrix}=\begin{bmatrix}+s&t-p&u\\0&-s&0\\0&-x&0\end{bmatrix}\,,$$ and $$f\cdot\begin{bmatrix}p&q&r\\s&t&u\\x&y&z\end{bmatrix}=\begin{bmatrix}-q&0&0\\p-t&+q&r\\-y&0&0\end{bmatrix}\,,$$ for all $p,q,r,s,t,u,x,y,z\in\mathbb{C}$.
For $i,j=1,2,3$, let $E_{i,j}\in\mathfrak{gl}_3(\mathbb{C})=N$ denote the matrix with $1$ at the $(i,j)$-entry and $0$ everywhere else. Note that $$N=M\oplus\text{span}_\mathbb{C}\left\{E_{1,1}+E_{2,2}+E_{3,3}\right\}\,.$$ Observe that $$h\cdot E_{i,i}=0\text{ for }i=1,2,3\,,$$ $$e\cdot\left(E_{1,1}+E_{2,2}\right)=0=f\cdot\left(E_{1,1}+E_{2,2}\right)$$ and $$e\cdot E_{3,3}=0=f\cdot E_{3,3}\,.$$ Therefore, $$M_1:=\text{span}_\mathbb{C}\left\{E_{1,1}+E_{2,2}\right\}\text{ and }M_2:=\text{span}_\mathbb{C}\left\{E_{3,3}\right\}$$ are $1$-dimensional simple submodules of $N$ isomorphic to the trivial module $V(0)$.
The $2$-dimensional simple module $M_3\cong V(1)$ of $M$ that the OP describes is given by $$M_3:=\text{span}_{\mathbb{C}}\left\{E_{1,3},E_{2,3}\right\}\,.$$ There is another $2$-dimensional simple module $M_4\cong V(1)$ of $M$. It is given by $$M_4:=\text{span}_{\mathbb{C}}\left\{E_{3,1},E_{3,2}\right\}\,.$$
The $3$-dimensional simple module $M_5\cong V(2)$ of $M$ that the OP describes is given by $$M_5:=\text{span}_{\mathbb{C}}\left\{E_{1,2},E_{1,1}-E_{2,2},E_{2,1}\right\}\,.$$ Clearly, $$N=M_1\oplus M_2\oplus M_3\oplus M_4\oplus M_5\cong\big(V(0)\big)^{\oplus 2}\oplus \big(V(1)\big)^{\oplus2 }\oplus V(2)$$ as an $\mathfrak{sl}_2(\mathbb{C})$-module. We now have to restrict to $M$ by removing one $1$-dimensional submodule. We have $$M=M_{1,2}\oplus M_3\oplus M_4\oplus M_5\cong V(0)\oplus \big(V(1)\big)^{\oplus2 }\oplus V(2)\,,$$ where $$M_{1,2}:=\text{span}_{\mathbb{C}}\left\{E_{1,1}+E_{2,2}-2\,E_{3,3}\right\}\,.$$
Edit: I forgot that the matrices of $\mathfrak{sl}_3(\mathbb{C})$ are traceless. Therefore, there is only one $1$-dimensional simple submodule of $M$.