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In the book it says,the rank of a matrix in echelon form is the maximum number of linearly independent rows and that is equal to the number of non zero rows in the matrix,

I am aware of echelon form of a matrix but have no ideas how to get the maximum number of linearly independent rows and how that is equal to the number of non zero rows,

I will be grateful if anyone can help me undeestand it by giving simple examples,

thanks.

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  • $\begingroup$ If a matrix is in row echelon form, then linearly dependent rows will be 0. This is because, by definition, linearly dependent rows can be written as a linear combination of the other rows, meaning that when we are reducing it to row echelon form, the row can be removed. Since a row is only 0 if it is dependent, the maximum number of linearly independent rows is equal to the number of non-zero rows. $\endgroup$
    – mode_er
    Jan 31, 2022 at 0:10
  • $\begingroup$ can u explain by giving an example,for instance if the three rows are (1,0,3),(4,5,6) and (0,0,0),how the first two rows are lindependent,thanks. $\endgroup$ Jan 31, 2022 at 0:55
  • $\begingroup$ It's worth pointing out that the matrix $\begin{pmatrix} 1 & 0 & 3 \\ 4 & 5 & 6 \\ 0 & 0 & 0 \end{pmatrix}$ is not in row-echelon form, as the leading $1$ in the first column does not have $0$s below it. This feature of REF is important in guaranteeing the linear independence of the non-zero rows. $\endgroup$ Jan 31, 2022 at 1:00
  • $\begingroup$ m sorry,if the rows are (1,3,4),(0,6,8) and (0,0,0),the matrix is in the echelon form,how to go about wih the first two rows for independene,thanks. $\endgroup$ Jan 31, 2022 at 1:06
  • $\begingroup$ The first two rows are linearly independent because one cannot be written as a scalar multiple of the other $\endgroup$
    – mode_er
    Feb 1, 2022 at 1:16

1 Answer 1

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$\begin{bmatrix} 1& 1\\2&2\end{bmatrix}$. How many rows do you see in the matrix ? 2 right? But is the second row really of any use? Its just $2$ times the first row, i.e., it depends on some other row for its formation. Just to show it mathematically, you do $RREF$ to get
$\begin{bmatrix} 1& 1\\0&0\end{bmatrix}$.

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  • $\begingroup$ are u meaning a(1,1) = b(2,2) is possible for a= 2 and b= 1 and not necessarily for a=b=0,so they are dependent $\endgroup$ Jan 31, 2022 at 10:32
  • $\begingroup$ can we conclude that two vectors will be linearly independent only if one is not a scalar multiple of the other,every other pair not involving a zero vector is linearly independent. $\endgroup$ Jan 31, 2022 at 11:36
  • $\begingroup$ Yes you are correct. $\endgroup$
    – Upstart
    Jan 31, 2022 at 11:48

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