Null space, basis

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 $ N$ of general solutions

$\displaystyle N = \{\mathbf{x} = t_1 \mathbf{v}_1 + \cdots + t_k \mathbf{v}_k:$   $\displaystyle \mbox{ $t_1,\ldots,t_k$ are any real numbers }$$\displaystyle \}.$ (11)

is spanned by $ \mathbf{v}_1,\ldots,\mathbf{v}_k$, and called the null space of $ A$. Furthermore, if $ \mathbf{v}, \ldots, \mathbf{v}_k$ are linearly independent, then the column vectors  $ \mathbf{v}, \ldots, \mathbf{v}_k$ are called a basis of the null space of $ A$.

In Matlab/Octave the function null(A) returns a matrix  $ [\mathbf{v}, \ldots, \mathbf{v}_k]$ containing column vector $ \mathbf{v}_i$'s which are a basis of the null space of $ A$. The choice of vectors  $ \mathbf{v}, \ldots, \mathbf{v}_k$ is not necessarily unique. Matlab/Octave produces the vectors $ \mathbf{v}_i$'s of unit norm (i.e., $ \Vert\mathbf{v}_i\Vert = 1$; see Section 6.1).

EXAMPLE 3. Find a basis for the null space of the matrix $ A = \begin{bmatrix}
-3 & 6 &-1 & 1 &-7 \\
1 &-2 & 2 & 3 &-1 \\
2 &-4 & 5 & 8 &-4
\end{bmatrix}$



Department of Mathematics
Last modified: 2005-10-21