# Summation simplification in the definition of composition of linear transformation

I'm reading a definition of composition of linear transformation, and I don't understand the second line in it. And I think that the part I don't understand is not directly related to linear transformation but the simplification of summation mark.

Say $\textsf{T}:\textsf{V}\to\textsf{W}, \textsf{U}:\textsf{W}\to\textsf{Z},A=[\textsf{U}]_{\beta}^{\gamma},B=[\textsf{T}]_{\alpha}^{\beta}$, and $\alpha,\beta,\gamma$ each has a cardinality $n,m,p$ respectively, and they're ordered bases. The work I've done is:

\begin{align}{} \large\sum_{k=1}^{m}{B_{kj}(\sum_{i=1}^{p}{A_{ik}z_i})}\\ \large=\sum_{k=1}^{m}{(\sum_{i=1}^{p}{B_{kj}A_{ik}z_i})} \end{align}

But I don't know how the swap between the two summation marks work: • So you meant that my $B_{kj}A_{ik}z_i$ is kind of your $A_{p',m'}$? – Ning Wang Apr 7 '18 at 2:20
• You cannot swipe $\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{i}$ as $\displaystyle\sum_{j=1}^{i}\sum_{i=1}^{n}$. – user284331 Apr 7 '18 at 2:24
• Cool! And this rule also works for multiple-independent $\prod$ right? – Ning Wang Apr 7 '18 at 2:29