Sum of rejection matrices of non-parallel vectors is invertible

Question

I need to prove or disprove that if I have n>1 non-parallel unit vectors: $$\{V_1, V_2, ..., V_n\}$$ then the sum of the rejection matrices of those vectors is invertible, i.e.: $$\sum_{i=1}^n{(I-V_iV_i^t)}$$is invertible.

Answer 1

You don't need all vectors to be non-parallel to each other. As long as some two of them are not parallel, the sum is invertible.

Suppose the sum is singular. Then $\sum_i(I-v_iv_i^t)x=0$ for some unit vector $x$. Hence $$ 0=x^t \sum_i(I-v_iv_i^t)x=\sum_i\left(1-\langle x,v_i\rangle^2\right).\tag{1} $$ However, by Cauchy-Schwarz inequality $1-\langle x,v_i\rangle^2$ is always nonnegative, and it is zero if and only if $x$ is parallel to $v_i$. Thus $(1)$ implies that $x$ is parallel to every $v_i$. Yet, this is impossible because there is a pair of linearly independent vectors among $v_1,v_2,\ldots,v_n$. Hence $\sum_i(I-v_iv_i^t)$ must be invertible.