Where does the summation convention break?

Question

In my work I have spotted something "weird". Consider the matrix

\begin{align*} M_{ij} =1-\delta_{ij}. \end{align*} I want to find $ M^{2} $ in terms of its components. So \begin{align*} (M^{2})_{ij} =(M M)_{ij}& =M_{ia}M_{aj} \tag{1}\\ & = (1-\delta_{ia})(1-\delta_{aj}) \tag{2}\\ & =1+\delta_{ia}\delta_{aj}-\delta_{ia}-\delta_{aj} \tag{3}\\ & =1+\delta_{ij}-1 -1 \tag{4}\\ & =-1+\delta_{ij},\tag{5} \end{align*} however, this is clearly wrong, as consider $ n=2 $ , in which case we have \begin{align*} \begin{pmatrix} 0 & 1 \\ 1 &0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} =\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} . \end{align*} A guess where this goes wrong is from $ (1) $ to $ (2) $ as expanding $ (3) $ gives $ k $ as a free index, over which I have summed over. I did so because I couldn't leave it since it is a dummy variable. My best guess is that I am simply not allowed to go from $ (1) $ to $ (2) $ but I am not sure as to why that is.

If anybody could shine some light on this I would greatly appriciate it.

Answer 1

The $1$ in the definition is a matrix, not a number, and should be written as $1_{ij}$. ($J_{ij}$ would be better because the matrix of all ones doesn't act like the multiplicative identity, but I will leave it as $1_{ij}$.) Not writing the indices here causes problems later.

Following your derivation, we get \begin{align*} M_{ij} =1_{ij}-\delta_{ij}, \end{align*} and computing $M^2$, we get \begin{align*} (M^{2})_{ij} =(M M)_{ij}& =M_{ia}M_{aj} \tag{1}\\ & = (1_{ia}-\delta_{ia})(1_{aj}-\delta_{aj}) \tag{2}\\ & =1_{ia}1_{aj}+\delta_{ia}\delta_{aj}-\delta_{ia}1_{aj}-1_{ia}\delta_{aj} \tag{3}\\ & =n 1_{ij}+\delta_{ij}-1_{ij} -1_{ij} \tag{4}\\ & =(n-2) 1_{ij} +\delta_{ij},\tag{5} \end{align*} where we see the important error occurred in equation 4, where $1_{ia}1_{aj} = n 1_{ij} \neq 1_{ij}$

Answer 2

Let $A=(1-\delta_{ij})$, a matrix of order $n$. Then an easy calculation show that $A$ looks like a matrix which have all the entries with $1$ except when $i=j$ and it least case the matrix have only $0$ in the diagonal.

If you calculate $A^2$ you can notice that we always sum the $1$ by each entry except when we multiply the entry with $i=j$. Hence we always have in all the entries with $i=j$the sum of $n-1$ ones. Otherwise we always sum $1$ in all the entries except when $i=j$ twice (in the row and in the column of $A$), which give the sum of $n-2$ ones. As result, if $A$ is a matrix of order $n$ and $A$ is written as above.

$$A^2=(a_{ij})=\begin{cases}a_{ij}=n-1 \, \text{if i=j } \\ a_{ij}=n-2 \, \text{if $i\neq j$} \end{cases}.$$

Answer 3

(1) is already wrong—the right-hand side should be $\displaystyle\sum_{a=1}^n M_{ia}M_{aj}$, with a summation sign. This will eventually expand to $$ \sum_{a=1}^n 1 - \sum_{a=1}^n \delta_{ia} - \sum_{a=1}^n \delta_{aj} + \sum_{a=1}^n \delta_{ia}\delta_{aj}, $$ from which you should be able to recover the correct answer. (In your method, you guessed to sum certain parts of the formula but not all of them; writing the sums explicitly takes the guesswork away.)