How do I show that a tensor product representation of $L(SU(2))\equiv su(2)$ is reducible? So I have been reading about the irreducible representations of the Lie algebra $L(SU(2))$ and came across the Cartan-Weyl basis:
$$ H = \sigma_3 $$
$$ E_+ = \frac{1}{2}(\sigma_1+i \sigma_2) $$
$$ E_- = \frac{1}{2}(\sigma_1-i\sigma_2) $$
where $\sigma_1,\sigma_2,\sigma_3 $ are the Pauli matrices. 

My understanding so far
If I choose to represent $L(SU(2))$ using the Cartan-Weyl basis with a representation $R$, I find that the eigenvectors of $R(H)$ form an $n$ dimensional basis of my representation space $V$, and the $R(E_{\pm})$ transition between these eigenvectors. There are no invariant subspaces of my representation space because each eigenvector can be transformed to another via repeated action of $R(E_{\pm})$ and therefore I have an $n$-dimensional irreducible representation of $L(SU(2))$. The eigenvectors $\{ v_\lambda \} $ have weights (eigenvalues) $\lambda \in \{ -\Lambda, -\Lambda + 2, ... , \Lambda \} $,  where the dimension of $V$ is $n = \Lambda + 1 $.

My confusion with tensor product representations
If I took the tensor product of two different representations $R_\Lambda$ and $R_{\Lambda '}$ of the Lie algebra with highest weights $\Lambda$ and $\Lambda'$ respectively, I will get a represnetation space $V_\Lambda \otimes V_{\Lambda'}$ which is spanned by $ \{ v_\lambda \otimes v'_{\lambda'} \} $. The representatives of $L(SU(2))$ are given by
$$ R_{\Lambda \otimes \Lambda'}(X) = R_{\Lambda}(X) \otimes I_{\Lambda'}+ I_\Lambda \otimes R_{\Lambda '}(X).$$
for $X \in L(SU(2)) $. Therefore, we find that $ \{ v_\lambda \otimes v'_{\lambda'} \} $ are eigenvectors of $ R_{\Lambda \otimes \Lambda'}(H)$ with eigenvalues $\lambda + \lambda' $. This is all fine, but at this point I have read that you can decompose our representation into irreps. I cannot wrap my head around this. My understanding is that a representation is reducible if all the representation matrices can be written in a block diagonal form, or equivalently that the representation space has invariant subspaces w.r.t. the representation elements $R(X)$. How can I show this?
I have seen that the highest weight of $ R_{\Lambda \otimes \Lambda'}$ is $\Lambda + \Lambda ' $, and it has multiplicity 1, and we can write:
$$ R_{\Lambda \otimes \Lambda'} = R_{\Lambda + \Lambda'} \oplus \widetilde{R}_{\Lambda,\Lambda' } $$
for some remainder $\widetilde{R}_{\Lambda,\Lambda' }$. This is the part I do not understand at all. Why can I just do that? If I have one eigenvector $ v_\Lambda \otimes v'_{\Lambda '}$ with weight $\Lambda + \Lambda'$, why can I just split the representation space up and say we have a representation $R_{\Lambda + \Lambda'}$ on an invariant subspace? Do I need to show that there are multiple invariant subspace w.r.t. the ladder operators $R(E_\pm)$?

I think this question is suitable for Maths stack exchange however I am a physicist studying a particle physics symmetry module so I have asked it here as I sometimes struggle to understand rigorous mathematical proofs.
 A: Let's work this out explicitly for the case of adding two spins of $1/2$. In your notation, the weights we’re adding are $\{-1, 1\}$ and $\{-1, 1\}$. Upon taking the tensor product, we get a four-dimensional vector space with a vector of weight $2$, a vector of weight $-2$, and a two-dimensional plane of vectors with weight $0$.
You want to know how we can generate an irreducible subspace by starting with the vector of weight $2$. The irreducible subspace containing a vector is the set of vectors we can reach from that vector by applying raising and lowering operators. So we start with the vector of weight $2$, lower to get a vector of weight $0$, and lower again to get a vector of weight $-2$. Further lowering can't get us anything, because there's nothing of weight $-4$. 
You might wonder why we can't get the other vector of weight $0$. Well, we already know all the irreducible representations of $\mathfrak{su}(2)$, and all of them have nondegenerate weights, so $0$ can't show up twice. But more explicitly, the only way to get a different $0$ is to start at $-2$ and apply the raising operator, and you can use the commutation relations to show you end up with the exact same $0$ weight vector as before.
Thus three vectors with weights $\{-2, 0, 2\}$ form an irreducible subrepresentation. We can then forget about them and proceed to the remaining weight $0$ which also forms an irreducible subrepresentation; the reasoning is similar for arbitrary spins. 
A: I) What OP calls weights 
$$\lambda ~\in~ \{ -\Lambda, -\Lambda + 2, \ldots ,\Lambda-2, \Lambda \},\qquad  \Lambda~\in~\mathbb{N}_0,\tag{1} $$ 
of an irreducible $sl(2,\mathbb{C})$ representation (irreps) are twice as big as physicists' traditional half-integer notation 
$$m ~\in~ \{ -j, -j + 1, \ldots ,j-1, j \},\qquad  j~\in~\frac{1}{2}\mathbb{N}_0. \tag{2}$$
A tensor product of $sl(2,\mathbb{C})$ irreps decomposes into a direct sum of irreps via the Clebsch-Gordan fusion rule
$$\underline{2j_1+1} \otimes \underline{2j_2+1} 
~=~ \oplus_{j=|j_1-j_2|}^{j_1+j_2} \underline{2j+1}.\tag{3}$$
Here $\underline{1}$, $\underline{2}$, $\underline{3}$, $\ldots$, refer to the singlet irrep, doublet irrep, triplet irrep, $\ldots$, respectively.  It is a nice exercise to check that the dimensions on the right- and left- hand sides of (3) match. See also this & this Phys.SE questions.
