Kern and Image of Linear Transformations

Let \(V,W\) be vector spaces over the field \(K\) and \(f:V\longrightarrow W\) a linear transformation.

Kernel of \(f\) is a subspace of \(V\)

The kernel of \(f\) also denoted as \(\mathrm{ker} f\) is a set of vectors which are mapped to zero by \(f\). The zero vector of \(V\) is in the kernel since \(f(0_V)=f(0_K0_V)=0_Kf(0_V)=0_W\) so \(\mathrm{ker} f\) is never  empty. Let us see if the kernel is complete regarding vector addition. To see that just take any two vectors \(v,w\in\mathrm{ker}f\) and check if their sum is also in \(\mathrm{ker}f\).

\[u:=v+w; f(f+w)=f(v)+f(w)=0+0=0=f(u)\Rightarrow u\in\mathrm{ker}f\] The same idea we apply to scalar vector multiplaction. \[\lambda\in K;u:=\lambda v; f(\lambda v)=\lambda f(v)=\lambda 0 = 0 = f(u) \Rightarrow u\in\mathrm{ker}f\] So the kernel of f is a subspace of \(V\).

Image of \(f\) is a subspace of \(W\) 

The image of our linear transformation,  denoted as \(\mathrm{img}f\) or  \(\mathrm{im}f\) , is the set \[\mathrm{img}f:=\{w\in W|\exists v\in V:f(v)=w\}\] Any vector taken from the kern of \(f\) is transformed to the zero vector in \(W\). So the image of \(f\) is not empty since \(\mathrm{ker}f\neq \{\}\). \[\forall w,w'\in \mathrm{img}f:\exists v,v'\in V:w=f(v)\wedge w'=f(v')\] \[w+w'=f(v)+f(v') = f(v+v')\] \[\lambda\in K; \lambda w = \lambda f(v)=f(\lambda v)\] The image is complete regarding vector sums and scalar vector multiplications which makes it a subspace of \(W\).

Dimensional Relationship

The linear transformation \(f:\mathbb{R}^3\rightarrow\mathbb{R}^3:(x,y,z)\mapsto (0,y,z)\)  implies that \[\mathrm{dim}\;\mathrm{ker}f +\mathrm{dim}\;\mathrm{img}f=\mathrm{dim}\mathbb{R}^3\] and also suggests that the more general formula for any linear transformation \(f:V\rightarrow W\)

\[ \mathrm{dim}\;\mathrm{ker}f +\mathrm{dim}\;\mathrm{img}f=\mathrm{dim}V \]

is valid as well.

To verify such a relationship we need to examine the properties of the kern and image space regarding \(f\) and to do that we can use quotient spaces. These spaces are useful when you want to examine a subspace by making use of the properties of it's superspace. In this case we want to look at the space \(V/\mathrm{ker}f\) for which we have the dimensional relationship given as \[\mathrm{dim}V=\mathrm{dim} V/\mathrm{ker}f + \mathrm{dim}\;\mathrm{ker}f\] A way to prove the initial guess is to show that \( \mathrm{dim} V/\mathrm{ker}f  = \mathrm{dim}\;\mathrm{img}f\) or in other words that \( V/\mathrm{ker}f \cong \mathrm{img}f \). One way to do so is to construct an isomorphism \(f^*: V/\mathrm{ker}f\rightarrow \mathrm{img}f\). As a first guess of such a linear transformation one can suggest for all  \(v+ \mathrm{ker}f\)\[v+ \mathrm{ker}f\mapsto \bar{f}(\{v+u|u\in \mathrm{ker}f \}) :=\{f(v)+f(u)|u\in \mathrm{ker}f \}=\{f(v)\}\] When we define the sum of vectors and a scalar vector multiplication on \(\mathrm{img}\bar{f}\)  for all \(v,w\in\mathrm{img}\bar{f}\) and \(\lambda\in K\) as follows \[\{v\}+\{w\}:=\{v+w\}\;\wedge\;\lambda\{v\}:=\{\lambda v\}\] the set \( \mathrm{img}\bar{f}\) becomes a vector space and also it is easy to see that \(\bar{f}\) is a linear transformation. Further we define a trivial linear transformation \(\hat{f}:\mathrm{img}\bar{f}\rightarrow W:\{v\}\mapsto v\) and set \[f^*:=\hat{f}\circ\bar{f}\] At first we check if \(\bar{f}\) is injective and take any two vectors \(v+\mathrm{ker}f, w+\mathrm{ker}f\in V/ \mathrm{ker}f:v+ \mathrm{ker}f \neq w+ \mathrm{ker}f \). If we apply \(\bar{f}\) we get \[\bar{f}(v+ \mathrm{ker}f-w+ \mathrm{ker}f )=\{f(v-w)\}=\{f(v)-f(w)\}\] and since the equivalence classes \(v+ \mathrm{ker}f\) and \(w+ \mathrm{ker}f\) are not equal the expression \(v-w\notin\mathrm{ker}f\) is valid, which yields \[f(v-w)=f(v)-f(w)\neq 0\Longrightarrow f(v)\neq f(w)\] or in other words the linear transformation \(f^*\) is injective.

\(f^*\) is also a surjection. Suppose there is a vector \(v\in V\) for which a \(w+ \mathrm{ker}f\in V/ \mathrm{ker}f\) can't be found such that \(f^*(w+ \mathrm{ker}f)=f(v)\). If \(v\in\mathrm{ker}f\) then \[f(v)=0=f^*(v+ \mathrm{ker}f)=f^*(0+ \mathrm{ker}f)=0f^*(0+ \mathrm{ker}f)\] implies that such a \(v\) does not exist since we can simply find a \(w=0\) (remember that any member of an equivalence class can be a representative). The case \(f(v)\neq 0\) leads directly to \(f^*(v+ \mathrm{ker}f)=f(v)\) if we set \(w=v\). As a result we can say:

For vector spaces \(V\) and \(W\) and a linear mapping \(f:V\longrightarrow W\) the following relation is true. \[ \mathrm{dim}\;\mathrm{ker}f +\mathrm{dim}\;\mathrm{img}f=\mathrm{dim}V \]