From the definition of $k$-leading zeros, the first $k$ elements of the row are all zeros and the $(k+1)$th element of the row is not zero.

From the above definition, can zero row be interpreted as $n$-leading zeros in $n \times n$ matrices?

To make my question clear,


$ A = \begin{pmatrix} 1&0&1\\ 0&-1&0 \\ 0&0&0\end{pmatrix}$

and let $R_i$ be the $i$th row matrix of A. Then, $R_1$ is $0$-leading zero, and $R_2$ is $1$-leading zero. This is evident from the above definition.

My question is that can zero row $R_3$ be regarded as $3$-leading zeros. I read an informal text that treat zero row as $n$-leading zeros in $n\times n$ matrices and I could not find any explicit statement about this.


closed as unclear what you're asking by Morgan Rodgers, N. F. Taussig, Shailesh, user228113, Harish Chandra Rajpoot Jan 11 '16 at 13:43

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  • $\begingroup$ As reviewers suggestion, I added an example to make my question clear. $\endgroup$ – z0nam Jan 13 '16 at 3:01

Actually, from the definition you wrote down now, technically, an all-zero row is not a row with $k$ leading zeroes for any $k$.

This is because the statement

The first, second, third, $\dots$, $k$-th element are equal to $0$ and the $(k+1)$-th element is not equal to $0$

is only well defined for $k< n$, but it is false for all $k < n$.

  • $\begingroup$ Your answer is sufficient for me. Thanks! $\endgroup$ – z0nam Jan 11 '16 at 12:53

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