Linear independence in 3-space

Question

The orthogonal orthonormal basis vectors $i,j,k$ in $3$-space are generally accepted as linearly independent.

However $j$ can be derived from $i$, by the process of differentiation, and likewise vector $k$ can be derived as a binormal vector to both vectors $i$ and $j$.

This then questions the notion of independence between orthonormal vectors. What is it that is missing from this seemingly contradictory situation. Is there something in the definition of "linearity" which is the crucial factor? Can someone please clarify?

Many thanks.

Answer 1

You seem to be slighly confuzed when you are writing about the vectors. You say that $j$ can be derived from $i$ by the process of differentiation, which is not true. In fact, $i$ is a constant vector, not a function, you can only have the differential of a function.

As far as the definition of independency goes, you have $2$ equivalent versions:

1. The vectors $i,j,k$ are linearly independent if for any $\alpha, \beta, \gamma\in \mathbb R$, for which $\alpha i + \beta j + \gamma k = 0$ holds, $\alpha=\beta=\gamma=0$ must also hold.
2. The vectors $i,j,k$ are linearly independent if none of the vectors can be written as a linear combination of the other two vectors, that is, there are no scalars $\alpha, \beta\in\mathbb R$ for which either $i=\alpha j + \beta k$, $j=\alpha i + \beta k$ or $k=\alpha i + \beta j$ would hold.

It is easy to see that the definitions are equivalent. It is also easy to see that the orthonormal basis vectors $i,j,k$ are linearly independend, especially using the first definition.