6.2: Null spaces

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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.

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.
The identity map \(I:V\to V \) is injective.
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 \(\kernel(T) = \{0\} \).
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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