Orthonormal Vectors are Linearly Independent

Apr 13, 2023

For a set of orthonormal vectors $(v_{1}, v_{2}, . . ., v_{m})$ in an m-dimensional vector space , and set of constants $r_{1}, r_{2}, . . . r_{m} \in R$ , if we consider the equation $r_{1} v_{1} + r_{2} v_{2} + . . . + r_{m} v_{m} = 0$

with the objective of demonstrating that the vectors in $(v_{1}, v_{2}, . . ., v_{m})$ are linearly independent, then we must show that $r_{1} = r_{2} = . . . = r_{m} = 0$ . If we take the inner product on both sides of the equation , we have that

$< r_{1} v_{1} + r_{2} v_{2} + \dots + r_{m} v_{m}, v_{1} >=< 0, v_{1} >$

Using the properties of inner products, we have that

$r_{1} < v_{1}, v_{1} > + r_{2} < v_{2}, v_{1} > \dots + r_{m} < v_{m}, v_{1} >= 0 ⟹ r_{1} = 0$

Repeating the same process but isolating $v_{2}, v_{3}, . . .$ on the right hand side using inner products, we will deduce that $r_{2} = r_{3} = . . . = r_{m} = 0$ . Thus, since

$r_{1} v_{1} + r_{2} v_{2} + \dots + r_{m} v_{m} ⟺ r_{1} = r_{2} = \dots = r_{m} = 0$

, we conclude that $(v_{1}, v_{2}, . . ., v_{m})$ is linearly independent.

Geometry