Elements of vector space bases can be substituted with linear combinations? I have trouble proving (or at least understanding) following theorem.
Let $ B = \{{\bf v_1},\dots,{\bf v_n}\} $ be some basis of vector space $ V $. Assume that $ \textbf{x} = \sum_{i=1}^{n} \alpha_i {\bf v_i} $ where $ \alpha_i \in \mathbb{R} $ and $ \alpha_1 \neq 0 $. Then $ B' = \{{\bf x}, {\bf v_2},\dots,{\bf v_n}\} $ is a basis of $ V $.
As I see it, this theorem states that an element of a basis can be substituted with a linear combination of itself. But does that not contradict the properties of bases (ie linearly independence)?
 A: No, it does not contradict it, because you removed the $v_1$. 
Let us shows that $\{x, v_2, \dots, v_n\}$ is indeed linearly independent. 
Suppose $bx + \sum_{i=2}^n b_i v_i=0$. We need to show that $b=b_2 = \dots =b_n=0$.
We substitute for $x$, to get 
$b(\sum_{i=1}^n \alpha_i v_i ) + \sum_{i=2}^n b_i v_i=0$
We can rewrite this as
$b\alpha_1 v_1 + \sum_{i=2}^n (b\alpha_i+ b_i ) v_i ) =0$. 
Since the $v_1, \dots, v_n$ is linearly independent it follows that 
$b \alpha_1 = b\alpha_2+ b_2 = \dots = b\alpha_n+ b_n =0$. 
Since $\alpha_1 \neq 0$ (this is important!) one gets $b=0$, and then 
$b_2 = \dots = b_n  =0$, showing the claim. 
A: You substitute $v_1$ with another vector. The resulting set may or may not be a basis. Let's see some ways to show that, in this case, the resulting set is indeed a basis.
Method 1 — coordinates
The matrix having as columns the coordinates of the vectors in $B'$ with respect to the basis $B$ is
\begin{bmatrix}
\alpha_1 & 0 & 0 & \dots & 0 & 0 & 0 \\
\alpha_2 & 1 & 0 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
\alpha_{n-1} & 0 & 0 & \dots & 0 & 1 & 0 \\
\alpha_n & 0 & 0 & \dots & 0 & 0 & 1
\end{bmatrix}
which is lower triangular having nonzero entries on the diagonal, so it has rank $n$. A set is linearly independent if and only if the coordinate vectors of its elements are linearly independent.
Method 2 — adding elements to a linearly independent set

If $\{w_1,\dots,w_k\}$ is a linearly independent set and $y$ does not belong to the span of $\{w_1,\dots,w_k\}$, then $\{y,w_1,\dots,w_k\}$ is also linearly independent.

Proof. Suppose $\beta y+\beta_1w_1+\dots+\beta_kw_k=0$. Then either $\beta=0$ or $\beta\ne0$. If $\beta=0$, then linear independence of $\{w_1,\dots,w_k\}$ yields that also $\beta_1=\beta_2=\dots=\beta_k=0$. If $\beta\ne0$, then
$$
y=(-\beta)^{-1}\beta_1w_1+\dots+(-\beta)^{-1}\beta_kw_k
$$
belongs to the span of $\{w_1,\dots,w_k\}$, contradicting the hypothesis on $y$. □
Apply the result with $\{w_1,\dots,w_k\}=\{v_2,\dots,v_n\}$ and $y=x$. Since $\alpha_1\ne0$, the vector $x$ does not belong to the span of $\{v_2,\dots,v_n\}$.
Method 3 — spanning set

If $\{w_1,w_2,\dots,w_n\}$ is a spanning set for a space of dimension $n$, then $\{w_1,w_2,\dots,w_n\}$ is a basis.

This is a very basic result. Apply it to $w_1=x$ and $w_i=v_i$ for $i=2,\dots,n$. All it's needed is to show that $v_1$ belongs to the span of $\{x,v_2,\dots,v_n\}$ which is true because
$$
v_1=\alpha_1^{-1}(x-\alpha_2v_2-\dots-\alpha_nv_n)
$$
