Joining orthogonal linindep sets

Dependencies:

  1. Inner product space
  2. Linear independence
  3. Span
  4. Gram-Schmidt Process
  5. A set of mutually orthogonal vectors is linearly independent
  6. Inner product is anti-linear in second argument

Let $A$ and $B$ be sets of vectors from an inner-product space such that $A$ is linearly independent, $B$ is linearly independent, and $\forall u \in A$, $\forall v \in B$, $\langle u, v \rangle = 0$. Then $A \cup B$ is linearly independent.

Proof

Apply the Gram-Schmidt process to $A$ to get orthogonal vectors $A'$. Apply the Gram-Schmidt process to $B$ to get orthogonal vectors $B'$. Then $\Span(A) = \Span(A')$ and $\Span(B) = \Span(B')$. Hence, $\Span(A \cup B) = \Span(A' \cup B')$. Also, $|A' \cup B'| = |A \cup B|$. We will see that $A' \cup B'$ is linearly independent, which will imply $A \cup B$ is linearly independent.

Let $u' \in A'$ and $v' \in B'$. Let $A \defeq \{a_1, a_2, \ldots, a_m\}$ and $B \defeq \{b_1, b_2, \ldots, b_n\}$. Since $u' \in \Span(A') = \Span(A)$, we get that $u' = \sum_{i=1}^m \alpha_ia_i$ for some $\alpha_1, \ldots, \alpha_m$. Similarly, $v' = \sum_{j=1}^n \beta_jb_j$. Hence, \[ \langle u', v' \rangle = \left\langle \sum_{i=1}^m \alpha_ia_i, \sum_{j=1}^n \beta_jb_j \right\rangle = \sum_{i=1}^m \sum_{j=1}^n \alpha_i\overline{\beta_j} \langle a_i, b_j \rangle = \sum_{i=1}^m \sum_{j=1}^n \alpha_i\overline{\beta_j} 0 = 0. \] Hence, $A' \cup B'$ is orthogonal. Hence, $A' \cup B'$ is linearly independent. Hence, $A \cup B$ is linearly independent.

Dependency for:

  1. Pointing a polyhedron

Info:

Transitive dependencies:

  1. /complex-numbers/conjugation-is-homomorphic
  2. Group
  3. Ring
  4. Field
  5. Vector Space
  6. Linear independence
  7. Span
  8. Inner product space
  9. Inner product is anti-linear in second argument
  10. Orthogonality and orthonormality
  11. Gram-Schmidt Process
  12. A set of mutually orthogonal vectors is linearly independent