linear independent rows of a matrix If I have a matrix
$$
\begin{bmatrix}
1 &amp; 2 &amp; 1 \\
1 &amp; 2 &amp; 1 \\
1 &amp; 2 &amp; 1 \\
\end{bmatrix}
$$

I can reduce it to 

$$
\begin{bmatrix}
1 &amp; 2 &amp; 1 \\
0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 \\
\end{bmatrix}
$$  

Do I have one linearly independent row here or zero independent rows?
$$
\begin{bmatrix}
1 &amp; 2 &amp; 1 \\
0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 \\
\end{bmatrix}
$$ 

First think about what it means to be linearly independent. Linearly independent means that every row/column cannot be represented by the other rows/columns. Hence it is independent in the matrix.

When you convert to RREF form, we look for "pivots"

Notice that in this case, you only have one pivot. 

A pivot is the first non-zero entity in a row.

Here is your pivot:

$$
\begin{bmatrix}
\color{red}{1} &amp; 2 &amp; 1 \\
0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 \\
\end{bmatrix}
$$

Therefore, since you only have one pivot, then two of your columns are not linearly independent, and one of them is. Therefore, we say that the $\dim(A)=1$, where $A$ is the matrix.
We say that the rows are not independant.

The space spanned by the rows has dimension 1.