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

Quotient Spaces

This text is about the construction of quotient spaces. Mainly to test Blogger in conjunction with MathJax.

Given a vector space \(V\) over a field \(K\) and a subspace \(U\subseteq V\) one can define the relation 

\[\forall v,w\in V: v\sim w\Longleftrightarrow v-w\in U\]

which actually is an equivalence relation since it is

reflexive:

\[v\in V:v\sim v\Leftrightarrow v-v=0\in U\]

symmetric:

\[v,w\in V:v\sim w \Leftrightarrow \exists! u\in U:v-w=u \Leftrightarrow w-v=-u\in U\Leftrightarrow w\sim v\]

transitive:

\[v,w,q\in V: v\sim q\wedge q\sim w \Leftrightarrow \exists!u\in U:v-q=u \wedge \exists!u'\in U:w-q=u'\]

\[\Rightarrow u-u'=v-q-w+q=v-w\in U\Leftrightarrow v\sim w\]

The equivalence class \([v]\) of \(v\in V\) is given by 

\[[v]=\{w\in V|v\sim w\}\]

\(w\in[v]:v\sim w\Leftrightarrow \exists!u\in U:v-w=u\Rightarrow w = v-u\) which leads us to assume that

\[v+U:=\{v+u|u\in U\}=[v]\]

First we check if \(v+U\subseteq [v]\), so let \(w\in v+U\) then there exists \(u\in U\) such that \(w=v+u\). The last equation leads to \(w-v=u\Leftrightarrow w\sim v\Leftrightarrow w\in[v]\). Next we take a look at \([v]\subseteq v+U\). For every \(w\in[v]\) there is \(v\sim w\Leftrightarrow \exists! u\in U:v-w=u\Rightarrow w=v-u \in v+U\). So the assumption

\[v+U=[v]\]

is valid.

Quotient Space

The equivalence classes \(v+U\) of \(V\) define a vector space \(V/U\), a so called quotient space, if we define the vector addition

\[[v]+[w]:=[v+w]\forall v,w\in V\]

and the scalar multiplication 

\[\lambda [v] := [\lambda v] \forall \lambda \in K;\forall v \in V\]

Since an equivalence class can be represented by several vectors, the first step should be to show that the above operations are actually well defined. So we take vectors \(v,v',w,w'\in V:[v]=[v']\wedge [w]=[w']\) which should yield \([v+w]=[v'+w']\) or in other words

\[v+w\sim v'+w'\Leftrightarrow v-v'+w-w' \in U\] 

Because there are \(u,u'\in U\) such that \(v-v' = u\) and \(w-w'=u'\) we get

\[ v-v'+w-w'\Leftrightarrow u+u'\in U\]

which just shows that \(v+w\) and \(v'+w'\) describe the same equivalence class. The same idea we can apply to show that the scalar multiplication is well defined as well. What is left is the proof that the addition and scalar multiplication satisfy a vector space which is just long winding and technical, but quite easy.

What about the dimension of a quotient space? Lets assume the vector space \(V\) is of finite dimension \(n\) then \(dimU=:m\leq n\). So we can take a basis sequence \((b_i)_{i=1,2,...,m}\) of \(U\) and extend it to be sequence of \(V\) with another vector sequence \((b_i)_{i=m+1,m+2,...,n}\).  The equivalence class of any vector taken from \(U\) is actually the subspace \(U\) itself. We can partition the vector space \(V-U\) with vector subspaces \(V',W'\). Then we get for any non zero \(v\in V',w\in W'\)

\[v+U \neq w+U\]

since \(v\sim w\) does not exist. We can extend this idea and realize that the equivalence classes \(b_i+U\) for \(i=m+1,m+2,...,n\) are pairwise disjoint and for \(i=1,2,...,m\) they are equal \(0+U\). We can assume that 

\[B:=(b_i + U)_{i=m+1,m+2,...,n}\]

is a basis sequence of the quotient space. So this set should be linearly independent as well as a generating system of the quotient space. When we take a look at the following equation

\[0+U=\sum_{i=m+1}^{n}\lambda_ib_i+U\]

whereby the \(\lambda_i\) are field elements of \(K\), then we can almost directly see that the presumed basis of the quotient space is indeed linearly independent, just note that \(\sum_{i=m+1}^n \lambda_ib_i = 0\Rightarrow \lambda_i = 0\forall i\). Further we can't express any element of \(U\) as a linear combination of the vectors \((b_i)_{i=m+1,m+2,...,n}\). What's left is to show that \(B\) spans \(V/U\). So we take a vector \(v+U\in V/U\). To avoid any triviality we let \(v\notin U\). So there should exist field elements \(\lambda_i\) such that

\[v+U=\sum_{i=m+1}^n\lambda_ib_i+U\]

Such elements do exist since \(v\notin U\), so the same field elements we used above also satisfy 

\[v=\sum_{i=0}^n\lambda_ib_i\]

\(B\) is shown to be linearly independent as well as a generating system of \(V/U\) and therefore a basis system. As a direct result we also get the dimension formula of quotient spaces

\[dim V= dim (V/U) -dim U\]