2
$\begingroup$

The definition is given below:

enter image description here

But I do not understand what is $s(i)$ and how to know it, could anyone give me a numerical example to explain the definition,please?

$\endgroup$
  • 1
    $\begingroup$ $s\left(i\right)$ is just a dummy variable. He doesn't write $s$ because there is one such integer for each $i$, and you cannot call them all $s$; you have to put the $i$ somewhere to be able to distinguish them from one another. Most people would call them $s_i$ instead, but it's just a matter of notation. $\endgroup$ – darij grinberg Sep 27 at 21:53
  • $\begingroup$ This may help to clarify the definition:math.stackexchange.com/questions/3355553/… $\endgroup$ – NoChance Sep 28 at 12:38
1
$\begingroup$

It means that for each row $i$, all elements before a certain column $j=s(i)$ are $0$, and that the function $i\longmapsto j=s(i)$ is (strictly) increasing, i.e. the number of $0$s at the beginning of a row is increasing.

This condition is not satisfied in the counterexample they give, as $s(1)=1, s(2)=3, s(3)=3, s(4)=5$.

$\endgroup$
  • $\begingroup$ But how is $s(i) \leq n+1?$ should not it be strictly less than $n+1$? $\endgroup$ – Mathstupid Sep 28 at 5:47
  • 1
    $\begingroup$ If $s(i)=n+1$, it means that you have a row of $0$s (so all further rows will be $0$ rows). $\endgroup$ – Bernard Sep 28 at 9:08
  • $\begingroup$ But $n$ is the number of columns not rows. $\endgroup$ – Mathstupid Sep 28 at 23:52
  • $\begingroup$ Oops! I thought this was a square matrix. However, $\mathcal M_{k\times n}$, in the usual conventions, denotes the set of matrices with $k$ rows and $n$ columns. $\endgroup$ – Bernard Sep 29 at 0:03
  • $\begingroup$ So what does it mean $s(i) = n +1 $ in this case? $\endgroup$ – Mathstupid Sep 29 at 0:20
1
$\begingroup$

$s(i)$ is the position (column number) of the first nonzero entry in row $i$. In the last matrix in your example $s(2) = s(3) = 3$.

If row $i$ is all $0$ then $s(i)=n+1$ even though there are only $n$ columns.

$\endgroup$
  • $\begingroup$ But how $s(i) \leq n+1$? is not it should be strictly less than $n+1$? $\endgroup$ – Mathstupid Sep 28 at 5:30

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.