I know what a free variable and a basic variable is. I understand free variables show up because there is a lack of a pivot. I understand a pivot represents a solution in a subspace that's not a nullspace.
So my question is do free variables represent a solution in a nullspace?
Is there some connection between that and a reduced row echelon form that would be easy to understand?
On
Identifying free variables from the row echelon form is merely a convenient way of computing a basis for the null space:
Bingo, you end up with as many linearly independent vectors as there are free variables and these vectors are a basis for the null space.
However, this is merely a convenient algorithm. There are other ways...
So, a free variable does NOT represent a solution in the null space. The null space consists of vectors. A free variables is, well, a free variable.
The free variables in a system of linear equations don’t represent the null space of the coordinate matrix, but they are the coordinates of elements of the null space relative to a basis derived from the RREF matrix. I’ll illustrate this with an example.
Suppose that in the process of solving a system of homogeneous linear equations you end up with the RREF matrix $$R = \begin{bmatrix}1&0&-2&-3 \\ 0&1&-1&4 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix}.$$ This is equivalent to the pair of equations $$x_1-2x_3-3x_4 = 0 \\ x_2-x_3+4x_4=0.$$ This system is obviously underdetermined and has two free variables. For convenience, we choose as the bound variables the ones that have corresponding pivots in the RREF. This ensures that their coefficients in the reduced system are $1$. Solving for $x_1$ and $x_2$ above, we find that every solution of $R\mathbf x = 0$ is of the form $$\mathbf x = \begin{bmatrix}2x_3+3x_4 \\ x_3-4x_4 \\ x_3 \\ x_4\end{bmatrix} = x_3\begin{bmatrix}2\\1\\1\\0\end{bmatrix} + x_4\begin{bmatrix}3\\-4\\0\\1\end{bmatrix},$$ that is, the null space of $R$ is the span of the two vectors at right. They’re obviously linearly independent, so they form a basis for the null space. Now recall what coordinates of a vector $\mathbf v$ relative to some basis are: they are the coefficients of the unique linear combination of basis vectors that produces $\mathbf v$. So, the free variables $x_3$ and $x_4$ represent coordinates of elements of the null space relative to the above basis.
If we’d chosen a different pair of variables to be our free variables, we would’ve ended up with a different basis for the null space. For example, taking $x_1$ and $x_2$ as free, we solve for the other two variables: $$x_3 = \frac4{11}x_1+\frac3{11}x_2 \\ x_4 = \frac1{11}x_1-\frac2{11}x_2$$ and so the null space basis that corresponds to this choice of free variables is $\left(1,0,\frac4{11},\frac1{11}\right)^T$ and $\left(0,1,\frac3{11},-\frac2{11}\right)^T$. This basis can be found mechanically by moving the last two columns of $R$ to the left to correspond to $x_3$ and $x_4$ being bound and row-reducing again.