Kernel (matrix)

In mathematics, the null space (also nullspace) of an operator A is the set of all operands v which solve the equation Av = 0. It is also called the kernel of A. In set-builder notation,

${\displaystyle {\hbox{Null}}(A)=\{\mathbf {v} \in V:A\mathbf {v} =\mathbf {0} \}.}$

While the term kernel is used more often, the term null space is sometimes used in the context in which one wants to avoid confusion with integral kernel. One should not confuse the null space with the zero vector space, which is the space consisting only of the zero vector.

If the operator is a linear operator on a vector space, the null space is a linear subspace. Hence the null space is a vector space and can uniquely be represented by a basis of this space.

The remainder of this article focuses on the special case of a finite dimensional operator, which can be represented by a matrix.

In linear algebra, the null space (also nullspace) of a matrix A is the set of all vectors x for which Ax = 0. The null space of a matrix with n columns is a linear subspace of n-dimensional Euclidean space.^[1]

If we regard the matrix as a linear transformation, then the null space is precisely the kernel of the mapping (i.e. the set of vectors that map to zero). For this reason, the kernel of a linear transformation between abstract vector spaces is sometimes referred to as the null space of the transformation.

The null space of an m × n matrix A is the set

${\displaystyle {\text{Null}}(A)=\left\{{\textbf {x}}\in {\textbf {R}}^{n}:A{\textbf {x}}={\textbf {0}}\right\}{\text{,}}}$^[2]

where 0 denotes the zero vector with m components. The matrix equation Ax = 0 is the same as a homogeneous system of linear equations:

${\displaystyle A{\textbf {x}}={\textbf {0}}\;\;\Leftrightarrow \;\;{\begin{alignedat}{6}a_{11}x_{1}&&\;+\;&&a_{12}x_{2}&&\;+\cdots +\;&&a_{1n}x_{n}&&\;=0&\\a_{21}x_{1}&&\;+\;&&a_{22}x_{2}&&\;+\cdots +\;&&a_{2n}x_{n}&&\;=0&\\\vdots \;\;\;&&&&\vdots \;\;\;&&&&\vdots \;\;\;&&\vdots \,&\\a_{m1}x_{1}&&\;+\;&&a_{m2}x_{2}&&\;+\cdots +\;&&a_{mn}x_{n}&&\;=0&\end{alignedat}}{\text{.}}}$

From this viewpoint, the null space of A is the same as the solution set to the homogeneous system.

Consider the matrix

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

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

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

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

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

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

The null space of an m × n matrix is a subspace of Rⁿ. That is, the set Null(A) has the following three properties:

1. Null(A) always contains the zero vector.
2. If x ∈ Null(A) and y ∈ Null(A), then x + y ∈ Null(A).
3. If x ∈ Null(A) and c is a scalar, then cx ∈ Null(A).

Here are the proofs:

1. A0 = 0.
2. If Ax = 0 and Ay = 0, then A(x + y) = Ax + Ay = 0 + 0 = 0.
3. If Ax = 0 and c is a scalar, then A(cx) = cAx = c0 = 0.

The null space of a matrix is not affected by elementary row operations. This makes it possible to use row reduction to find a basis for the null space:

Input An m × n matrix A.

Output A basis for the null space of A

1. Use elementary row operations to put A in reduced row echelon form.
2. Interpreting the reduced row echelon form as a homogeneous linear system, determine which of the variables x₁, x₂, ..., x_n are free. Write equations for the dependent variables in terms of the free variables.
3. For each free variable x_i, choose a vector in the null space for which x_i = 1 and the remaining free variables are zero. The resulting collection of vectors is a basis for the null space of A.

For example, suppose that the reduced row echelon form of A is

${\displaystyle \left[{\begin{alignedat}{6}1&&0&&-3&&0&&2&&-8\\0&&1&&5&&0&&-1&&4\\0&&0&&0&&1&&7&&-9\\0&&\;\;\;\;\;0&&\;\;\;\;\;0&&\;\;\;\;\;0&&\;\;\;\;\;0&&\;\;\;\;\;0\end{alignedat}}\,\right]{\text{.}}}$

Then x₃, x₅, and x₆ are free, with

${\displaystyle {\begin{alignedat}{7}x_{1}&&\;=\;&&3x_{3}&&\;-\;&&2x_{5}&&\;+\;&&8x_{6}&\\x_{2}&&\;=\;&&-5x_{3}&&\;+\;&&x_{5}&&\;-\;&&4x_{6}&\\x_{4}&&\;=\;&&&&\;-\;&&7x_{5}&&\;+\;&&9x_{6}&\end{alignedat}}{\text{.}}}$

