There is a pretty standard exercise on $\mathfrak{sl}_2 (\mathbb{C}$) representations that consists in decomposing the representation given by $\mathfrak{sl}_3(\mathbb{C})$ via $\operatorname{ad}$, where $\mathfrak{sl}_2(\mathbb{C})$ is identified with $e= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 &0 \end{pmatrix}, \quad h= \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 &0 \end{pmatrix}, \quad f= \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 &0 \end{pmatrix}$
so $e.v = [e,v]$ would be the action that defines the representation. This is, for example, exercise 7.3 on Humphreys's book.
Now, I did all the math and found out that, after choosing $\{ e, f, h, E_{13}, E_{31}, E_{23}, E_{32}, E_{22}-E_{33}\}$ as a basis of $\mathfrak{sl}_3(\mathbb{C})$, you have three subrepresentations (that is, subspaces that are $\operatorname{ad}e$, $\operatorname{ad}f$, $\operatorname{ad}h$ stable).
$V_2 = \langle e, f, h \rangle$
$V_1 = \langle E_{13}, E_{23}\rangle$
$V_1'= \langle E_{32}, -E_{31}\rangle$
Where the matrices of the three $\operatorname{ad}$s are exactly the ones you expect for irreducible subrepresentations.
Now the thing is I'm almost done, and only the matrix $D = E_{22} - E_{33}$ is left outside my decomposition. The thing is... the subspace generated by this matrix is not $\operatorname{ad}$-stable! In fact
$$\operatorname{ad}e (D) = e \quad \operatorname{ad}h (D)=0 \quad \operatorname{ad}f(D) = -f $$
So $\mathbb{C}D$ is not a subrepresentation, but it's the only thing left... and I'm not able to finish my decomposition. Now since there is a theorem stating this decomposition has to exist, and since I even found solutions online that simply say this $\mathbb{C}D$ is $V_0$, and $$\mathfrak{sl}_3(\mathbb{C}) \simeq V_0 \oplus V_1 \oplus V_1' \oplus V_2$$ what am I missing? What's the structure of that $\oplus$, if $V_0$ isn't a subrepresentation?
Did I get the $\mathfrak{sl}_2(\mathbb{C})$ representations decomposition theorem wrong? Or what?
The line that's missing is not necessarily spanned by $D$, but by some element in $D + V_1 + V_1^{\prime} + V_2$, so you have to look for some ${\mathfrak sl}_2$-invariant element in there: $D + \frac{1}{2}h$ works.