Kernel (matrix) - Example

Example

In this section, we show on a very simple example how the null space of a matrix may be computed. However the method which is sketched here is not practical for effective computations. A more efficient method is presented below.

Consider the matrix

The null space of this matrix consists of all vectors (x, y, z) ∈ R3 for which

This can be written as a homogeneous system of linear equations involving x, y, and z:

\begin{alignat}{7} 2x &&\; + \;&& 3y &&\; + \;&& 5z &&\; = \;&& 0, \\ -4x &&\; + \;&& 2y &&\; + \;&& 3z &&\; = \;&& 0.\\ \end{alignat}

This can be written in matrix form as:

\left[\begin{array}{ccc|c} 2 & 3 & 5 & 0 \\ -4 & 2 & 3 & 0 \end{array}\right].

Using Gauss–Jordan elimination, this reduces to:

\left[\begin{array}{ccc|c} 1 & 0 & 1/16 & 0 \\ 0 & 1 & 13/8 & 0 \end{array}\right].

Rewriting yields:

\begin{alignat}{7} x = \;&& -\frac{1}{16}z\,\,\, \\ y = \;&& -\frac{13}8z. \end{alignat}

Now we can write the null space (solution to Ax = 0) in terms of c, where c is scalar:

Since c is a free variable this can be simplified to

\begin{bmatrix} x\\ y\\ z \end{bmatrix} = c\begin{bmatrix} -1\\ -26\\ 16 \end{bmatrix}.

The null space of A is precisely the set of solutions to these equations (in this case, a line through the origin in R3).

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