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:

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\begin{align*} &(A_{1,1},A_{1,2},...,A_{1,m})\\ &(A_{2,1},A_{2,2},...,A_{2,m})\\ &\vdots\\ &(A_{p,1},A_{p,2},\cdots,A_{p,m}) \end{align*} if you want to add them all together, one starts to add row by row, or you can do it through column by column.

  • $\begingroup$ So you meant that my $B_{kj}A_{ik}z_i$ is kind of your $A_{p',m'}$? $\endgroup$ – Ning Wang Apr 7 '18 at 2:20
  • $\begingroup$ If there are more than three summation marks, would it still be true to randomly swap any two? It's hard to imagine/draw it and think... $\endgroup$ – Ning Wang Apr 7 '18 at 2:23
  • 1
    $\begingroup$ As long as the indices of summation are independent. $\endgroup$ – user284331 Apr 7 '18 at 2:23
  • $\begingroup$ You cannot swipe $\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{i}$ as $\displaystyle\sum_{j=1}^{i}\sum_{i=1}^{n}$. $\endgroup$ – user284331 Apr 7 '18 at 2:24
  • $\begingroup$ Cool! And this rule also works for multiple-independent $\prod$ right? $\endgroup$ – Ning Wang Apr 7 '18 at 2:29

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