Let $X:=\{x \in R^n |Ax=b \}$ and $I \subset \{1,...,n\}$ such that, $b\in C([\{a_i\}_{i \in I}$]). Prove that for every set $B \subset \{1,...,n\}$, such that $\{a_i\}_{i \in B}$ is linearly independent and $I \subset B$, implies that $x_i=0 , \forall i \notin I$.

Obs: $x=(x_1,x_2,...,x_n)$

Regarding notation , the set $C([\{a_i\}_{i \in I}$]) stands for the cone generated by the convex hull of the columns of the matrix A.

I'm curretnly having trouble with this question due to the fact that I'm not sure how to use the fact that I is a subset of B, neither how to use the linear independence over this set of indices.

Thanks in advance

  • $\begingroup$ What is $x_i$? ${}$ $\endgroup$ – copper.hat Nov 17 '19 at 22:01
  • $\begingroup$ Do you mean for all $x \in X$, ... $\endgroup$ – copper.hat Nov 17 '19 at 22:05
  • $\begingroup$ In the last line, where is $i$ contained in? If we are saying $\forall i \in B$ such that $i\not\in I$ then the statement follows from linear independence. $\endgroup$ – Tom Nov 17 '19 at 22:07
  • $\begingroup$ There's no information regarding where the index i is actually contained, just that $i \notin I$ $\endgroup$ – Guilherme takata Nov 17 '19 at 22:12

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