In geometric algebra, what is the grade of a k-vector in an n-dimensional space？

Question

It's a stupid question, but I am not figure it out. In the book I read, a k-vector in an n-dimensional space is defined as a linear combination of this space k-blade. I'm curious what is the grade of the k- vector, is it also k?

Answer 1

Yes, grade selection is a linear operation.

${\left\langle{{A + B}}\right\rangle}_{{k}} = {\left\langle{{A}}\right\rangle}_{{k}} + {\left\langle{{B}}\right\rangle}_{{k}}$

If the grade-k selection of a multivector equals that multivector, it has grade-k.

Example: $\mathbf{e}_1 \mathbf{e}_2$, and $\mathbf{e}_3 \mathbf{e}_4$ are both 2-blades. Their sum $\mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_3 \mathbf{e}_4$ cannot be expressed as a product, so is a 2-vector, not a 2-blade, and

${\left\langle{{\mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_3 \mathbf{e}_4}}\right\rangle}_{{2}} = {\left\langle{{\mathbf{e}_1 \mathbf{e}_2}}\right\rangle}_{{2}} + {\left\langle{{\mathbf{e}_3 \mathbf{e}_4}}\right\rangle}_{{2}} = \mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_3 \mathbf{e}_4,$

This sum, a 2-vector, has grade-2, as it equals it's grade-2 selection.

The dimension $n$ of the underlying vector space is not relevant.