I have a theorem stating, that the non-zero-row-vectors of the row reduced echelon form of $A$ form the basis for the row-space of $A$.
As can be seen in the image I posted, the row vectors of $A$ contain $5$ coordinates. As an example the first row-vector: $(1,-3,2,2,1)$.
How come then, after I've $RREF(A)$, I only get $3$ non-zero-row-vectors as the basis for these? This means, that the vectors are 3D.
This goes against what I've been taught. That if we have a set of vectors S in $R^n \implies dim(S) = n$.
What is the explanation for this?
From our discussion in chat, it seems like the key idea missing is the following:
Suppose you have a matrix, like so: $$M=\begin{bmatrix}1&0\\2&0\end{bmatrix}$$ (I'm going to use $2\times2$ so that we can draw pictures.) It's natural to want to study matrix multiplication involving $M$; indeed, $M$ defines a function $L^*_M$ sending a row vector $\vec{v}^*$ to $\vec{v}^*M$.Fn. 1
Since $L^*_M$ is defined on two-vectors and produces two-vectors, we can summarize it as sending $\mathbb{R}^2\to\mathbb{R}^2$. But this notation doesn't mean that $L^*_M$ hits everything in $\mathbb{R}^2$. A useful analogy is $(\cdot)^2\in\mathbb{R}\to\mathbb{R}$. Taking squares sends real numbers to real numbers, but you can't produce a negative number this way. We said $(\cdot)^2$ sends things to $\mathbb{R}$, because it produces real numbers — $\mathbb{R}$ intuitively seems like "a natural category of objects produced by my function." The name for that sort of set is a codomain.
Conversely, the space of things $L^*_M$ does hit is called the range of $L^*_M$.
Indeed, if you compute $\vec{v}^*M$, you'll see that any such output must have second coordinate $0$. So it isn't really a 2D object. Instead, we define the dimension of a vector space to be the size of the any basis.Fn. 2 A basis for "elements of $\mathbb{R}^2$ with second coordinate $0$" is the single element $\begin{bmatrix}1&0\end{bmatrix}$, so we say that the range is 1D, even though the codomain (all of $\mathbb{R}^2$) is 2D.
Interestingly, the range of $L^*_M$ is also the row space of $M$. This is not a concidence! The row space of $M$ is the set of linear combinations of rows of $M$ and $$\begin{bmatrix}\lambda_1&\lambda_2\end{bmatrix}M=\lambda_1\begin{bmatrix}1&0\end{bmatrix}+\lambda_2\begin{bmatrix}2&0\end{bmatrix}$$ You can always get the row space of a matrix this way. So, to figure out the dimension of the row space, you need to do a little more than just "count the number of coordinates I wrote down." One technique is to find a basis of the range of $L^*_M$ and (it turns out) RREF does this for you.Fn. 3, 4
In your example, $A$ is $5\times5$, so $L_A^*\in\mathbb{R}^5\to\mathbb{R}^5$. Your codomain is 5-dimensional, but the range of $L_A^*$ is much smaller. Wolfram|Alpha confirms your RREF calculation: the row space of $A$ — which is also the range of $L_A^*$ — is 3D, because it has a $3$-element basis.
Fn. 1: $M$ also defines a function $L_M$ sending a column vector $\vec{v}$ to $M\vec{v}$. Most books use that definition first, but it isn't useful for studying the row space.
Fn. 2: It might not be obvious to you that any two bases have the same size. The proof takes a couple lines, and the key fact is the Steinitz exchange lemma.
Fn. 3: If you're confused about why, ask your linear algebra instructor.
Fn. 4: There are other techniques you'll learn later too. The rank-nullity theorem is key, because it lets you count related objects instead.