In an assignment given in a numerical analysis class, the following notation is used to define an $n\times n$ matrix:

$$A = (a_{ij})^{n}_{i,j=1} = \left\lbrace \begin{array}{ll}0, & i>j,\\1, & i=j,\\j-i+1,&i<j\\\end{array}\right.$$

The very left and very right part is perfectly clear to me. But I do not understand what the $(\ldots)_{i,j=1}$ part is supposed to mean. I only know $a_{ij}$ to denote the element at the position $ij$.

So what does the subscript $i,j=1$ and superscript $n$ mean here?


It is just a notation to denote that $A$ is an $n\times n$ matrix with coefficients $a_{ij}$, for $i$ and $j$ ranging from $1$ to $n$.

That is, the notation $A=(a_{ij})_{1\leq i,j\leq n}$, or $A=(a_{ij})_{i,j=1}^n$, identifies the matrix $A$ to the tuple of its $n^2$ coefficients $(a_{11}, a_{12}, \dots, a_{1n}, a_{21}, \dots, a_{nn})$.

Then, the RHS is actually an abuse of notation: it is the definition of $a_{ij}$ (for any fixed $1\leq i,j\leq n$), not of $A=(a_{ij})_{ij}$. A correct definition should read

$A = (a_{ij})^{n}_{i,j=1}$, where for $1\leq i,j\leq n$ we have $$ a_{ij} = \left\lbrace \begin{array}{ll}0, & i>j,\\1, & i=j,\\j-i+1,&i<j\\\end{array}\right.$$

| cite | improve this answer | |
  • $\begingroup$ That does actually make a lot of sense. Thank you for your very comprehensive answer, especially for pointing out the abuse of notation. $\endgroup$ – JackWhiteIII Nov 15 '17 at 21:54
  • $\begingroup$ @JackWhiteIII You're welcome -- glad this helped. $\endgroup$ – Clement C. Nov 15 '17 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.