Matrix Algebra

Null Space

Null space. Let

be an $m\times n$ -matrix. Suppose that the homogeneous equation $A\mathbf{x} = \mathbf{0}$ has nontrivial solutions. Since general solutions are expressed in the parametric vector form, the collection of general solutions

$\displaystyle \mathrm{null}\,A = \{\mathbf{x} = t_1 \mathbf{v}_1 + \cdots + t_k \mathbf{v}_k: t_1,\ldots,t_k\in\mathbb{R}\}.$

becomes a subspace spanned by $\mathbf{v}_1, \ldots, \mathbf{v}_k$ in $\mathbb{R}^n$ , and is called the null space of

. When the homogeneous equation $A\mathbf{x} = \mathbf{0}$ has no nontrivial solutions, we write $\mathrm{null} A = \{\mathbf{0}\}$ .

EXAMPLE 3. Find a basis for the null space of the matrix

$\displaystyle A = \begin{bmatrix} -3 & 6 &-1 & 1 &-7 \\ 1 &-2 & 2 & 3 &-1 \\ 2 &-4 & 5 & 8 &-4 \end{bmatrix}$

Matlab/Octave. The function null(A) returns a matrix $[\mathbf{v}_1,\ldots,\mathbf{v}_k]$ containing column vector $\mathbf{v}_i$ 's which are a basis of the null space of . The choices of basis vectors $\mathbf{v}_1, \ldots, \mathbf{v}_k$ are not unique. Matlab/Octave produces a orthonormal basis which is not obtained from the parametric vector form by solving the homogeneous equation $A\mathbf{x} = \mathbf{0}$ .

Dimension of null space. If the column vectors $\mathbf{v}, \ldots, \mathbf{v}_k$ are constructed from an REF, then it is easily observed that they are linearly independent. Therefore, the collection $\{\mathbf{v}, \ldots, \mathbf{v}_k\}$ becomes a basis of the null space of . This implies that the number of free variables in the matrix equation $A\mathbf{x} = \mathbf{0}$ determines $\dim$ null. The rank of , denoted by $\mathrm{rank}\,A$ , is identified as the number of pivot columns in . Thus, we obtain

$\displaystyle \dim(\mathrm{null}\,A) = n\, - \mathrm{rank}\,A$

is an $m\times n$ -matrix.