# Projecting into non-orthogonal basis

I've 2 pairs of vectors, lets say $v_1 = (\sqrt3/2, 1/2)$, $v_2 = (1/2, \sqrt3/2)$ and they are supposed to span a space $V$. While having $x_0 = (1,1)$ By letting $Pv_1 . Pv_2 . x$ the projection of $x$ into space $V$, My question is what happens if I project multiple times onto that space, will n th projection converge (taking into account that the space basis are not orthogonal)? Also how can I prove that $Pv_1 . Pv_2 . x = \langle v_1,v_2\rangle \langle v_1,x \rangle v_2$. I tried to start with $pv_1 = \langle v_1,x\rangle v_1$, $pv_2 = \langle v_2,x\rangle v_2$ and go from there but I didn't end up having the same result.

I should first point out that $v_1$ and $v_2$ in fact span the whole $\mathbb R^2$. So you have that $V=\mathbb R^2$, $x\in V$ and projection of $x$ onto $V$ is $x$. However, the iterative process you describe will lead to the zero vector.
Since the vectors $v_1$ and $v_2$ have unit length, the number $$\langle v_1,v_2 \rangle = \frac{\sqrt3}2$$ is the length of the projection of $v_1$ into the direction of $v_2$ and, at the same time, the length of the projection of $v_2$ into the direction of $v_1$.
This means when you start with a vector which is a multiple of $v_2$, after projecting it into the direction of $v_1$ the length will be $\sqrt3/2$-multiple of the original length. So after taking $n$ projections the new length is $(\sqrt3/2)^n \|x_0\|$, so this process converges to the zero vector.