Null Space

Definition

The null space of a matrix $\mathrm{A}$, denoted $\mathrm{Null}(\mathrm{A})$, is the set of all column vectors $\mathrm{x}$ satisfying

$$\mathrm{Ax}=0.$$

$\mathrm{Null}(\mathrm{A})$ is a vector subspace: if $\mathrm{x},\mathrm{y}\in \mathrm{Null}(\mathrm{A})$, then $a\mathrm{x}+b\mathrm{y}\in \mathrm{Null}(\mathrm{A})$. For an $m\times n$ matrix $\mathrm{A}$, the null space is a subspace of ${\mathbb{R}}^n$.

Intuition

The null space answers: “what inputs does $\mathrm{A}$ completely destroy (map to zero)?” It captures all the “invisible” directions — perturbations in the input that produce no change in the output. A large null space means the matrix has many directions it ignores, implying high redundancy or underdetermination. A trivial null space ($\{0\}$ only) means the mapping is injective.

Formal Description

Finding a basis. Bring $\mathrm{A}$ to reduced row echelon form. The variables associated with pivot columns are basic variables; those with non-pivot columns are free variables. Express each basic variable in terms of the free variables and collect vectors.

Example. For

$$\mathrm{A}=\left(\begin{array}{ccccc}-3& 6& -1& 1& -7\\ 1& -2& 2& 3& -1\\ 2& -4& 5& 8& -4\end{array}\right),\mathrm{rref}(\mathrm{A})=\left(\begin{array}{ccccc}1& -2& 0& -1& 3\\ 0& 0& 1& 2& -2\\ 0& 0& 0& 0& 0\end{array}\right),$$

basic variables are $x_1,x_3$; free variables are $x_2,x_4,x_5$. Solving:

$$x_1=2x_2+x_4-3x_5,x_3=-2x_4+2x_5.$$

The general null-space vector is

$$x_2\left(\begin{array}{c}2\\ 1\\ 0\\ 0\\ 0\end{array}\right)+x_4\left(\begin{array}{c}1\\ 0\\ -2\\ 1\\ 0\end{array}\right)+x_5\left(\begin{array}{c}-3\\ 0\\ 2\\ 0\\ 1\end{array}\right),$$

giving a basis for $\mathrm{Null}(\mathrm{A})$. The null space is a 3-dimensional subspace of ${\mathbb{R}}^5$.

• $\dim (\mathrm{Null}(\mathrm{A}))$ equals the number of non-pivot columns of $\mathrm{rref}(\mathrm{A})$.
• If $\mathrm{A}$ is square and invertible, $\mathrm{Null}(\mathrm{A})=\{0\}$.
• By the rank-nullity theorem: $\dim (\mathrm{Col}(\mathrm{A}))+\dim (\mathrm{Null}(\mathrm{A}))=n$.

Applications

Underdetermined systems. If $\mathrm{u}$ is the general vector in $\mathrm{Null}(\mathrm{A})$ and $\mathrm{v}$ is any particular solution to $\mathrm{Av}=\mathrm{b}$, then the general solution is $\mathrm{x}=\mathrm{u}+\mathrm{v}$, since $\mathrm{A}(\mathrm{u}+\mathrm{v})=0+\mathrm{b}=\mathrm{b}$.

Regularisation. In ill-posed inverse problems, solutions are constrained to be orthogonal to the null space to ensure uniqueness.

Trade-offs

Computing the full null space via RREF is exact but numerically unstable for ill-conditioned matrices. The SVD provides a numerically stable null-space basis via the right singular vectors corresponding to zero (or near-zero) singular values.

Links

• Column Space
• Span, Basis and Dimension
• Row Space and Rank
• Vector Space