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In $\mathbb{R}^2$, given one-dimensional disjoint subspaces $V_1$ and $V_2$, and orthogonal projections $P_1$ and $P_2$, I can write the decomposition for $v$ in the conic combination of $V_1$ and $V_2$ as

$$v = \frac{\| v-P_2v\|_2}{c_1\|P_1 v\|_2}P_1 v + \frac{\| v-P_1v\|_2}{c_2\|P_2 v\|_2}P_2 v $$

where $$c_1=c_2 = \frac{1}{\sqrt{1-\frac{(\langle P_1 v,P_2 v\rangle)^2}{\|P_1 v\|_2^2\|P_2v\|_2^2}}}$$

This was obtained by doing some basic high school geometry. I have a few questions about this.

When $v$ is not in the conic combination of $V_1$ and $V_2$ it seems like $c_i$ is replaced by $-c_i$ in one of the two summands of the decomposition. Clearly this is from $c_i$ looking like the $\sin$ of the angle $V_1$ and $V_2$ make. I wanted an expression with $P_i$ and $v$ only so I used the cosine expression in terms of the dot product. $c_i$ is replaced in the $V_i$ for which $v$ is "farther" away from and gets projected to the negative part. Is there any way to quickly check that a vector $v$ would be in the conic combination or outside of it?

I also am curious about ways to generalize this decomposition to higher dimensions.

Thank you, I have some quick and dirty matlab code if anyone wants to test this out.

b1 = [2;3]; %basis for V_1
b2 = [-1;-3]; %basis for V_2
v = [-1;0];
P1 = b1*inv(b1'*b1)*b1';
P2 = b2*inv(b2'*b2)*b2';
c = sqrt(1- (dot(P1*v,P2*v)/norm(P1*v)/norm(P2*v))^2);
v1 = norm(v-P2*v)/norm(P1*v)/c*P1*v;
v2 = norm(v-P1*v)/norm(P2*v)/c*P2*v;

1 Answers 1

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Compute the following angles:

$$ \theta_1 = \arccos\left(\frac{\langle v_1,v\rangle}{\|v_1\| \|v\|}\right)\\ \theta_2 = \arccos\left(\frac{\langle v_2,v\rangle}{\|v_2\| \|v\|}\right)\\ \theta = \arccos\left(\frac{\langle v_1,v_2\rangle}{\|v_1\| \|v_2\|}\right) $$

If $|\theta_1| + | \theta_2 | = |\theta|$, then $v \in \text{Cone}(v_1,v_2)$.