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Kernel of a linear map

by , PhD

A linear map (or transformation, or function) $f:S ightarrow T$ transforms elements of a vector space $S$ called domain into elements of another vector space $T$ called codomain.

The kernel (or null space) of a linear transformation is the subset of the domain that is transformed into the zero vector.

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Definition of kernel

Let us provide a more formal definition of kernel.

Definition Let $S$ and $T$ be two vector spaces. Let $f:S ightarrow T$ be a linear map. The set[eq1]is called the null space (or kernel) of $f$.

Let us see some examples.

Example Let $S=T$ be the space of all $2 imes 1$ column vectors having real entries. Let $f:S ightarrow T$ be the linear map defined by the matrix product[eq2]where [eq3]For any $sin S$, denote by $s_{1}$ and $s_{2}$ the two entries of $s$, so that[eq4]As a consequence, $As=0$ whenever $s_{1}=s_{2}$. Therefore, the kernel of $f$ is formed by all the vectors of $S$ whose two entries are equal to each other:[eq5]

Example Let $S$ and $T$ respectively be the spaces of all $2 imes 1$ and $3 imes 1$ column vectors having real entries. Let $f:S ightarrow T$ be the linear map defined by the matrix product[eq6]where [eq7]For any $sin S$, denote by $s_{1}$ and $s_{2}$ the two entries of $s$, so that[eq8]By looking at the three entries of $As$, it is apparent that $As=0$ only when $s_{1}=s_{2}=0$. Therefore the kernel of $f$ is formed by a single vector, the zero vector:[eq9]

The kernel is a linear subspace of the domain

An interesting property of the null space is that it is a subspace of the domain, that is, it is closed with respect to taking linear combinations.

Proposition Let $S$ and $T$ be two vector spaces. Let $f:S ightarrow T$ be a linear map. Then, the null space $QTR{rm}{null}f$ is a subspace of $S$.

Proof

By the definition of subspace, $QTR{rm}{null}f$ is a subspace of $S$ if and only if any linear combination of elements of $QTR{rm}{null}f$ belongs to $QTR{rm}{null}f$. Let us check that this condition is verified. Arbitrarily choose two vectors [eq10] and two scalars $lpha _{1}$ and $lpha _{2}$. Then,[eq11]where: in step $ rame{A}$ we have used the fact that $f$ is a linear map; in step $ rame{B}$ we have used the fact that [eq12] and, as a consequence, [eq13]. Thus, the linear combination[eq14]belongs to the kernel (because the function $f$ maps it into the zero vector). This is exactly what we needed to prove.

Note that the zero vector always belongs to the kernel. In fact, the linearity of $f$ implies that[eq15]for any $sin S$ and any scalar $lpha $. Thus, when we set $lpha =0$, the previous equation becomes[eq16]

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let $S=T$ be the space of all $2 imes 1$ column vectors having real entries. Let $f:S ightarrow T$ be the linear map defined by [eq17]where [eq18]

Find the null space of $f$.

Solution

For any $sin S$, denote by $s_{1}$ and $s_{2}$ the two entries of $s$, so that[eq19]Hence, $As=0$ for $s_{1}=0$ and any value of $s_{2}$. Thus, the null space of $f$ is:[eq20]

Exercise 2

Remember that a linear transformation $f:S ightarrow T$ can be defined by specifying the values taken by $f$ in correspondence of a basis of $S$ (see the lecture on linear maps).

Let [eq21] be a basis for $S$. Let [eq22] be a basis for $T$. Suppose that $f$ is defined by[eq23]Find the kernel of $f$.

Solution

Any vector $sin S$ can be represented in terms of the basis $B$ as[eq24]where [eq25] are scalars. Then, by the linearity of $f$, we have that[eq26]So, we have that [eq27] whenever[eq28]Thus, [eq29]

Exercise 3

Let $S$ and $T$ respectively be the spaces of all $3 imes 1$ and $4 imes 1$ column vectors having real entries. Let $f:S ightarrow T$ be the linear map defined by the matrix product[eq30]where A is a $4 imes 3$ matrix.

Find the null space of $f$ under the hypothesis that the columns of A are linearly independent.

Solution

Denote by $A_{1},A_{2},A_{3}$ the three columns of A. For any $sin S$, denote by $s_{1},s_{2},s_{3}$ the three entries of $s$. The product $As$ can be written as a linear combination of the columns of A with coefficients taken from the vector $s$:[eq31]Since the three columns of A are linearly independent, the only linear combination that gives[eq32]is the combination with coefficients [eq33]. Therefore,[eq34]

How to cite

Please cite as:

Taboga, Marco (2021). "Kernel of a linear map", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/kernel-of-a-linear-map.

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