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I never got introduced to how, formally, indexing of elements forms a matrix – I suppose it works via mapping of ordered pairs. Now, dealing with transposed matrices, I am slightly confused:

Let $A=(a_{ij})_{i=1:m,j=1:n}$ be any matrix. Now, I have two definitions of the transpose $A^t$:

1) $A^{t,1} := (a_{ji})_{i=1:m, j=1:n}$

2) $A^{t,2} := (a_{ij})_{j=1:n, i=1:m}$

I get the first: When pointing on the second entry of the third row of $A^{t,1}$ the first index – $j$ in my presentation – is $3$ and the second, $i$, is $2$, that is, I get $a_{23}$, the third entry in the second row of $A$. That is how the transpose should work.

However, I cannot work with the second. How does the order in the subscript affect the matrix? In any way that I could imagine the first definition would not make sense then and produce $A$? Whats the right way to define the transpose?

Thanks in advance.

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    where did you take the second one?2017-01-05
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    From a german textbook on numerical linear algebra and the german wikipedia page on the transpose.2017-01-05

1 Answers 1

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  1. The second entry of the third row of $A^{t,1}$ is $(a_{ji})_{i=3,j=2}=a_{2,3}$.

  2. The second entry of the third row of $A^{t,2}$ is $(a_{ij})_{j=3,i=2} = a_{2,3}$. The key is that outside the parentheses, the subscript "$j=1:n,i=1:m$" implies that $j$ now counts rows and $i$ counts columns (in contrast to "$i=1:m,j=1:n$").

If you want an intuitive understanding of the transpose, an animation (Wikipedia) is much more helpful.

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    Sorry to dig this out, but should it not be $a_{2,3}$ in both cases in your answer?2017-01-19
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    @Ramen Yes, sorry about that2017-01-19