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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.

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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.

I will also add that in general if you have a subspace spanned by orthogonal vectors, you can get the projection onto the subspace as the sum of projections into directions of these vectors. (So you have to add the projections, not apply them consecutively as suggested in the question.) But this does not work (in general) if the spanning vectors are not orthogonal.

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  • $\begingroup$ Thanks alot I see it now! $\endgroup$ – Assem Jul 30 '18 at 17:48

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