Editing Kernel (linear algebra) (section)

==Illustration==
The following is a simple illustration of the computation of the kernel of a matrix (see {{slink||Computation by Gaussian elimination}}, below for methods better suited to more complex calculations). The illustration also touches on the row space and its relation to the kernel.

Consider the matrix
<math display="block">A = \begin{bmatrix} 2 & 3 & 5 \\ -4 & 2 & 3 \end{bmatrix}.</math>
The kernel of this matrix consists of all vectors {{math|(''x'', ''y'', ''z'') ∈ [[real coordinate space|'''R'''<sup>3</sup>]]}} for which
<math display="block">\begin{bmatrix} 2 & 3 & 5 \\ -4 & 2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix},</math>
which can be expressed as a homogeneous [[system of linear equations]] involving {{mvar|x}}, {{mvar|y}}, and {{mvar|z}}:
<math display="block">\begin{align}
 2x + 3y + 5z &= 0, \\
-4x + 2y + 3z &= 0.
\end{align}</math>

The same linear equations can also be written in matrix form as:
<math display="block">
  \left[\begin{array}{ccc|c}
    2 & 3 & 5 & 0 \\
    -4 & 2 & 3 & 0
  \end{array}\right].
</math>

Through [[Gauss–Jordan elimination]], the matrix can be reduced to:
<math display="block">
  \left[\begin{array}{ccc|c}
    1 & 0 & 1/16 & 0 \\
    0 & 1 & 13/8 & 0
  \end{array}\right].
</math>

Rewriting the matrix in equation form yields:
<math display="block">\begin{align}
x &= -\frac{1}{16}z \\
y &= -\frac{13}{8}z.
\end{align}</math>

The elements of the kernel can be further expressed in [[Parametric equations|parametric vector form]], as follows:
<math display="block">\begin{bmatrix} x \\ y \\ z\end{bmatrix} = c \begin{bmatrix} -1/16 \\ -13/8 \\ 1\end{bmatrix}\quad (\text{where }c \in \mathbb{R})</math>

Since {{mvar|c}} is a [[Free variables (system of linear equations)|free variable]] ranging over all real numbers, this can be expressed equally well as:
<math display="block">
\begin{bmatrix} x \\ y \\ z \end{bmatrix}
 = c \begin{bmatrix} -1 \\ -26 \\ 16 \end{bmatrix}.
</math>
The kernel of {{mvar|A}} is precisely the solution set to these equations (in this case, a [[line (geometry)|line]] through the origin in {{math|'''R'''<sup>3</sup>}}). Here, the vector {{math|(−1,−26,16)<sup>T</sup>}} constitutes a [[Basis (linear algebra)|basis]] of the kernel of {{mvar|A}}. The nullity of {{mvar|A}} is therefore 1, as it is spanned by a single vector.

The following dot products are zero:
<math display="block">
 \begin{bmatrix} 2 & 3 & 5 \end{bmatrix}
 \begin{bmatrix} -1 \\ -26 \\ 16 \end{bmatrix}
= 0
\quad\mathrm{and}\quad
 \begin{bmatrix} -4 & 2 & 3 \end{bmatrix}
 \begin{bmatrix} -1 \\ -26 \\ 16 \end{bmatrix}
= 0 ,
</math>
which illustrates that vectors in the kernel of {{mvar|A}} are orthogonal to each of the row vectors of {{mvar|A}}.

These two (linearly independent) row vectors span the row space of {{mvar|A}}—a plane orthogonal to the vector {{math|(−1,−26,16)<sup>T</sup>}}.

With the rank 2 of {{mvar|A}}, the nullity 1 of {{mvar|A}}, and the dimension 3 of {{mvar|A}}, we have an illustration of the rank-nullity theorem.