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I am very confused by something our lecturer said today:

We were given two matrices:

$B=\begin{pmatrix}2 & 3\\ 2 &0 \\ 0&3\end{pmatrix}$

C=$\begin{pmatrix}6 &3&4\\6&6&0\end{pmatrix}$

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And we were supposed to find: $B_{ij}+C_{ji}$

To me, this seemed like another way of writing $B+C^T \implies \begin{pmatrix}2 & 3\\ 2 &0 \\ 0&3\end{pmatrix}+ \begin{pmatrix}6 & 6\\ 3 &6 \\ 4&0\end{pmatrix}=\begin{pmatrix}8 & 9\\ 5 &6 \\ 4&3\end{pmatrix}$

Our lecturer said that when adding the two matrices I don't get a matrix, I actually get an entry. She also added that the correct way of writing the answer would be: $$\begin{pmatrix}8 & 9\\ 5 &6 \\ 4&3\end{pmatrix}_{ij}$$

How is that not a matrix? Can anyone explain what she means by this?

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    $\begingroup$ It's analogous to distinguishing an egg from a carton of eggs $\endgroup$ Commented Apr 30, 2015 at 13:15
  • $\begingroup$ I'd be contrarian and write the answer as $$\begin{pmatrix}8 & 5 & 4\\ 9 &6 & 3 \end{pmatrix}_{ji}$$ Or maybe $$\begin{pmatrix}9 & 8\\ 6 &5 \\ 3&4\end{pmatrix}_{i,3-j}$$ These are all equivalent. $\endgroup$ Commented Apr 30, 2015 at 16:38

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You lecturer referred to the $(i,j)$-th entry of the matrix $(B+C^T)$, i.e. $(B+C^T)_{i,j}$, which is indeed a number. You are referring to the sum itself, $(B+C^T)$ which is a matrix indeed -- a matrix whose $(i,j)$-th entry you are asked to find.

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  • $\begingroup$ But if no entry like $(B+C^T)_{22}$ is specified and I am just asked to calculate $(B+C^T)_{ij}$ then doesn't that just equal all of the entries of my new matrix i.e. a matrix? $\endgroup$ Commented Apr 30, 2015 at 13:11
  • $\begingroup$ If it wasn't stated by your lecturer, she seems to have meant "for a fixed $i$ and $j$..." $\endgroup$ Commented Apr 30, 2015 at 13:15
  • $\begingroup$ Mo, that's the "general term." Think of it like functions: when asked for the number $f(x)$, you give for instance an expression like $x^2-3$, which is a number. ($x$ is the variable, playing the role of $(i,j)$.) When asked for $f(0)$, you can give $3$ -- which is also a number (corresponding to "(i,j)=(2,2)"). $\endgroup$ Commented Apr 30, 2015 at 13:15
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Ugh, this is a case of terrible notation.

I would agree with your interpretation. Absent anyone telling you what $i, j$ actually are (for example, $i = 1$ and $j = 2$), then I would argue that it is an accepted convention that writing $B_{ij}$ refers to the matrix, not to any specific entry. Writing $(B_{ij})_{i=1,2,3; j=1,2}$ may be a little clearer, but a lot more cumbersome.

So given that, I would say that your interpretation is exactly correct.

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  • $\begingroup$ Good I am not the only one who would read it this way. She actually got quite fed up with me because I didn't understand how $B_{ij}+C_{ji}$ is not a matrix. $\endgroup$ Commented Apr 30, 2015 at 13:16
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    $\begingroup$ My guess is that your lecturer is a mathematician and not a physicist. It is quite common in physics to use $X_{ijk\ldots}$ to refer to the entire object. $\endgroup$ Commented Apr 30, 2015 at 13:18
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    $\begingroup$ @Rzeta: I don't really agree. To me, $(B_{i,j})_{i,j}$ (or better, $(B_{i,j})_{1\leq i\leq m, 1\leq j\leq n}$) is a matrix, but $B_{i,j}$ is an entry (where $i,j$ are assumed to be fixed). Same as $(a_n)_{n\geq 0}$ being a sequence, and $a_n$ being the general term of the sequence. $\endgroup$ Commented Apr 30, 2015 at 13:18
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    $\begingroup$ In my experience $A_{ij}$ is an entry and $(A_{ij})$ is a matrix. However, there is also the convention to denote by $A_{ij}$ the matrix which is generated by deleting the $i$-th row and $j$-th column from $A$. I always thought this was quite confusing... $\endgroup$ Commented Apr 30, 2015 at 13:20
  • $\begingroup$ As I said, a case of bad/ambiguous notation. Different people mean different things with the exact same common notation... $\endgroup$ Commented Apr 30, 2015 at 13:49
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Usually when one writes $A_{ij}$, one means the entry of $A$ at position $(i,j)$. For example

$$\begin{pmatrix}6 &3&4\\6&6&0\end{pmatrix}_{2,2} = 6$$

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