Given vectors v and w, prove |v*w| =< |v|*|w|

[JoeTrumpet]

Homework Statement

Given vectors v and w, show that |v*w| =< |v|*|w|

Homework Equations

I know that |v| = sqrt((x1)^2+(x2)^2+...+(xn)^2)
and that |w| = sqrt((y1)^2+(y2)^2+...+(yn)^2)

also |v*w| = x1y1 + x2y2 + ... + xnyn

We were also told
Hint: use |v+c*w|^2
and (v+c*w)(v+c*w)

The Attempt at a Solution

I am not at all sure what to do. I distributed the second expression in the hint to get
w^2 + 2vwc + c^2*v^2, but that doesn't tell me much.

Thanks!

 

[Dick]

(v+cw).(v+cw)>=0, right? Now pick a special value of c=(-v.w)/(w.w). It's basically a trick. It's the value of c that extremizes (v+cw).(v+cw) as a function of c.

 

[JoeTrumpet]

How come (v+cw).(v+cw) >=0? And after finding that value of c, I'm still not sure how that relates back to the original question. Sorry if I'm missing something obvious!

 

[Dick]

The dot product of any vector with itself is >=0. Look at your formula, it's a sum of squares. Put the special value of c in and expand it out.

 

[JoeTrumpet]

Hm, I think I get it. (Vectors are fairly new to me so forgive me if I misunderstand and misuse these operations.)

knowing c = (v.w)/(w.w)
in my expansion I had

v.v - 2*(v.w)*c + (w.w)*c^2 >= 0

a vector times itself must be the magnitude of itself squared, right? So knowing that and plugging in c gets me

|v|^2 - 2(v.w)(v.w)/(w.w) + (|w|^2)*(v.w)(v.w)/((w.w)(w.w)) >= 0

multiplying everything by |w|^2

|w|^2*|v|^2 - 2(v.w)(v.w) + (v.w)(v.w) >= 0
and
|w|^2*|v|^2 >= (v.w)(v.w)

and I suppose I can take the square root of both sides to get

|w|*|v| >= (v.w)

Is this correct?

 

[Dick]

Yes, but what you actually get after the square root is |w||v|>=|v.w|. When you take square roots of squares you get absolute values. Good job.

 

[JoeTrumpet]

Ah, that makes sense. Thank you very much!

 

[Dick]

Ah, that makes sense. Thank you very much!

No problem. The inequality is called Cauchy-Schwarz, if you didn't already know that.