Null Spaces of Linear Transformations

Why

When is a linear transformation between $V$ and $W$ one-to-one? In other words, when does

\[ Tx = Ty \Rightarrow x = y \]

Rearranging, and using additivity, we ask when

\[ T(x - y) = 0 \Rightarrow x - y = 0 \]

Clearly we are interested in vectors $z$ for which $Tz = 0$.

Definition

Suppose $T \in \mathcal{L} (V, W)$. The null space (or kernel) of $T$ is the set of vectors in $V$ which are mapped to $0$ under $T$. In symbols, the null space of $T$ is the set

\[ \null T = \Set{v \in V}{Tv = 0} \]

The word “null” means “zero” in German.

A subspace

Why use the term space? Well, $\null T$ is a subspace of $V$.

Suppose $T \in \mathcal{L} (V, W)$. Then $\null(T)$ is a subspace of $V$.

We verify that $\null(T)$ contains $0$ and is closed under vector addition and scalar multiplication. First, $0 \in \null T$ since $T0 = 0$ by homogeneity. Second, by additivity, if $x, y \in \null T$, then

\[ T(x + y) = Tx + Ty = 0 + 0 = 0 \]

Third, if $u \in \null T$ and $\alpha \in \F $, then

\[ T(\lambda u) = \lambda (Tu) = \lambda 0 = 0 \]

Characterization of injectivity

Suppose $T \in \mathcal{L} (V, W)$. Then

\[ \null T = \set{0} \iff T \text{ is one-to-one} \]

If $\null T = \set{0}$ we say that $T$ has zero nullspace or trivial nullspace.

Examples

Zero map. Suppose $T$ is the zero map from $V$ to $W$. In other words,

\[ Tv = 0 \quad \text{for all } v \in V \]

Then $\null T = V$. I.e., the null space is the whole space.

Simple function on $\C ^3$. Define $\phi \in \mathcal{L} (\C ^3, \C )$ by

\[ \phi (z_1, z_2, z_3) = z_1 + 2z_2 + 3z_3 \]

Then $\null \phi $ is

\[ \Set{(z_1,z_2,z_3) \in \C ^3}{z_1 + 2z_2 + 3z_3 = 0} \]

This is the solution set of a linear equation.

Polynomial differentiation. Suppose $D \in \mathcal{L} (\mathcal{P} (\R ), \mathcal{P} (\R ))$ is the linear map defined by

\[ Dp = p' \quad \text{for all } p \in \mathcal{P} (\R ) \]

In other words, $Dp$ is the derivative of the polynomial $p$. Then $\null(D)$ is the set of constant functions.

Multiplication by $x^2$. Define $T \in \mathcal{L} (\mathcal{P} (\R ), \mathcal{P} (\R ))$ by

\[ (Tp)(x) = x^2p(x) \quad \text{for all } x \in \R \text{ and } p \in \mathcal{P} (\R ) \]

Then $\null(T) = \set{0}$, since no other polynomial satisfies $x^2p(x) = 0$ for all $x \in \R $.

Backward shift. Define $T \in \mathcal{L} (\F ^\N , \F ^\N )$ by

\[ T(x_1, x_2, x_3, \dots , ) = (x_2, x_3, \dots , ) \]

so that $T$ is the backward shift Then $T(x_1, x_2, x_3, \dots ) = 0$ if and only $x_2 = x_3 = \cdots = 0$. So

\[ \null T = \Set{(\alpha , 0, 0, \dots ,)}{\alpha \in \F } \]