A set of mutually orthogonal vectors is linearly independent

Dependencies:

  1. Inner product space
  2. Linear independence
  3. Orthogonality and orthonormality

Let $S = \{v_1, v_2, \ldots, v_n\}$ be a set of orthogonal vectors from an inner product space $V$. Then $S$ is linearly independent.

Proof

Assume that $S$ is linearly dependent. Without loss of generality, assume that $v_n$ is a linear combination of the rest of the vectors. Let $v_n = \sum_{i=1}^{n-1} a_iv_i$.

\begin{align} & \|v_n\|^2 = \langle v_n , v_n \rangle \\ &= \left\langle \sum_{i=1}^{n-1} a_iv_i , v_n \right\rangle \\ &= \sum_{i=1}^{n-1} a_i \langle v_i, v_n \rangle \tag{by linearity} \\ &= \sum_{i=1}^{n-1} a_i 0 \\ &= 0 \end{align}

By the positive-definiteness property of inner-product spaces, $\|v_n\|^2 = 0 \Rightarrow v_n = 0$. This contradicts the fact that all vectors in $S$ are non-zero. Therefore, $S$ is linearly independent.

Dependency for:

  1. Symmetric operator on V has a basis of orthonormal eigenvectors
  2. Joining orthogonal linindep sets

Info:

Transitive dependencies:

  1. Group
  2. Ring
  3. Field
  4. Vector Space
  5. Linear independence
  6. Inner product space
  7. Orthogonality and orthonormality