I think the above property is symmetric. For example, for 2 unit vectors v and u separated by 30 degree; projection of v on u will be cos30 whereas projection of u on v will also be cos30. So the property is symmetric.
keenan
Good observation. Now, can you make the same argument without appealing to the use of cosine (which we didn't really define, starting from first principles)? In other words, what does "cosine" mean geometrically, and why then should there be symmetry in this case?
There is also another, simpler argument for why you get symmetry here that doesn't rely on any argument about cosines or angles.
(Perhaps someone else wants to jump in?)
stutiRastogi
I am just trying to guess a simpler argument here. As we know, the inner product just says how aligned two vectors are. So how much ever u is aligned with v, v would also be aligned with u.
So in the case of unit vectors, the inner product is denoting the length of one vector along another which is still fundamentally how inline or apart they are from each other, and that remains symmetric.
ninkamat
I guess a simpler argument would be that since the vectors have equal magnitude and only direction varies, the way in which you name them should not affect the inner product of the vectors. So if in the above image, if we swap u and v, the result would still be the same. In that case it would be a projection of u on v, which shows that the property is symmetric.
yizhous
I am wondering if we take the inner product of two non-unit vectors, does it have any geometric meaning? In this case it appears the inner product no longer measures the alignment anymore because the result also depends on the norm of the vectors. For example, We can have two vectors going quite different directions but have large norms, and their inner product may be greater than two vectors pretty aligned together but have smaller norms.
bpx
@yizhous Typically, we need to normalize one of the vectors (make 'u' a unit vector) and then the inner product <u, v> measures the length of v along u (or in that direction). Still, the inner product between two non-unit vectors CAN be useful. One example is that if <u, v> = 0 (and neither is 0), then u and v are perpendicular. From this and a bit of observation, we also know if <u, v> < 0 the "small" angle between u and v is 90 < theta <= 180 in degrees. If <u, v> > 0, the small angle is 0 <= theta < 90 in degrees. This could be useful if you only want to know the general difference in direction between two vectors (and it saves computation from not having to normalize). However, the inner product is most useful if one vector is normalized.
keenan
@stutiRastogi For unit vectors, the simplest argument is basically to observe that, by symmetry, there is no way to distinguish $\mathbf{u}$ and $\mathbf{v}$. So everything about $\mathbf{u}$ must be true about $\mathbf{v}$, including properties about projecting $\mathbf{u}$ onto $\mathbf{v}$ and so forth.
@yizhous For the Euclidean inner product there is a clear geometric meaning even for vectors that are not unit: $\langle \mathbf{u}, \mathbf{v} \rangle = |\mathbf{u}||\mathbf{v}|\cos(\theta)$, where $\theta$ is the angle between the two vectors. In other words, the inner product always tells you about alignment and magnitude, as you've observed.
I think the above property is symmetric. For example, for 2 unit vectors v and u separated by 30 degree; projection of v on u will be cos30 whereas projection of u on v will also be cos30. So the property is symmetric.
Good observation. Now, can you make the same argument without appealing to the use of cosine (which we didn't really define, starting from first principles)? In other words, what does "cosine" mean geometrically, and why then should there be symmetry in this case?
There is also another, simpler argument for why you get symmetry here that doesn't rely on any argument about cosines or angles.
(Perhaps someone else wants to jump in?)
I am just trying to guess a simpler argument here. As we know, the inner product just says how aligned two vectors are. So how much ever u is aligned with v, v would also be aligned with u.
So in the case of unit vectors, the inner product is denoting the length of one vector along another which is still fundamentally how inline or apart they are from each other, and that remains symmetric.
I guess a simpler argument would be that since the vectors have equal magnitude and only direction varies, the way in which you name them should not affect the inner product of the vectors. So if in the above image, if we swap u and v, the result would still be the same. In that case it would be a projection of u on v, which shows that the property is symmetric.
I am wondering if we take the inner product of two non-unit vectors, does it have any geometric meaning? In this case it appears the inner product no longer measures the alignment anymore because the result also depends on the norm of the vectors. For example, We can have two vectors going quite different directions but have large norms, and their inner product may be greater than two vectors pretty aligned together but have smaller norms.
@yizhous Typically, we need to normalize one of the vectors (make 'u' a unit vector) and then the inner product <u, v> measures the length of v along u (or in that direction). Still, the inner product between two non-unit vectors CAN be useful. One example is that if <u, v> = 0 (and neither is 0), then u and v are perpendicular. From this and a bit of observation, we also know if <u, v> < 0 the "small" angle between u and v is 90 < theta <= 180 in degrees. If <u, v> > 0, the small angle is 0 <= theta < 90 in degrees. This could be useful if you only want to know the general difference in direction between two vectors (and it saves computation from not having to normalize). However, the inner product is most useful if one vector is normalized.
@stutiRastogi For unit vectors, the simplest argument is basically to observe that, by symmetry, there is no way to distinguish $\mathbf{u}$ and $\mathbf{v}$. So everything about $\mathbf{u}$ must be true about $\mathbf{v}$, including properties about projecting $\mathbf{u}$ onto $\mathbf{v}$ and so forth.
@yizhous For the Euclidean inner product there is a clear geometric meaning even for vectors that are not unit: $\langle \mathbf{u}, \mathbf{v} \rangle = |\mathbf{u}||\mathbf{v}|\cos(\theta)$, where $\theta$ is the angle between the two vectors. In other words, the inner product always tells you about alignment and magnitude, as you've observed.