Let #V# and #W# be vector spaces. Because a linear mapping #V\to W# is of course just a special case of a general mapping, we can talk for example about the image of a vector or the image of a subset of #V# and about the full inverse image of a subset of #W#.

Let # L : V \rightarrow W# be a mapping, #D# a subset of #V#, and #E# a subset of #W#.

- By # L (D)# we denote the
**image** of #D# under #L#: the set #\left\{L(\vec{d})\mid \vec{d}\in D\right\}#.
- By #L^{-1}(E)# we denote the
**full inverse image** of #E# under #L#: the set #\left\{\vec{x}\in V\mid L(\vec{x})\in E\right\}#.

Let the mapping #L:\mathbb{R}\to\mathbb{R}# be given by #L(x) = a\, x+ b#.

The image of #\mathbb{R}# under #L# is equal to #\mathbb{R}# if #a\ne0# and equal to #\{b\}# if #a=0#.

The full inverse image of #\{c\}# under #L# consists of the solution of #a\,x+b=c# and is thus equal to

- #\left\{\frac{c-b}{a}\right\}# if #a\ne0#
- #\mathbb{R}# if #a=0# and #b=c#
- #\emptyset# if #a=0# and #b\ne c#

We denote the full inverse image of #E# under #L# by #L^{-1}(E)# or # L^{\leftarrow}(E)#. The former notation is most common in mathematics, but may lead to confusion with the notation for the inverse of a mapping. To avoid confusion when using this notation we often add the meaning in words, such as "full inverse image #L^{-1}(E)# of #E#".

Two important linear subspaces can be associated with linear mappings.

Let # L :V \rightarrow W# be a mapping. Define

\[\begin{array}{rclcl}

\im{L}&=&L(V) &=& \{ L (\vec{v}) \in W\mid\vec{v}\in V\} \\ \text{and}&&&&\\ \ker{L}&=& L^{-1}(\{\vec{0}\}) &=& \{\vec{v}\in V \mid L( \vec{v})=\vec{0}\}

\end{array}

\]

#\im{L}# is called the **image** or the **image space** of #L# and #\ker{L}# is called the **null space** or **kernel** of #L#.

The image space is the image of #V# under the mapping #L# and the null space is the full inverse image of #\{ \vec{0}\}# under #L#.

The first definition generalizes the column space of the coefficient matrix (that is to say: the *span* of the columns of the matrix) of a system of linear equations.

The second definition generalizes the solution space of a *homogeneous system of linear equations*.

As indicated *before*, we an #(m\times n)#-matrix #A# is also used to refer to the linear map #L_A:\mathbb{R}^n\to\mathbb{R}^m# *determined by it* (given by \( L_A(\vec{x}) = A\vec{x}\)). This way image and kernel of #A# als also defined: #\im{A}=\im{L_A}# en #\ker{A} = \ker{L_A}#.

As the name indicates, the image space and the null space of a linear mapping are linear subspaces:

Let # L :V \rightarrow W# be a linear mapping.

- The image #\im{L}# of #L# is a linear subspace of #W#.
- The kernel #\ker{L}# of #L# is a linear subspace of #V#.

Let the mapping #L:\mathbb{R}\to\mathbb{R}# be given by #L(x) = a\, x+ b#. The image space #\im{L}# equals #\mathbb{R}# if #a\ne0#. But if #a=0# and #b\ne0# (so #L# is a constant mapping distinct from #0#), then the image space is #\{b\}#, not a linear subspace. The kernel #\ker{L}# consists of the solution of #a\,x+b=0# and is thus equal to

- #\left\{\frac{-b}{a}\right\}# if #a\ne0#
- #\mathbb{R}# if #a=0# and #b=0#
- #\emptyset# if #a=0# and #b\ne 0#

In particular, #\ker{L}# is not a linear subspace of #\mathbb{R}# if #b\ne0#. We see again that #L# is linear only if #b=0#.

