>>2967So I will just go through the proof from basics.
We seek to prove that
|<u,v>| =< |u||v|
If we have either vector=0 then this holds trivially.
Now to prove the case where v is non-zero we set
z=u-u_v
Where u_v = <u,v>/<v,v> * v
It is then showed that z is such that it is orthogonal to v (by the property of dot products).
So we now have 3 vectors. u and v, which are any nontrivial vector, and z which is orthogonal to v. From our definition of z, we have from a simple rearrangement
u = u_v + z
Basically we can then square the norms of everything on both sides from the Pythagorean theorem which is pic related. Basically we technically have a sum on both sides. So we take the norm and square it (left hand side of the equation), then from the right hand side (usually we call it RHS) we can just expand it out.
Using various properties of dot products we can rearrange this to get Cauchy-Schwarz. Do let me know if you'd like me to clarify any of this. :)
So we can see that u subscript v is just a vector that is useful for us to get u in a form that is nice to play with. In a more intuitive sense u_v is the magnitude of the dot product between u and v, in the direction of v.