Definition 4.4.1.
Let \(V,W\) be vector spaces over a field \(F\text{,}\) and let \(T\colon V\to W\) be an \(F\)-linear transformation. The kernel of \(T\) is
\begin{equation*}
\ker(T)=\{v\in V:T(v)=0\}.
\end{equation*}
Section 4.4 Kernel and image of a linear homomorphism
We define the kernel of a linear transformation.
Remark 4.4.2.
Note that \(\ker(T)=T^{-1}\{0\}.\)Lemma 4.4.3.
We keep the notations of Definition 4.4.1. The kernel of an \(F\)-linear transformation is a subspace of \(V\text{.}\)Lemma 4.4.4.
We keep the notations of Definition 4.4.1. The map \(T\) is injective if and only if \(\ker(T)=\{0\}\text{.}\)Proof.
Suppose that \(T\) is injective. If \(v\in V\) is such that \(T(v)=0\) then, we have \(T(v)=T(0)=0\) (see Checkpoint 4.1.3). Injectivity of \(T\) implies that \(v=0\text{.}\)
Now assume that \(\ker(T)=0\text{.}\) If \(x,y\in V\) are such that \(T(x)=T(y)\) then, by \(F\)-linearity of \(T\text{,}\) we have \(T(x-y)=T(x)-T(y)=0\text{.}\) Therefore, \(x=y\text{.}\)
We now define the image of a linear map.
Definition 4.4.5.
Let \(T\colon V\to W\) be an \(F\)-linear map between \(F\)-vector spaces \(V\) and \(W\text{.}\) The image of \(T\) is defined by
\begin{equation*}
\Im(T)=\{T(v):v\in V\}=T(V).
\end{equation*}