*The image space #\im{L}# is a linear subspace of #W#:* It is a subset of #W#. The zero vector of #W# is the image of the zero vector of #V# and thus belongs to #\im{L}#. If #\vec{u}#, #\vec{v}# belong to #\im{L}# and #\alpha# and #\beta# are scalars, then there are vectors #\vec{x}#, #\vec{y}# in #V# such that #L(\vec{x}) = \vec{u}# and #L(\vec{y}) = \vec{v}#, so linearity of #L# implies:

\[\begin{array}{rcl}\alpha \vec{u}+\beta \vec{v} &=& \alpha L(\vec{x})+\beta L(\vec{y})\\ &=&L(\alpha \vec{x})+L(\beta \vec{y})\\&=&L(\alpha \vec{x}+\beta \vec{y})\end{array}\] We conclude that \(\alpha \vec{u}+\beta \vec{v}\) belongs to #\im{L}#, from which we deduce that #\im{L}# is a linear subspace of #W#.

*The kernel #\ker{L}# is a linear subspace of #V#:* It is a subset of #V# which always contains the zero vector #\vec{0}#. Further, it follows from the linearity of #L# that if #\vec{x}# and #\vec{y}# belong to #\ker{L}# and #\alpha# and #\beta# are scalars, we have

\[L (\alpha \vec{x}+\beta \vec{y})=\alpha L(\vec{x})+\beta L(\vec{y})=\vec{0} + \vec{0}=\vec{0}\] so #\alpha \vec{x}+\beta \vec{y}\in{\ker{L}}#.

We determine the kernel and the image space of #L_A:\mathbb{R}^3\to\mathbb{R}^2#, the *linear mapping determined by* the #(2\times 3)#-matrix \[ A=\matrix{ 1 & -1 & 2 \\ 1 & -1 & 2 }\] This means that, if we use column vectors, the mapping rule is \[

L_A \matrix{ x_1 \\ x_2 \\ x_3 } = \matrix{ 1 & -1 & 2 \\ 1 & -1 & 2 }

\matrix{ x_1 \\ x_2 \\ x_3 }

\] The null space of #L_A# consists of all vectors # \vec{x}# that satisfy \[

\matrix{ 1 & -1 & 2 \\ 1 & -1 & 2 }

\matrix{ x_1 \\ x_2 \\ x_3 } =

\matrix{ 0 \\ 0}

\] This is a homogeneous system of linear equations with #A# as a coefficient matrix. The null space is the plane with equation #x_1 -x_2 +2x_3 =0#.

The image space #L_A# consists of all vectors of the form

\[

\matrix{ 1 & -1 & 2 \\ 1 & -1 & 2 }

\matrix{ x_1 \\ x_2 \\ x_3 }

\] that is, vectors of the form

\[

x_1 \matrix {1 \\ 1 } +

x_2 \matrix{ -1 \\ -1 } +

x_3 \matrix{ 2 \\ 2 }

\] This describes exactly the span of the columns of #A#, that is, the column space. We conclude that the image space equals \[\linspan{\cv{1\\ 1}}\]

We also determine the full inverse image #L_A^{-1}(\ell)# of the straight line #\ell# with parametric representation

\[

\ell : \quad\vec{x} = \left(\begin{array}{c} 3\\ 2 \end{array} \right) +

\lambda \left(\begin{array}{c} 2\\1 \end{array} \right)

\] So we are looking for vectors #\vec{x}# which satisfy

\[

A\vec{x} = \left( \begin{array}{c} 3+2\lambda \\ 2 + \lambda \end{array} \right)

\] for some #\lambda#. This means that we must solve the system with augmented matrix

\[

\left(\begin{array}{ccc|c}

1 & -1 & 2 & 3+2\lambda\\ 1 & -1 & 2 & 2+\lambda \\ \end{array} \right)

\]By row reduction it is easy to deduce that this system has only solutions for #\lambda =-1#. The solutions form the plane with equation #x_1-x_2+2x_3=1#.

Can you see on the basis of the relative position of #\ell# and the image space #\im{L_A}# that the calculation of the full inverse image can be limited to the calculation of the full inverse image of the vector #\cv{1\\1}#?

If #A# is a matrix, then the kernel of #L_A#, the *linear map determined by* #A#, is the solution space of a homogeneous system of linear equations with coefficient matrix #A#, and the image of #L_A# is the column space (which is the *span* of the columns of the matrix) of #A#, as shown in the previous example.

The null space of #L_A# consists of all vectors # \vec{x}# that satisfy #L_A( \vec{x})=\vec{0}#, that is to say, all solutions of the homogeneous system #A\vec{x}=\vec{0}#.

