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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,

Suppose

$ 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.

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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

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

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