- #1
jbusc
- 211
- 0
This is an interesting question that my friend had...
Is the zero vector perpendicular to all other vectors?
We know that the zero vector is orthogonal to all other vectors; I verified this in both "Linear Algebra and it's Applications" by Strang and "Calculus" by Stewart; both are exceedingly clear about this.
Now while this may be pedantic, does this translate into "perpendicularity"?
If perpendicularity means the same as orthogonality (vanishing of the inner or dot product) then the argument is over, and the zero vector is perpendicular to all vectors. Mathworld (http://mathworld.wolfram.com/Perpendicular.html) agrees with this sentiment.
While we typically think of perpendicularity as equivalent to orthogonality, however, there's the argument that orthogonality is some kind of generalization of perpendicularity. In this argument perpendicularity can be defined as vectors which form 90 degree angles (and thus only makes sense in Euclidean n-space R^n), whereas orthogonality is the vanishing of the inner product and so is valid in any inner product space. Since we cannot claim that the zero vector forms any angle at all, it cannot form a 90 degree angle, and then could not be perpendicular to anything.
However, even if the case that "perpendicularity" is defined as forming 90 degree angles, if it is to be a generalization of orthogonality, then in regular R^n space, vectors should be perpendicular iff they are orthogonal as well (otherwise, it is not much of a generalization if the generalized version does not degenerate fully into the less general version in the basic case). Therefore, we again have the zero vector perpendicular to all vectors.
It seems like part of the problem is lack of a rigorous and standardized definition of perpendicular. Since perpendicularity is not a terribly important concept compared to orthogonality, Strang for example barely mentions perpendicularity at all.
So...comments? Etc?
Is the zero vector perpendicular to all other vectors?
We know that the zero vector is orthogonal to all other vectors; I verified this in both "Linear Algebra and it's Applications" by Strang and "Calculus" by Stewart; both are exceedingly clear about this.
Now while this may be pedantic, does this translate into "perpendicularity"?
If perpendicularity means the same as orthogonality (vanishing of the inner or dot product) then the argument is over, and the zero vector is perpendicular to all vectors. Mathworld (http://mathworld.wolfram.com/Perpendicular.html) agrees with this sentiment.
While we typically think of perpendicularity as equivalent to orthogonality, however, there's the argument that orthogonality is some kind of generalization of perpendicularity. In this argument perpendicularity can be defined as vectors which form 90 degree angles (and thus only makes sense in Euclidean n-space R^n), whereas orthogonality is the vanishing of the inner product and so is valid in any inner product space. Since we cannot claim that the zero vector forms any angle at all, it cannot form a 90 degree angle, and then could not be perpendicular to anything.
However, even if the case that "perpendicularity" is defined as forming 90 degree angles, if it is to be a generalization of orthogonality, then in regular R^n space, vectors should be perpendicular iff they are orthogonal as well (otherwise, it is not much of a generalization if the generalized version does not degenerate fully into the less general version in the basic case). Therefore, we again have the zero vector perpendicular to all vectors.
It seems like part of the problem is lack of a rigorous and standardized definition of perpendicular. Since perpendicularity is not a terribly important concept compared to orthogonality, Strang for example barely mentions perpendicularity at all.
So...comments? Etc?
