
# 6.2 Null spaces

Definition 6.2.1.  Let $$T:V\to W$$ be a linear map. Then the null space (a.k.a.~kernel) of $$T$$ is the set of all vectors in $$V$$ that are mapped to zero by $$T$$. I.e.,

$\begin{equation*} \kernel(T) = \{v\in V \mid Tv=0\}. \end{equation*}$

Example 6.2.2.  Let $$T\in \mathcal{L}(\mathbb{F}[z],\mathbb{F}[z])$$ be the differentiation map $$Tp(z)=p'(z)$$. Then

$\begin{equation*} \kernel(T) = \{ p \in \mathbb{F}[z] \mid p(z) \rm{~ is~ constant~}\}. \end{equation*}$

Example 6.2.3.  Consider the linear map $$T(x,y)=(x-2y,3x+y)$$ of Example 6.1.2. To determine the null space, we need to solve $$T(x,y)=(0,0)$$, which is equivalent to the system of linear equations

$\begin{equation*} \left. \begin{array}{rl} x-2y&=0\\ 3x+y&=0 \end{array} \right\}. \end{equation*}$

We see that the only solution is $$(x,y)=(0,0)$$ so that $$\kernel(T) =\{(0,0)\}$$.

Proposition 6.2.4. Let $$T:V\to W$$ be a linear map. Then $$\kernel(T)$$ is a subspace of $$V$$.

Proof.

We need to show that $$0\in \kernel(T)$$ and that $$\kernel(T)$$ is closed under addition and scalar multiplication. By linearity, we have

$\begin{equation*} T(0) = T(0+0) = T(0) + T(0) \end{equation*}$

so that $$T(0)=0$$. Hence $$0\in \kernel(T)$$. For closure under addition, let $$u,v\in\kernel(T)$$. Then

$\begin{equation*} T(u+v) = T(u) + T(v) = 0 + 0 = 0, \end{equation*}$

and hence $$u+v\in\kernel(T)$$. Similarly, for closure under scalar multiplication, let $$u\in \kernel(T)$$ and $$a\in \mathbb{F}$$. Then
$\begin{equation*} T(au) = aT(u) = a0=0, \end{equation*}$

and so $$au\in\kernel(T)$$.

Definition 6.2.5.  The linear map $$T:V \to W$$ is called injective if, for all $$u,v\in V$$, the condition $$Tu=Tv$$ implies that $$u=v$$. In other words, different vectors in $$V$$ are mapped to different vectors in $$W$$.

Proposition 6.2.6.  Let $$T:V\to W$$ be a linear map. Then $$T$$ is injective if and only if $$\kernel(T)=\{0\}$$.

Proof.

$$( "\Longrightarrow" )$$ Suppose that $$T$$ is injective. Since $$\kernel(T)$$ is a subspace of $$V$$, we know that $$0\in \kernel(T)$$. Assume that there is another vector $$v\in V$$ that is in the kernel. Then $$T(v)=0=T(0)$$. Since $$T$$ is injective, this implies that $$v=0$$, proving that $$\kernel(T)=\{0\}$$.

$$( "\Longleftarrow" )$$ Assume that $$\kernel(T)=\{0\}$$, and let $$u,v\in V$$ be such that $$Tu=Tv$$. Then $$0=Tu-Tv=T(u-v)$$ so that $$u-v\in \kernel(T)$$. Hence $$u-v=0$$, or, equivalently, $$u=v$$. This shows that $$T$$ is indeed injective.

Example 6.2.7.

1. The differentiation map $$p(z) \mapsto p'(z)$$ is not injective since $$p'(z)=q'(z)$$ implies that $$p(z)=q(z)+c$$, where $$c\in\mathbb{F}$$ is a constant.
2. The identity map $$I:V\to V$$ is injective.
3. The linear map $$T:\mathbb{F}[z] \to \mathbb{F}[z]$$ given by $$T(p(z)) = z^2 p(z)$$ is injective since it is easy to verify that $$\kernel(T) = \{0\}$$.
4. The linear map $$T(x,y)=(x-2y,3x+y)$$ is injective since $$\kernel(T)=\{(0,0)\}$$, as we calculated in Example 6.2.3.

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