It's important to note that this is not a decomposition of V as a representation of SU(2), only as a representation of T. In particular, if you act with a nondiagonal matrix in SU(2) then it will mix up these subspaces V_i.
Weight space decomposition
Weight spaces
We now embark on our study of the fine structure of SU(2) representations, aiming towards the classification theorem for irreducible representations that we stated earlier.
The first key observation is that SU(2) contains a subgroup T isomorphic to U(1): T is the group of diagonal matrices with diagonal entries e to the i theta, e to the minus i theta
If I have a complex representation R from S U 2 to G L V then I can restrict it to T and I get a representation R restricted to T from T to G L V. By the classification of irreps of U(1), we find that V is V_1 direct sum V_2 dot dot dot direct sum V_n where n is the dimension of V and, with respect to this splitting, R of e to the i theta, 0, 0, e to the minus i theta is the diagonal matrix with entries e to the i m_1 theta, dot dot dot, e to the i m_n theta. The m_1 up to m_n are integers called the weights, the V_i are called weight spaces, and the direct sum is called the weight space decomposition.
Example
Take the standard representation R from S U 2 to G L 2 C. We have R of e to the i theta, 0, 0, e to the minus i theta equals e to the i theta, 0, 0, e to the minus i theta. This is already diagonal. The weight spaces are spanned by the vectors (1,0) and (0,1) and the weights are 1 and minus 1 respectively.
We will depict the weight space decomposition as follows (the weight diagram). Consider the integer points sitting on the number line. Colour in each integer m which appears as a weight, and label it by the number of V_i having weight m if this number is strictly bigger than 1. In the example above, the weight diagram is a blob at minus 1 and a blob at 1.
Consider the representation R from S U 2 to G L of little s u 2 tensor C, R of g applied to M_v equals g M_v g inverse. We computed the associated Lie algebra homomorphism R star from little s u 2 to little g l 3 C. We found that R star of i x, y plus i z, minus y plus i z, minus i x equals the 3by3 matrix 0, minus 2 z, 2 y; 2 z, 0, minus 2 x; minus 2 y, 2 x, 0. Since we're interested in the diagonal subgroup T we can set y = z = 0. We're interested in R of exp of i x, 0, 0, minus i x equals exp of R star of i x, 0, 0, minus i x which gives exp of the 3by3 matrix 0, 0, 0; 0, 0, minus 2 x; 0, 2 x, 0, which equals the 3by3 matrix 1, 0, 0; 0, cos 2 x, minus sine 2 x; 0, sine 2 x, cos 2 x The weights are going to be minus 2, 0 and 2. This is because:

We calculated the weights of the representation U 1 to G L 2 C, sending e to the i theta to cos theta, minus sine theta, sine theta, cos theta: these were plus or minus 1 because with respect to a basis of eigenvectors this matrix diagonalised and became e to the i theta, 0, 0, e to the minus i theta.

We're looking at the same submatrix but with 2 theta everywhere instead of theta, so this submatrix can be diagonalised to give e to the i 2 theta, 0, 0, e to the minus i 2 theta.

What's left is the topleft entry which is 1 equals e to the i zero theta, so we also get a weight space with weight 0.
The weight diagram is therefore:
Take Sym 2 C 2 . If e_1 and e_2 form the standard basis for the standard representation then this has basis e_1 squared, e_1 e_2, e_2 squared. Since e to the i theta, 0, 0, e to the minus i theta acts as e_1 goes to e to the i theta e_1 and e_2 goes to e to the minus i theta e_2 in the standard representation, Sym 2 of e to the i theta, 0, 0, e to the minus i theta acts as e_1 squared goes to e to the i 2 theta e_1 squared, e_1 e_2 goes to e_1 e_2, and e_2 squared goes to e to the minus i 2 theta e_2 squared (each factor transforms under the standard representation and then you multiply them together and the factors of e to the i theta or e to the minus i theta either combine or cancel).
The weight spaces are therefore:

V_1 spanned by e_1 squared, with weight 2,

V_2 spanned by e_1 e_2, with weight 0,

V_3 spanned by e_2 squared, with weight 2.
The weight diagram is therefore the same as the weight diagram for the previous example:
We'll see later that this is enough to tell us that these representations are isomorphic.
Preclass exercise
Figure out the weight diagram for Sym n C 2.