# Null Space of a Linear Map

Definition: If $T \in \mathcal L (V, W)$ then the Null Space or Kernel of the linear transformation $T$ is the subset of $V$ defined as $\mathrm{null} (T) = \{ v \in V : T(v) = 0 \}$, that is, the null space of $T$ is the set of vectors from $V$ that are mapped to the zero vector in $W$ under $T$. |

Before we look at some examples of the null spaces of linear transformations, we will first establish that the null space can never be equal to the empty set, in other words, the null spaces always contains at least one element from $V$, namely, the zero vector from $V$.

Lemma 1: If $T \in \mathcal L (V, W)$ then the null space of $T$ contains at least one element from $V$, that is $\mathrm{null} (T) \neq \emptyset$. |

**Proof:**Since $V$ is a vector space, by definition $V$ contains an additive identity, call it $0_V$. Therefore $0_V = 0_V + 0_V$, and so $T(0_V) = T(0_V + 0_V)$. Since $T$ is a linear transformation, we note that $T(0_V + 0_V) = T(0_V) + T(0_V)$. Therefore $T(0_V) = T(0_V) + T(0_V)$ which implies that $T(0_V) = 0_W$. Therefore $0_V \in \mathrm{null} (T)$, and so $\mathrm{null} T \neq \emptyset$. $\blacksquare$

From the lemma above, we conclude that the zero vector (additive identity) in $V$ is always mapped to the zero vector in $W$. Now the following lemma will tell us that the null space of $T$ is also a subspace of $V$.

Lemma 2: If $T \in \mathcal L (V, W)$ then the subset $\mathrm{null} (T)$ is a subspace of $V$. |

**Proof:**Since $\mathrm{null} (T) \subseteq V$, all we must do is verify that $\mathrm{null} (T)$ is closed under addition, closed under scalar multiplication, and contains the zero vector.

- Let $u, v \in \mathrm{null} (T)$ and $a \in \mathbb{F}$.

- Since $u, v \in \mathrm{null} (T)$ we have that by the definition of the null space of $T$ that $T(u) = 0$ and $T(v) = 0$. Since $T$ is a linear transformation we also have that $0 + 0 = T(u) + T(v) = T(u + v) = 0$, and so $(u + v) \in \mathrm{null} (T)$. Therefore $\mathrm{null} (T)$ is closed under addition.

- Now since $T(u) = 0$, we have that $aT(u) = a0 = 0$. Since $T$ is a linear transformation, then $aT(u) = T(au) = 0$, and so $(au) \in \mathrm{null} (T)$. Therefore $\mathrm{null} (T)$ is closed under scalar multiplication.

- From Lemma 1, we verified that $0_V \in \mathrm{null} (T)$ and so $\mathrm{null} (T)$ contains the zero vector. Therefore $\mathrm{null} (T)$ is a subspace of $V$. $\blacksquare$

We will now look at some examples of null spaces.

## The Null Space of The Zero Map

If $0 \in \mathcal L (V, W)$ represents the zero map, then $\mathrm{null} (T) = V$, since all vectors $v \in V$ are mapped to the zero vector $0_W \in W$ by the definition of the zero map.

## The Null Space of The Identity Map

If $I \in \mathcal L (V, V)$ represents the identity map, then $\mathrm{null} (T) = \{ 0 \}$. We note that all vectors $v \in V$ are mapped back to themselves, and so the only vector in $V$ that mapped to zero is $0$ itself.

## The Null Space of The Differentiation of Polynomials Map

If $T \in \mathcal L (\wp (\mathbb{R}), \wp (\mathbb{R}))$ represents the transformation defined by $T(p(x)) = p'(x)$, then the null space of $T$ is the set of all polynomials whose first derivatives are equal to zero, that is, the set of constant functions $p(x) = a_0$, $a_0 \in \mathbb{R}$. So $\mathrm{null} (T) = \{ p(x) : p(x) = a_0, \: a_0 \in \mathbb{R} \}$.

## The Null Space of The Left Shift Operator

If $T \in \mathcal L (\mathbb{F}^{\infty}, \mathbb{F}^{\infty})$ represents the left shift operator, then the null space of $T$ is the set of infinite sequences, all of whose terms are zero. Thus any sequence in the form $(a_1, 0, 0, …)$ is contained in the null space since $T((a_1, 0, 0, …)) = (0, 0, …)$, so $\mathrm{null} (T) = \{ (a_1, 0, 0, …) : a_1 \in \mathbb{F} \}$.

## The Null Space of The Right Shift Operator

If $T \in \mathcal L (\mathbb{F}^{\infty}, \mathbb{F}^{\infty})$ represents the right shift operator, then the null space of $T$ is the set of infinite sequences, all of whose terms are zero once again. The only such sequence is the zero sequence in $\mathbb{F}^{\infty}$, since $T((a_1, a_2, …)) = (0, a_1, …)$ we must have that $a_1 = a_2 = … = 0$, and so $\mathrm{null} (T) = \{ (0, 0, … ) \}$.

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