I don't understand this concept about solution space and basis? "Solution space has a basis whose elements correspond to the columns of RRE form E which do not contain the leading entry of any row"
I don't understand this. Why is this true? Could you please give me an example?
Thanks in advance.
 A: The main reasons:


*

*Elementary row operations do not alter dependencies among columns of the matrix.

*In reduced row echelon form, columns without leading entries can be given as linear combinations of the columns with leading entries.

*The column space has dimension equal to the rank of the matrix (the number of leading entries).
To illustrate, consider the matrix $$\begin{bmatrix} 1 & 0 & -1 & 2 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ in reduced row echelon form.  The columns without leading entries (columns 3 and 4) can be formed as linear combinations of the columns with leading entries (columns 1 and 2).  For example, $$\overbrace{\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}}^{C_3}=\color{blue}{-1}\overbrace{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}}^{C_1}+\color{red}{1}\overbrace{\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}}^{C_2}.$$  Similarly, $C_4=2C_1$.
So any row equivalent matrix, such as $$\begin{bmatrix} 1 & 5 & 4 & 2 \\ 0 & 9 & 9 & 0 \\ 2 & 1 & -1 & 4 \\ \end{bmatrix}$$ also satisfies $C_3=-C_1+C_2$ and $C_4=2C_1$.
We need $\mathrm{rank}(A)$ linearly independent column vectors to form a basis of $\mathrm{col}(A)$, and the columns of $A$ corresponding to columns with leading entries in the reduced row echelon form give precisely that (and they're obviously linearly independent in the reduced row echelon form).
