Homework Help: Given vectors v and w, prove |v*w| =< |v|*|w|

1. Feb 5, 2009

JoeTrumpet

1. The problem statement, all variables and given/known data
Given vectors v and w, show that |v*w| =< |v|*|w|

2. Relevant 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)

3. 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!

2. Feb 5, 2009

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.

3. Feb 5, 2009

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!

4. Feb 5, 2009

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.

5. Feb 5, 2009

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)

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?

6. Feb 5, 2009

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.

7. Feb 5, 2009

JoeTrumpet

Ah, that makes sense. Thank you very much!

8. Feb 5, 2009

Dick

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