The image space #L_A# consists of all vectors of the form #L_A( \vec{x})#. If #A# has columns #\vec{a}_1,\ldots,\vec{a}_n#, then this is exactly the subset

\[

\left\{ x_1 \vec{a}_1+\cdots+x_n\vec{a}_n \mid x_1, \ldots , x_n

\in \mathbb{R}\right\}

\] of #A#. This is the column space of #A#.

Null space and image space thus generalize two concepts from the world of matrices.

Consider the orthogonal projection #{ P}# in #\mathbb{R}^2# on a line #\ell = \langle\vec{a}\rangle # through the origin. If we take the length of #\vec{a}# to be equal to #1#, we can describe #P# algebraically by the mapping rule \[{ P}\vec{x} =( \dotprod{\vec{x}}{\vec{a}})\cdot \vec{a}\]

- The image space of #P# is equal to #\ell#. This is geometrically obvious (each point is projected onto #\ell#), and can be derived algebraically from the mapping rule. This does in fact show that each vector in the image is a scalar multiple of #\vec{a}#. Because the image is a linear subspace and #P(\vec{a}) = \vec{a}#, the image space should coincide with #\linspan{\vec{a}} = \ell#.
- The null space is the line #\ell^{\perp}# that is perpendicular to #\ell# and passes through the origin. Indeed, the null space consists of all vectors # \vec{x}# for which #(\dotprod{\vec{x}}{\vec{a}})\cdot \vec{a} =\vec{0}#, that is, #\dotprod{\vec{x}}{\vec{a}} =0#, which is precisely the orthoplement of #\vec{a}#.

We discuss *The link between systems of linear equations and affine subspaces* again; this time from the point of view of linear mappings.

Let # L :V \rightarrow W# be a linear mapping and consider the vector equation # L (\vec{x})=\vec{b}#.

The solution of the equation is equal to the full inverse image #L^{-1}(\vec{b}) # of #\vec{b}# under #L#. In particular:

- If #\vec{b} \not\in \im{L}#, then #L^{-1}(\vec{b}) # is empty and the equation has no solution.
- If #\vec{b} \in \im{L}#, then there is a vector #\vec{p}# such that # L( \vec{p})=\vec{b}#. The vector #\vec{p}# is a
*particular solution* of the vector equation # L \vec{x}=\vec{b}#. All solutions (that is, the *general solution*) of this vector equation form the *affine subspace*

\[

\vec{p}+\ker{L} = \left\{\vec{p}+\vec{n}\mid\vec{n}\in{ \ker{L}}\right\}

\]with *support vector* #\vec{p}# and *direction space* #\ker{L}#.

In general, the particular solution is not unique: each solution can act as particular solution. Likewise, each vector in the affine subspace #\vec{p}+\ker{L} # can act as support vector.

The theorem *General and particular solution* states that we we can find all solutions of the vector equation # L(\vec{x})=\vec{b}# by adding a particular solution to the solutions of the corresponding homogeneous equation. The corresponding homogeneous equation is (by definition) the equation # L (\vec{x})=\vec{0}#. The set of all solutions of this equation is the null space #\ker{ L} #. In particular, the equation # L( \vec{x})=\vec{b}# has at most one solution if #\ker{L}=\{\vec{0}\}#.

The first part of the statement is trivial. For the proof of the second part, we let #\vec{p}# be a particular solution. For each #\vec{n}#, we have # L (\vec{p}+\vec{n})= L (\vec{p})+ L( \vec{n})=\vec{b}+\vec{0}=\vec{b}#. Therefore, #\vec{p}+\vec{n}# is also a solution. Conversely, if #\vec{q}# is a solution, then # L (\vec{q}-\vec{p})= L( \vec{q})- L( \vec{p})=\vec{b}-\vec{b}=\vec{0}#, so #\vec{q}-\vec{p}\in \ker{L}#. Since #\vec{q}=\vec{p}+(\vec{q}-\vec{p})# the vector #\vec{q}# is indeed the sum of #\vec{p}# and a vector from the kernel. We conclude that the set of all solutions is the affine subspace #\vec{p}+\ker{L}#.

This property is often used in solving linear differential equations.