Null space （零空间）的简单实例分析，参考wiki

We give here a simple illustration of computing the kernel of a matrix (see the section Basis below for methods better suited to more complex calculations.) We also touch on the row space and its relation to the kernel.

Consider the matrix

{\displaystyle A={\begin{bmatrix}\,\,\,2&3&5\\-4&2&3\end{bmatrix}}.}

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

{\displaystyle {\begin{bmatrix}\,\,\,2&3&5\\-4&2&3\end{bmatrix}}{\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}},}

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

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

which can be written in matrix form as:

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

Gauss–Jordan elimination reduces this to:

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

Rewriting yields:

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

Now we can express an element of the kernel:

{\displaystyle {\begin{bmatrix}x\\y\\z\end{bmatrix}}=c{\begin{bmatrix}-1/16\\-13/8\\1\end{bmatrix}}.}

for c a scalar.

Since c is a free variable, this can be expressed equally well as,

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

The kernel of A is precisely the solution set to these equations (in this case, a line through the origin in R³); the vector (−1,−26,16)^T constitutes a basis of the kernel of A. Thus, the nullity of A is 1.

Note also that the following dot products are zero:

{\displaystyle \left[{\begin{array}{ccc}2&3&5\end{array}}\right]\cdot {\begin{bmatrix}-1\\-26\\16\end{bmatrix}}=0\quad \mathrm {and} \quad \left[{\begin{array}{ccc}-4&2&3\end{array}}\right]\cdot {\begin{bmatrix}-1\\-26\\16\end{bmatrix}}=0\mathrm {,} }

which illustrates that vectors in the kernel of A are orthogonal to each of the row vectors of A.

These two (linearly independent) row vectors span the row space of A, a plane orthogonal to the vector (−1,−26,16)^T.

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