Reduced row echelon form and linear independence

Yes, but since you are considering the column vectors, you know that they are all independent if each column has a leading one in the reduced row-echelon form.

If we look at an explicit example:

$A = \begin{bmatrix} 1 & 3 & -1 & 0 \\ 4 & 1 & 7 & 11 \\ 0 & 4 & -4 & -4 \\ 2 & 0 & 4 & 6 \end{bmatrix}$

Then we have

$RREF(A)=\begin{bmatrix} 1 & 0 & 2 & 3 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

Notice that since columns $3$ and $4$ in $RREF(A)$ do not have leading $1$'s, they are dependent on the first two columns. Specifically:

$\begin{bmatrix} -1 \\ 7 \\ -4 \\ 4 \end{bmatrix}=2\begin{bmatrix} 1 \\ 4 \\ 0 \\ 2 \end{bmatrix}-\begin{bmatrix} 3 \\ 1 \\ 4 \\ 0 \end{bmatrix}$

$\begin{bmatrix} 0 \\ 11 \\ -4 \\ 6 \end{bmatrix}=3\begin{bmatrix} 1 \\ 4 \\ 0 \\ 2 \end{bmatrix}-\begin{bmatrix} 3 \\ 1 \\ 4 \\ 0 \end{bmatrix}$

Notice that the entries in column $3$ of $RREF(A)$ respectively correspond to the scalars for columns $1$ and $2$, such that column $3$ can be written as a linear combination of those columns. And the same goes for column $4$.

But if each column of $RREF(A)$ has a leading $1$, then each column is linearly independent.

Sidenote: The number of rows with all zeros in $RREF(A)$ tell only how many rows are independent/dependent (unless it is a square matrix, then the number of (in)dependent rows/columns will be the same) Looking at the leading 1's is much more helpful in determining linear dependence.

The "row rank" of $A$ is the number of linearly independent rows it has, and the "column rank" the number of its linearly independent columns. The key facts are (for any matrix $A$) that:

• The row rank is equal to the column rank.

• The row (equiv. column) rank is unchanged by elementary row operations.

Therefore you can get the rank of $A$ (as we say for simplicity) by counting the leading ones of its row echelon form. Since $S$ has $n$ vectors, we need the rank of $A$ to be $n$ (it cannot be more) in order for $S$ to be a linearly independent set.

Yes, if you can convert the matrix into reduced row echelon form(or even just row echelon form) without a row of $0$s,then the vectors are linearly independent.