Numerically, we get <u,v> == <v,u>, but are they the same idea in projection? like <u,v> projects v onto u, shouldn't <v,u> projects u onto v? Are they still the same?
rmvenkat
I am not sure, but is something wrong with this definition. Projection of v onto u is |v|cos(theta) and the inner product is |u||v|cos(theta). So, I think the labelling of <u,v> as |v|*cos(theta) in first figure is true only if u is a unit vector right ? There seems to be some inconsistency here.
ryanmelon
I am thinking we need a definition of "the length along a vector". For example, there're two vectors u(?3,1) and v(?3,0). If we project u onto v, the length would be ?3. If we project v onto u, the length would be 3/2. They are different. I do not know is this the definition of projection or not.
Sera
I know we said in class that the property is symmetric but how do we prove that it is?
Numerically, we get <u,v> == <v,u>, but are they the same idea in projection? like <u,v> projects v onto u, shouldn't <v,u> projects u onto v? Are they still the same?
I am not sure, but is something wrong with this definition. Projection of v onto u is |v|cos(theta) and the inner product is |u||v|cos(theta). So, I think the labelling of <u,v> as |v|*cos(theta) in first figure is true only if u is a unit vector right ? There seems to be some inconsistency here.
I am thinking we need a definition of "the length along a vector". For example, there're two vectors u(?3,1) and v(?3,0). If we project u onto v, the length would be ?3. If we project v onto u, the length would be 3/2. They are different. I do not know is this the definition of projection or not.
I know we said in class that the property is symmetric but how do we prove that it is?