Assume that we construct a representation for a group $G$ which is reducible. Then to block diagonlize it (decompose it to irreducible one), we first calculate the frequency of each irreps using (here the representation with superscript means the irreducible representaion)
$$ n_\alpha = \frac{1}{|G|}\sum_g \chi^{\alpha}(g)^* \chi(g) \\ D(g) = \sum_{\alpha} n_{\alpha}D^{\alpha}(g). $$
Then we use the projection operator to get the basis for each irreps.
$$ P_{ij}^{\alpha} = \frac{1}{|G|}\sum_g D_{ij}^{\alpha}(g)^* \hat{g} $$
Using this projection operator for fixed $j$, if the $P_{ij}^{\alpha} f \neq 0$ for an initial basis function $f$, we can generate the set of basis functions by varying $i$ for irreps $D^{\alpha}$.
My question is that "Is is possible to generate all the basis function using this projection oeprator?"
To be precise, let's assume that we first constructed our representaion $D(g)$ with dimension $20$. This means that we start with 20 basis functions. Now assume we got $n_{\alpha}=3$ and $dim(D^{\alpha}(g))=2$
Since the dimension of $D^{\alpha}$ is two, we have two projection operator onto this representation, $P_{i1}^{\alpha}, P_{i2}^{\alpha} $. Also, since we start from 20 basis functions, the number of possible set of projected functions we could try is
$$ \{ P_{i1}^{\alpha}, P_{i2}^{\alpha} \} \times \{f_1, f_2, ... , f_{20} \} \rightarrow 2 \times 20 = 40 $$
Thus, we can construct $40$ set of functions using projection. But they might be linearly dependent or unitarily equivalent or could be zero. Since we know that this representation contains $n_{\alpha}=3$ irreps $D^{\alpha}$, we need three set of independet basis. Does our $40$ functions got from the projection operaotr always contain these $3$ basis? I am not sure that this is always guaranteed.
The way the projection operator can be used to decompose a group rep into irreps when there are copies of certain irreps is as follows.
Let the complete vector space be spanned by basis vectors $|\alpha,i,u\rangle$ where $\alpha$ labels the inequivalent irreps, $i=1,2,\ldots,m_{\alpha}$ labels the states within irrep $\alpha$ and $u=1,2,\ldots,n_{\alpha}$ labels the copies of irrep $\alpha$. Of course, we don't know these basis vectors. All we have are the block diagonal group matrices $\langle \alpha,i,u|D(g)|\beta,j,v\rangle=[D^{\alpha}(g)]^{i}_{\ j}\delta^{\alpha}_{\ \beta}\delta^{u}_{\ v}$ which are used to construct the projection operator. Using the block-diagonal formula for the group matrices, the projection operator can be written in terms of the basis vectors and dual vectors as, \begin{equation} P^{\alpha}_{ij}=\sum_{u}|\alpha,i,u\rangle\langle\alpha,j,u| \end{equation} This form of the projection operator makes it easier (for me) to see what is going on in the decomposition into irreps when there are copies.
Pick $j$ and form the projection operator $P^{\alpha}_{jj}$ with no summation on repeated $j$. Pick random vectors $|\psi_{u}\rangle$ for $u=1,2,\ldots,n_{\alpha}$ and project $P^{\alpha}_{jj}|\psi_{u}\rangle$. These vectors are all in the subspace spanned by vectors $|\alpha,j,v\rangle$ for fixed $\alpha,j$ and $v=1,2,\ldots,n_{\alpha}$. They are used as our known basis for the $\alpha,j$ subspace. Let's use the notation $|\alpha,j;u\rangle=P^{\alpha}_{jj}|\psi_{u}\rangle$ for these basis vectors. They are just linear combinations of the $|\alpha,j,v>$ for fixed $\alpha,j$ and variable $v$. Each of these vectors are in only one irrep $\alpha$. If we have two vectors $|\alpha,j;u\rangle$ and $|\alpha,j;v\rangle$ for $u\neq v$ they are in different copies of irrep $\alpha$.
Now project $P^{\alpha}_{ij}|\alpha,j,u\rangle$. For fixed $\alpha,u$ and variable $i=1,2,\ldots,m_{\alpha}$, these vectors span the copy $u$ of irrep $\alpha$. Let's call these vectors $|\alpha,i;u\rangle=P^{\alpha}_{ij}|\alpha,j,u\rangle$. We've now got basis vectors for all the copies of irrep $\alpha$. The group matrices are block-diagonal in this basis having the form, \begin{equation} \langle \alpha,i;u|D(g)|\alpha,j;v\rangle=[D^{\alpha}(g)]^{i}_{\ j}\delta^{u}_{\ v} \end{equation}