Therefore, the three vectors

${\displaystyle {\begin{array}{r}\\\\x_{3}\rightarrow \\\\x_{5}\rightarrow \\x_{6}\rightarrow \end{array}}\left[\!\!{\begin{array}{r}3\\-5\\\mathbf {1} \\0\\\mathbf {0} \\\mathbf {0} \end{array}}\right],\;\left[\!\!{\begin{array}{r}-2\\1\\\mathbf {0} \\-7\\\mathbf {1} \\\mathbf {0} \end{array}}\right],\;\left[\!\!{\begin{array}{r}8\\-4\\\mathbf {0} \\9\\\mathbf {0} \\\mathbf {1} \end{array}}\right]}$

are a basis for the null space of A.

The product of a matrix A and a vector x can be written in terms of the dot product of vectors:

${\displaystyle A{\textbf {x}}={\begin{bmatrix}{\textbf {a}}_{1}\cdot {\textbf {x}}\\{\textbf {a}}_{2}\cdot {\textbf {x}}\\\vdots \\{\textbf {a}}_{m}\cdot {\textbf {x}}\end{bmatrix}}{\text{.}}}$

Here a₁, ..., a_m denote the row vectors of the matrix A. It follows that x is in the null space of A if and only if x is orthogonal (or perpendicular) to each of the row vectors of A.

The row space of a matrix A is the span of the row vectors of A. By the above reasoning, the null space of A is the orthogonal complement to the row space. That is, a vector x lies in the null space of A if and only if it is perpendicular to every vector in the row space of A.

The dimension of the row space of A is called the rank of A, and the dimension of null space of A is called the nullity of A. These quantities are related by the equation

${\displaystyle {\text{rank}}(A)+{\text{nullity}}(A)=n{\text{.}}\,}$

Here n denotes the number of columns of the matrix A. The equation above is known as the rank-nullity theorem.

The null space also plays a role in the solution to a nonhomogeneous system of linear equations:

${\displaystyle A{\textbf {x}}={\textbf {b}}\;\;\;\;\;\;{\text{or}}\;\;\;\;\;\;{\begin{alignedat}{7}a_{11}x_{1}&&\;+\;&&a_{12}x_{2}&&\;+\cdots +\;&&a_{1n}x_{n}&&\;=\;&&&b_{1}\\a_{21}x_{1}&&\;+\;&&a_{22}x_{2}&&\;+\cdots +\;&&a_{2n}x_{n}&&\;=\;&&&b_{2}\\\vdots \;\;\;&&&&\vdots \;\;\;&&&&\vdots \;\;\;&&&&&\;\vdots \\a_{m1}x_{1}&&\;+\;&&a_{m2}x_{2}&&\;+\cdots +\;&&a_{mn}x_{n}&&\;=\;&&&b_{m}\\\end{alignedat}}}$

If u and v are two possible solutions to the above equation, then

${\displaystyle A({\textbf {u}}-{\textbf {v}})=A{\textbf {u}}-A{\textbf {v}}={\textbf {b}}-{\textbf {b}}={\textbf {0}}\,}$

Thus, the difference of any two solutions to the equation Ax = b lies in the null space of A.

It follows that any solution to the equation Ax = b can be expressed as the sum of a fixed solution v and an arbitrary element of the null space. That is, the solution set to the equation Ax = b is

${\displaystyle \left\{{\textbf {v}}+{\textbf {x}}\,:\,{\textbf {x}}\in {\text{Null}}(A)\,\right\}{\text{,}}}$

where v is any fixed vector satisfying Av = b. Geometrically, this says that the solution set to Ax = b is the translation of the null space of A by the vector v.

The left null space of a matrix A consists of all vectors x such that x^TA = 0^T, where T denotes the transpose of a column vector. The left null space of A is the same as the null space of A^T. The left null space of A is the orthogonal complement to the column space of A, and is the cokernel of the associated linear transformation. The null space, the row space, the column space, and the left null space of A are the four fundamental subspaces associated to the matrix A.

• Kernel (linear algebra)
• Matrix (mathematics)
• Euclidean subspace
• System of linear equations
• Row space
• Column space
• Row reduction
• Four fundamental subspaces

Template:Linear algebra references

• MIT Video Lecture on Column Space and Nullspace at Google Video, from MIT OpenCourseWare