Only 3 Basis Vectors for $\mathbb{R}^4$. I've heard that for $\mathbb{R}^n$, there can only be $n$ number of basis vectors for it. Is that really the case though? Or am I having a misconception somewhere.
Let's take the example of $\mathbb{R}^4$. It seems that I can pick only 3 basis vectors $(1,1,0, 0)$, $(0,1,1,0)$ and $(0,0,1,1)$.
Firstly, these basis vectors are linearly independent. For example, no linear combination of $(0,1,1,0)$ and $(0,0,1,1)$ yields $(1,1,0,0)$. This is because these linear combinations will still have a zero as the first number in the vector.
Secondly, it (seems) that these 3 vectors span $\mathbb{R}^4$. 
 A: Any vector in the span of the three vectors you named will be of the form
$$\left(\begin{matrix} a \\ a + b \\ b + c \\ c\end{matrix}\right)$$
Suppose we want to choose constants $a, b, c$ to make this equal to the vector $(1, 2, 3, 4)^T$. Then starting at the top and working down, we get $a = 1, b = 1, c = 2$, which clearly doesn't work.

This is an example of an overdetermined system, which can always appear when you have fewer basis vectors than coordinates.
A: Can $(0,0,0,1)$ be written as a linear combination of these vectors?
A: Well in order for a group of vectors to be considered a basis they need to be linearly independent and span the space. Does your basis span $R^4$?
A: The first statement of your question is absolutely correct that's what is referred to as the dimension of the vector space. 
But coming to the example, we can see that your assumed basis vector (1,1,0,0), (0,1,1,0), and (0,0,1,1) does not span the entire vector space. If you add a vector (1,0,0,0), this completes your basis vector which spans the entire 4 dimensional vector space.
