The definition is given below:
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?
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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
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$\begingroup$ This may help to clarify the definition:math.stackexchange.com/questions/3355553/… $\endgroup$ – NoChance Sep 28 at 12:38
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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 $s(1)=1, s(2)=3, s(3)=3, s(4)=5$.
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$\begingroup$ But how is $s(i) \leq n+1?$ should not it be strictly less than $n+1$? $\endgroup$ – Mathstupid Sep 28 at 5:47
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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
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$\begingroup$ 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
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$\begingroup$ So what does it mean $s(i) = n +1 $ in this case? $\endgroup$ – Mathstupid Sep 29 at 0:20
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$s(i)$ 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.
answered Sep 27 at 21:55
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$\begingroup$ But how $s(i) \leq n+1$? is not it should be strictly less than $n+1$? $\endgroup$ – Mathstupid Sep 28 at 5:30
