How is the square matrix $\langle a''|A|a'\rangle=\langle a'|A|a'\rangle\delta_{a'a''}$
Could it be that we apply the closure property to it like
$\langle a''|A|a'\rangle=\sum_{a'}\langle a''|A|a'\rangle|a'\rangle\langle a'|=\sum_{a'}\langle a''|a'\rangle A\langle a'|a'\rangle=\langle a'|A|a'\rangle\delta_{a'a''}$
Is it even a correct method. Please help.
You need to be careful with your summation indicies and bra/ket ordering. You can't sum over the index that you've already got in your expression for one \begin{align*} \langle \alpha | A |\beta\rangle = \langle \alpha | 1 A |\beta\rangle &= \sum_\gamma \langle \alpha | |\gamma\rangle\langle\gamma | A |\beta\rangle \\ &= \sum_\gamma \delta_{\alpha, \gamma} \langle \gamma | A | \beta\rangle \\ &= \langle \alpha | A |\beta\rangle \end{align*} Inserting the completeness relation here doesn't help you, it's more of a definition for a diagonal matrix. Only matrix elements on the diagonal i.e. $A_{ii}$ are non zero. All the other entries $A_{i\neq j}$ are zero.
This has nothing to do with the closure property, rather, the properties of kronecker delta function itself.
The kronecker delta function $\delta(x_0,x_1) = 1$ 'if and only if' $x_0 = x_1$.
What the statement $$\langle a''|A|a'\rangle = \langle a''|A|a''\rangle \delta_{a''a'} = \langle a'|A|a\rangle \delta_{a"a'}$$ implies is that the square matrix is diagonal. The off-diadgonal elements are zero i.e. $a' \neq a'' \implies \delta_{a''a'} =0$ and hence, the matrix element is also zero.
