I am reading a proof of Schur's Lemma and I am confused in the last reasoning of the proof. It says:
The element of the representation matrix, $D(A)_{kj}$(k, j are fixed index), is zero for all the group elements $A$, then the representation is reducible (can be brought into the block diagonal form).
I don't know how can we bring all the matrix into the diagonal form given all the matrix has a zero entry at the same place.
May be the reasoning in the previous chapter is relevant:
Consider the division of the rows and columns of a matrix system into two groups, say "marked" and "unmarked". Any matrix system for which it is possible to make this division in such a way that only zero elements occur at intersections of "marked" rows and "unmarked" columns, and at intersections of "unmarked" rows and "marked" columns, is either reducible or already in reduced form. To demonstrate this, we need only point out that we can transfer the "marked" rows and columns to the top and left of the matrices and obtain the block diagonal form.
Place use the basic matrix reasoning, not the lemma or theorem from the group theory. After all, Schur's Lemma is the foundation of the latter theorems of group theory.
I agree that the wording is somewhat confusing. Here is my interpretation: suppose that $d_{11}, d_{22}, \dots, d_{nn}$ are not all the same. Then group them into sets of equal elements: if there are $N$ distinct values on the diagonal matrix $d$, $\lambda_1, \dots, \lambda_N$, then the indices $1 \dots n$ can be grouped into $N$ subsets $T_1, \dots, T_N$, where $T_j = \{ i \mid d_{i,i} = \lambda_j \}$.
Then if $i \neq j$, we have $\lambda_i \neq \lambda_j$, so for all $k \in T_i$ and $m \in T_j$ we have $A_{km} =0$. More relevant to the lemma, for every $k \in T_i, m \notin T_i$ we have $A_{km}=0$. Therefore, letting the rows and columns belonging to $T_i$ be "marked" and the rows and columns outside of $T_i$ be "unmarked" gives us the desired partition. We can see that there are $N$ different "blocks" in the reduced matrix.