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So in my linear algebra I. We proved the rank-nullity theorem.

As I understood the concept of image and kernel, the image does contain the kernel (since it`s a vektorspace, and i is defined as follow
{T(v) | v \in V} ), because according to the definition all the vectors are sent to zero are part of the Image as well).

We defined a basis for the kernel

B := { v_1,.., v_n }

And then we defined a set which should be the basis for the image

S := { T(v_k+1),..., T(V_n)}.

BUT doesn`t the image contain the kernel? So why are we taking them out?

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    The image and kernel of $T:V\to W$ are subspaces of different spaces, so in general one can’t be contained in the other. The kernel of $T$ is a subspace of the *domain* $V$: the vectors “sent to zero” are elements of $V$. The image, on the other hand, is a subspace of the *codomain* $W$—$T(v)\in W$ for any $v\in V$. Even in the case of $T:V\to V$, the kernel of $T$ need not be contained within its image. – 2017-01-03
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    Oh, yeah. That`s right. Although, even then I don`t understand why we take out the Image of those basis vectors, which are part of the kernel. Because if 0 is part of the image, then don`t we have to take the image of kernel to get the whole image? – 2017-01-03
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    You’re not taking $0$ out. $T(0)=0$, and $0\in V$ is in the span of $\{v_{k+1},\dots,v_n\}$. You do have to remove the (images of) the kernel basis vectors, though. If you include any $T(v)$ for $v\in\ker T$ in your proposed basis for the image, you end up with a set of vectors that’s not linearly independent—it includes $0$—so can’t be a basis. – 2017-01-03
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    Yes, but aren`t those 0 different 0 then the ones which are the image of the kernel? – 2017-01-03
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    No. There’s only one $0$ in any vector space. The kernel of $T$ gets mapped to $0\in W$, which is the same $0$ that you get with $T(0)$. – 2017-01-03
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    if we only had the zero vector, that would be linearly indipendent, so i don`t understand the arguement, why including 0 would make it linearly dependent. and we actually proove, that \{ v_1,..,v_n} are all linearly indipendent form each other in the proof. – 2017-01-03
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    yeah, sure. Totally right with the one zero in a vector space. – 2017-01-03
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    My question would just be: why do we take out those vectors which are the image of the basis of the kernel? – 2017-01-03

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(Collecting up comments into an answer. I’ll use subscripts to distinguish zero vectors in different vector spaces.)

For a linear map $T:V\to W$, the set of vectors that get “sent to zero” by $T$ is a subset (a subspace, in fact) of the domain $V$. The image of $T$, on the other hand, is a subspace of the codomain $W$. Thus in general the kernel and image of a linear map belong to different vector spaces and so the image can’t contain the kernel. Even in the case of an endomorphism $T:V\to V$ this need not be the case. For instance, the kernel of the zero map $T:v\mapsto 0_V$ is the entire space and contains $T$’s image $\{0_V\}$ instead of vice-versa. For a non-trivial example, consider an orthogonal projection onto a non-trivial proper subspace of $V$: the kernel and image have only the trivial intersection $\{0_V\}$.

The proof you’re asking about starts by choosing a basis $\{v_{r+1},\dots,v_n\}$ for $\ker(T)$ and then extending it to a complete basis $\{v_1,\dots,v_{r},v_{r+1},\dots,v_n\}$ for $V$. (Note that I’ve swapped the order of the kernel basis and the extension from that in the question, so the latter has the lower indices.) It should be pretty obvious that the set $S=\{Tv_1,\dots,Tv_n\}\subset W$ of images of these vectors under $T$ spans $\operatorname{im}(T)$. If $T$ has a nontrivial kernel, though, there’s a problem: this set can’t form a basis for the image because it’s linearly dependent. There’s at least one kernel basis vector, which means that $r\lt n$ and $Tv_i=0_W\in S$ for $i\gt r$.[1] So, we discard all of those kernel basis vectors and take the images of the remaining basis vectors of $V$: $S=\{Tv_1,\dots,Tv_r\}$. All that we’ve really done is to remove $0_W$ from $S$—by construction none of the remaining vectors are $0_W$—so we haven’t changed its span. We still have $0_W$ within the span of $S$, of course. It remains to be shown that this smaller set is linearly independent, but I think that’s beyond the scope of your question.

[1]: No set that contains the zero vector, including the singleton $\{0\}$, can be linearly independent. Since $k0=0$ for all scalars $k$, a non-trivial linear combination that equals $0$ can be created by choosing $k\ne0$ and multiplying all other vectors in the set by the scalar $0$.