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