6.2: Null spaces
- Page ID
- 272
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.
Contributors
- Isaiah Lankham, Mathematics Department at UC Davis
- Bruno Nachtergaele, Mathematics Department at UC Davis
- Anne Schilling, Mathematics Department at UC Davis
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