I am currently researching a QR-based root finding algorithm encountered two operations that I don't understand. I'd love to look them up, but I can't find the names of these operations/notations.

The first was an unfamiliar usage of a colons and commas in a possibly set related way - (1 : 2, 1). The term seemed to be expressed as the coefficient to another term (I.E. F(1 : 2, 1)). I later encountered a similar term (1 : 2, 2 : 3) that also seemed represented as a coefficient to a variable.

NOTE: After running a few more searches and rereading the article I encountered this on, I realized this notation may be matrix related rather than set related. F is actually a unitary matrix used to generate another matrix, it seems. I thought these might be matrix indices, but the typical notation for this used in the article is two comma separated subscript numbers.

Specifically, I encountered this in equation (21) under Theorem 2.8 of the paper A fast implicit QR eigenvalue algorithm for companion matrices. The equation describes a vector hk as the result of a formula Fk (1 : 2, 1) and a matrix Bk + 1 as the result of an equation Fk (1 : 2, 2 : 3).

The second is the one I had trouble running a search on. It seemed to be two terms, one over top of the other, in parenthesis as if functioning as a coefficient. I came up with the term "binomial coefficient" at some point, but I believe this might not be correct.

I'd settle for just the names of these two operations/notations, but if anyone is willing to explain what exactly they mean, it'd be appreciated. Thank you for taking the time to read this question. I apologize if this is not a suitable question for this forum or the details are inadequate.

  • $\begingroup$ I'm not sure why did you decide that [set-theory] was fitting. $\endgroup$ – Asaf Karagila Mar 12 '12 at 7:46
  • $\begingroup$ I assumed the (1 : 2, 1) was part of set notation. I apologize for this mistake. $\endgroup$ – Dizzy Mar 12 '12 at 22:39
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    $\begingroup$ Please provide some context. Where did you see F(1:2,1) and (1:2,2:3)? $\endgroup$ – Peter Phipps Mar 13 '12 at 0:07
  • $\begingroup$ If you can quote some context from the source where you found this notation, or better yet give a reference or link to the source itself, that would probably help. It's also quite possible that this is some non-standard notation invented by that author; did you check carefully whether the author defines it previously? $\endgroup$ – Nate Eldredge Mar 13 '12 at 0:47
  • $\begingroup$ I have updated the question to include the information requested. I'm sorry that I had not previously included this. $\endgroup$ – Dizzy Mar 13 '12 at 1:00

Thanks for linking the source. It looks to me like this notation is used to refer to submatrices. So if $A$ is a matrix, $A(i:j,k:l)$ is the $(j-i+1) \times (l-k+1)$ matrix consisting of rows $i$ through $j$ and columns $k$ through $l$ of $A$. Variations are $A(i, k:l)$ to refer to the $i$th row, columns $k$ through $l$, and $A(i, :)$ to refer to the $i$th row, all columns.

Also, it appears that notation like $$\begin{pmatrix} A \\ B \end{pmatrix}$$ refers to the block matrix whose first rows come from $A$ and whose last rows come from $B$.

  • $\begingroup$ Thank you (and everyone else) for your persistence in answering this question. Hopefully I can finally finish this algorithm soon. $\endgroup$ – Dizzy Mar 13 '12 at 1:23
  • $\begingroup$ The "colon" notation here is inherited from MATLAB; in fact, books on numerical linear algebra like Golub and Van Loan's sometimes borrows idioms from MATLAB to express algorithms compactly. This paper seems to be no exception. $\endgroup$ – J. M. is a poor mathematician Mar 28 '18 at 15:29

Regarding your second one, I'm not sure what kind of search you tried: did you try Wikipedia for "binomial coefficient"?

The definition is $$ {n\choose{k} }=\frac{n!}{k!(n-k)!}, $$ usually refered in English as "$n$ choose $k$". They appear very naturally in combinatorics, but they get their name from the binomial formula $$ (a+b)^n=\sum_{k=0}^n{n\choose k}a^kb^{n-k} $$

  • $\begingroup$ Thank you for your answer. It seems I made a misconception about what exactly a binomial coefficient was. $\endgroup$ – Dizzy Mar 12 '12 at 3:54

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