When does equality hold in Cauchy-Schwarz inequality

[Hitman2-2]

Homework Statement

Prove that if V is a vector space over \mathbb{C}^n with the standard inner product, then

<br /> |&lt;x,y&gt;| = ||x|| \cdot ||y||<br />

implies one of the vectors x or y is a multiple of the other.

The Attempt at a Solution

Assume the identity holds and that y is not zero. Let

<br /> a = \frac {&lt;x,y&gt;} {||y||^2}<br />

and let z = x - ay. I've shown that y and z are orthogonal and want to show

<br /> |a| = \frac {||x||} {||y||}<br />

Well,

<br /> |a| \cdot ||y|| = \frac {|&lt;x,y&gt;|} {||y||} = \frac {|\sum_{i=1}^n a_i \overline{b_i} |} {\sqrt{\sum_{i=1}^n |b_i|^2}}<br /> = \sqrt{\frac {\left(\sum_{i=1}^n a_i \overline{b_i} \right) \left(\sum_{i=1}^n \overline{a_i} b_i \right)} {\sum_{i=1}^n |b_i|^2} }<br />

but now I don't see how to simply this further to get this equal to the norm of x.

 

[Hitman2-2]

<br /> |a| \cdot ||y|| = \frac {|&lt;x,y&gt;|} {||y||} = \frac {|\sum_{i=1}^n a_i \overline{b_i} |} {\sqrt{\sum_{i=1}^n |b_i|^2}}<br /> = \sqrt{\frac {\left(\sum_{i=1}^n a_i \overline{b_i} \right) \left(\sum_{i=1}^n \overline{a_i} b_i \right)} {\sum_{i=1}^n |b_i|^2} }<br />

but now I don't see how to simply this further to get this equal to the norm of x.

Never mind ... I think I've got it. I've totally neglected that by assumption,

<br /> \frac { | \langle x,y \rangle | } { \| y \| } = \| x \|<br />

so then it follows that

<br /> |a| \cdot \| y \| = \| x \|<br />

Then since

<br /> \| x \|^2 = \| ay + z \|^2 = \| ay \|^2 + \| z \|^2 \Rightarrow \| z \| = 0<br />

the result follows.

Embarassing.