Free variables and basis for \(N(A)\)

Let \(A \in \mathbb{F}^{m \times n}\) be a matrix in reduced row-echelon form. Recall that we can get all the solutions to \(Ax = 0\) by setting the free variables to distinct parameters. Then the set of solutions can be written as a linear combination of \(n\)-tuples where the parameters are the scalars. These \(n\)-tuples give a basis for the nullspace of \(A\). Hence, the dimension of the nullspace of \(A\), called the nullity of \(A\), is given by the number of non-pivot columns.

We now look at an example of finding a basis for \(N(A)\). Let \(A \in \mathbb{R}^{2\times 4}\) be given by \(\begin{bmatrix} 1 & -1 & -1 & 3\\2 & -2 & 0 & 4\end{bmatrix}\). We perform the following elementary row operations: \begin{eqnarray} & & \begin{bmatrix} 1 & -1 & -1 & 3\\2 & -2 & 0 & 4\end{bmatrix} \\ & \stackrel{R_2 \leftarrow R_2 - 2R_1}{\longrightarrow} & \begin{bmatrix} 1 & -1 & -1 & 3\\0 & 0 & 2 & -2\end{bmatrix} \\ & \stackrel{R_2 \leftarrow \frac{1}{2}R_2}{\longrightarrow} & \begin{bmatrix} 1 & -1 & -1 & 3\\0 & 0 & 2 & -2\end{bmatrix} \\ & \stackrel{R_2 \leftarrow \frac{1}{2}R_2}{\longrightarrow} & \begin{bmatrix} 1 & -1 & -1 & 3\\0 & 0 & 1 & -1\end{bmatrix} \\ & \stackrel{R_1 \leftarrow R_1 + R_2}{\longrightarrow} & \begin{bmatrix} 1 & -1 & 0 & 2\\0 & 0 & 1 & -1\end{bmatrix} \end{eqnarray}

To obtain all solutions to \(Ax = 0\), note that \(x_2\) and \(x_4\) are the free variables. Set \(x_2 = s\) and \(x_4 = t\). Then all the solutions are given by \[\begin{bmatrix} x_1\\x_2\\x_3\\x_4\end{bmatrix} = \begin{bmatrix} s-2t\\s\\t\\t\end{bmatrix} = s~\begin{bmatrix} 1\\1\\0\\0\end{bmatrix}+ t~\begin{bmatrix} -2\\0\\1\\1\end{bmatrix}.\]

Hence, a basis for \(N(A)\) is given by \(\left\{\begin{bmatrix} 1\\1\\0\\0\end{bmatrix}, \begin{bmatrix} -2\\0\\1\\1\end{bmatrix}\right\}\) and \(\dim(N(A)) = 2\).

Remark: Note that one could have obtained directly the first vector by setting \(x_2 = 1\), \(x_4 = 0\) and then solving for \(x_1\) and \(x_3\) and the second vector by setting \(x_2 = 0\), \(x_4 = 1\) and then solving for \(x_1\) and \(x_3\). In general, if \(A\) is in RREF, then a basis for the nullspace of \(A\) can be built up by doing the following: For each free variable, set it to 1 and the rest of the free variables to zero and solve for the pivot variables. The resulting solution will give a vector to be included in the basis.

Quick Quiz

Exercises

1. \(A \in \mathbb{R}^{3 \times 4}\) given by \(\begin{bmatrix} 1 & -1 & 0 & 1\\0 & 2 & 1 & -1\\1 & 1 &1 &0\end{bmatrix}.\)  

   There are many possible answers. One possible answer is \(\left\{ \begin{bmatrix} -\frac{1}{2} \\ -\frac{1}{2} \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -\frac{1}{2} \\ \frac{1}{2} \\ 0 \\ 1 \end{bmatrix} \right \}\).

2. \(A \in GF(2)^{2 \times 4}\) given by \(\begin{bmatrix} 1 & 1 & 0 & 1\\0 & 1 & 1 & 1\end{bmatrix}.\) Recall that \(GF(2)\) is the field consisting of only 0 and 1 with the usual integer addition and multiplication except that “\(1+1 =0\)”.  

   \(\left\{ \begin{bmatrix} 0\\ 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 1 \\ 1 \\ 0 \end{bmatrix} \right \}\)