# rank of a matrix by linearly independent rows

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.

• 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. Jan 31, 2022 at 0:10
• 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. Jan 31, 2022 at 0:55
• 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. Jan 31, 2022 at 1:00
• 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. Jan 31, 2022 at 1:06
• The first two rows are linearly independent because one cannot be written as a scalar multiple of the other Feb 1, 2022 at 1:16

$$\begin{bmatrix} 1& 1\\2&2\end{bmatrix}$$. 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
$$\begin{bmatrix} 1& 1\\0&0\end{bmatrix}$$.